Bài 3:

a: \(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\cdots\left|x+\frac{1}{2019\cdot2020}\right|=2020x\) (1)

=>2020x>=0

=>x>=0

Phương trình (1) sẽ trở thành:

\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+\cdots+x+\frac{1}{2019\cdot2020}=2020x\)

=>\(2020x=2019x+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2019\cdot2020}\right)\)

=>\(x=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2019\cdot2020}\)

=>\(x=1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2019}-\frac{1}{2020}\)

=>\(x=1-\frac{1}{2020}=\frac{2019}{2020}\)

b: \(\left|x+\frac{1}{1\cdot3}\right|+\left|x+\frac{1}{3\cdot5}\right|+\cdots+\left|x+\frac{1}{197\cdot199}\right|=100x\) (2)

=>100x>=0

=>x>=0

(2) sẽ trở thành: \(x+\frac{1}{1\cdot3}+x+\frac{1}{3\cdot5}+\cdots+x+\frac{1}{197\cdot199}=100x\)

=>\(100x=99x+\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{197\cdot199}\right)\)

=>\(x=\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{197}-\frac{1}{199}\right)=\frac12\left(1-\frac{1}{199}\right)\)

=>\(x=\frac12\cdot\frac{198}{199}=\frac{99}{199}\)

c: \(\left|x+\frac12\right|+\left|x+\frac16\right|+\left|x+\frac{1}{12}\right|+\cdots+\left|x+\frac{1}{110}\right|=11x\left(3\right)\)

=>11x>=0

=>x>=0

(3) sẽ trở thành:

\(11x=x+\frac12+x+\frac16+\ldots+x+\frac{1}{110}\)

=>\(11x=10x+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{10\cdot11}\)

=>\(x=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{10\cdot11}\)

=>\(x=1-\frac12+\frac12-\frac13+\cdots+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\) (nhận)

Bài 2:

a: \(\left|5-\frac23x\right|\ge0\forall x;\left|\frac23y-4\right|\ge0\forall y\)

Do đó: \(\left|5-\frac23x\right|+\left|\frac23y-4\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}5-\frac23x=0\\ \frac23y-4=0\end{cases}\Rightarrow\begin{cases}\frac23x=5\\ \frac23y=4\end{cases}\Rightarrow\begin{cases}x=5:\frac23=\frac{15}{2}\\ y=4:\frac23=6\end{cases}\)

b: \(\left|\frac23-\frac12+\frac34x\right|=\left|\frac34x+\frac16\right|\ge0\forall x\)

\(\left|1,5-\frac34-\frac32y\right|=\left|\frac34-\frac32y\right|\ge0\forall y\)

Do đó: \(\left|\frac34x+\frac16\right|+\left|\frac34-\frac32y\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}\frac34x+\frac16=0\\ \frac34-\frac32y=0\end{cases}\Rightarrow\begin{cases}\frac34x=-\frac16\\ \frac32y=\frac34\end{cases}\Rightarrow\begin{cases}x=-\frac16:\frac34=-\frac16\cdot\frac43=-\frac{4}{18}=-\frac29\\ y=\frac34:\frac32=\frac24=\frac12\end{cases}\)

c: \(\left|x-2020\right|\ge0\forall x;\left|y-2021\right|\ge0\forall y\)

Do đó: \(\left|x-2020\right|+\left|y-2021\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-2020=0\\ y-2021=0\end{cases}\Rightarrow\begin{cases}x=2020\\ y=2021\end{cases}\)

d: \(\left|x-y\right|\ge0\forall x,y\)

\(\left|y+\frac{21}{10}\right|\ge0\forall y\)

Do đó: \(\left|x-y\right|+\left|y+\frac{21}{10}\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-y=0\\ y+\frac{21}{10}=0\end{cases}\Rightarrow x=y=-\frac{21}{10}\)

Bài 1:

a: \(\left|\frac32x+\frac12\right|=\left|4x-1\right|\)

=>\(\left[\begin{array}{l}4x-1=\frac32x+\frac12\\ 4x-1=-\frac32x-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}4x-\frac32x=\frac12+1\\ 4x+\frac32x=-\frac12+1\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac52x=\frac32\\ \frac{11}{2}x=\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac32:\frac52=\frac35\\ x=\frac12:\frac{11}{2}=\frac{1}{11}\end{array}\right.\)

b: \(\left|\frac75x+\frac12\right|=\left|\frac43x-\frac14\right|\)

=>\(\left[\begin{array}{l}\frac75x+\frac12=\frac43x-\frac14\\ \frac75x+\frac12=\frac14-\frac43x\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac75x-\frac43x=-\frac14-\frac12\\ \frac75x+\frac43x=\frac14-\frac12\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac{1}{15}x=-\frac34\\ \frac{41}{15}x=-\frac14\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac34:\frac{1}{15}=-\frac34\cdot15=-\frac{45}{4}\\ x=-\frac14:\frac{41}{15}=-\frac14\cdot\frac{15}{41}=-\frac{15}{164}\end{array}\right.\)

c: \(\left|\frac54x-\frac72\right|-\left|\frac58x+\frac35\right|=0\)

=>\(\left|\frac54x-\frac72\right|=\left|\frac58x+\frac35\right|\)

=>\(\left[\begin{array}{l}\frac54x-\frac72=\frac58x+\frac35\\ \frac54x-\frac72=-\frac58x-\frac35\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac54x-\frac58x=\frac35+\frac72\\ \frac54x+\frac58x=-\frac35+\frac72\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac58x=\frac{41}{10}\\ \frac{15}{8}x=\frac{29}{10}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{41}{10}:\frac58=\frac{41}{10}\cdot\frac85=\frac{164}{25}\\ x=\frac{29}{10}:\frac{15}{8}=\frac{29}{10}\cdot\frac{8}{15}=\frac{116}{75}\end{array}\right.\)

d: \(\left|\frac78x+\frac56\right|-\left|\frac12x+5\right|=0\)

=>\(\left|\frac78x+\frac56\right|=\left|\frac12x+5\right|\)

=>\(\left[\begin{array}{l}\frac78x+\frac56=\frac12x+5\\ \frac78x+\frac56=-\frac12x-5\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac78x-\frac12x=5-\frac56\\ \frac78x+\frac12x=-5-\frac56\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac38x=\frac{25}{6}\\ \frac{11}{8}x=-\frac{35}{6}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{25}{6}:\frac38=\frac{25}{6}\cdot\frac83=\frac{200}{18}=\frac{100}{9}\\ x=-\frac{35}{6}:\frac{11}{8}=-\frac{35}{6}\cdot\frac{8}{11}=-\frac{140}{33}\end{array}\right.\)

lI dau la lI

mik bt giải r chờ tí

nhanh lên bạn mình cần gấp lém

✨ Bước 1: Rút gọn hai vế của phương trình

Vế phải:

\(8 , 5 - \frac{1}{2} = 8 , 0\)

Vậy phương trình trở thành:

\(2 \mid 5 - x \mid + \frac{1}{2} = 8\)

✨ Bước 2: Chuyển vế

Trừ \(\frac{1}{2}\) hai vế:

\(2 \mid 5 - x \mid = 8 - \frac{1}{2} = \frac{16}{2} - \frac{1}{2} = \frac{15}{2}\)

✨ Bước 3: Chia hai vế cho 2

\(\mid 5 - x \mid = \frac{15}{4}\)

✨ Bước 4: Giải giá trị tuyệt đối

Ta có:

\(\mid 5 - x \mid = \frac{15}{4} \Rightarrow \left{\right. 5 - x = \frac{15}{4} \\ 5 - x = - \frac{15}{4}\)

Giải từng phương trình:

1. \(5 - x = \frac{15}{4} \Rightarrow x = 5 - \frac{15}{4} = \frac{20}{4} - \frac{15}{4} = \frac{5}{4}\)
2. \(5 - x = - \frac{15}{4} \Rightarrow x = 5 + \frac{15}{4} = \frac{20}{4} + \frac{15}{4} = \frac{35}{4}\)

✅ Kết luận:

Vậy phương trình có 2 nghiệm:

\(\boxed{x = \frac{5}{4} \text{ho}ặ\text{c} x = \frac{35}{4}}\)
Tk

7​251​−x​+​x−51​​+851​=1,2⇒​251​−x​+​x−51​​=1,2−851​⇒​251​−x​+​x−51​​=−7

Nhận xét:

\(\left{\right. \mid 2 \frac{1}{5} - x \mid \geq 0 , \forall x \\ \mid x - \frac{1}{5} \mid \geq 0 , \forall x \Rightarrow \mid 2 \frac{1}{5} - x \mid + \mid x - \frac{1}{5} \mid \geq 0 , \forall x\)
Mà \(- 7 < 0\) nên:
Không tìm được giá trị \(x\) thỏa mãn đề bài
Vậy...

\(a,\frac{15^3.\left(-5\right)^4}{\left(-3\right)^5.5^6}\)\(=\frac{3^3.5^3}{\left(-3\right)^5.5^2}\)\(=-\frac{5}{\left(3\right)^2}=-\frac{5}{9}\)

\(b,\frac{6^3.2.\left(-3\right)^2}{\left(-2\right)^9.3^7}\)\(=-\frac{6^3}{2^8.3^5}\)\(=-\frac{2^3.3^3}{2^8.3^5}\)\(=-\frac{1}{2^5.3^2}=-\frac{1}{288}\)

\(c,\frac{3^6.7^2-3^7.7}{3^7.21}\)\(=\frac{3^6.7\left(7-3\right)}{3^7.21}\)\(=\frac{3^6.7.4}{3^7.7.3}\)\(=\frac{4}{3.3}=\frac{4}{9}\)

\(a,\left(x-1,2\right)^2=4\)

\(\Rightarrow x-1,2=2\)

\(\Rightarrow x=3,2\)

\(b,\left(x+1\right)^3=-125\)

\(\Rightarrow\left(x+1\right)^3=\left(-5\right)^3\)

\(\Rightarrow x+1=-5\Rightarrow x=-6\)

\(c,\left(x-5\right)^3=2^6\)

\(\Rightarrow\left(x-5\right)^3=4^3\)

\(\Rightarrow x-5=4\Rightarrow x=9\)

\(d,\left(2x+1\right)^{x+1}=5^{x+1}\)

\(\Rightarrow2x+1=5\Rightarrow x=2\)

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

~ Học tốt ~

Bài 1:

1) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)

\(=3^2.\left(\dfrac{1}{3}\right)^5.\left(3^4\right)^2.\dfrac{1}{3^3}\)

\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)

\(=3^2=9\)

2) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)

\(=\left(2^2.2^5\right):[2^3.\left(\dfrac{1}{2}\right)^4]\)

\(=2^7:2^3:\dfrac{1}{2^4}\)

\(=2^4.2^4=256\)

3)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)

\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}.1:2^3\)

\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2^4}\)

\(=\dfrac{43}{48}\)

4)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=-3-1+\dfrac{1}{4}.\dfrac{1}{2}\)

\(=-3-1+\dfrac{1}{8}\)

\(=-4+\dfrac{1}{8}\\ \)

\(=-\dfrac{31}{8}\)

5)\([\left(0,1\right)^2]^0+[\left(\dfrac{1}{7}\right)^{-1}]^2.\dfrac{1}{49}.[\left(2^2\right)^3:2^5]\\ =1+7^2.\dfrac{1}{7^2}.2^6:2^5\\ =1+1.2\\ =3\)

Chúc bạn học tốt

#)Giải :

a) \(\left(5x+1\right)^2=\frac{36}{49}\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\Leftrightarrow5x+1=\frac{6}{7}\Leftrightarrow5x=-\frac{1}{7}\Leftrightarrow x=-\frac{1}{35}\)

b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Leftrightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\Leftrightarrow x=\frac{2}{3}\)

c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\Leftrightarrow8x-1=5\Leftrightarrow8x=6\Leftrightarrow x=\frac{6}{8}\)

a) \(\left(5x+1\right)^2=\frac{36}{49}\)

 \(\left(5x+1\right)^2=\frac{6^2}{7^2}\)

\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)

\(\Leftrightarrow5x+1=\frac{6}{7}\)

\(5x=\frac{6}{7}-1\)

\(5x=\frac{6}{7}-\frac{7}{7}\)

\(5x=-\frac{1}{7}\)

\(x=-\frac{1}{7}\div5\)

\(x=-\frac{1}{7}\times\frac{1}{5}\)

\(x=-\frac{1}{35}\)

Vậy \(x=-\frac{1}{35}\)

làm bài 3 BĐT

theo bảng xét dấu

còn bài 1,2 ở trên là 1.1 và 1.2 đều trg bài 1.2

bài 1.2 (tức bài 2 ở trên )làm a,b,c,d

\còn bài 2( tức bài 2 ở trên) làm hết

thanks

|2x-1|=1,5

TH(1)2x-1=1,5

2x =1,5+1

2x =2,5

x =2,5 :2

x =1,25

TH(2) 2x-1=-1,5

2x =-1,5+1

2x =-0,5

x =-0,5:2

x =-0,25

các câu khác cứ tương tự bạn nhé

b) \(7,5-\left|5-2x\right|=-4,5\)

\(\left|5-2x\right|=7,5+4,7\)

\(\left|5-2x\right|=12\)

th1 :\(5-2x=12\)

\(2x=5-12\)

\(2x=-7\)

\(x=-7:2\)

\(x=-3,5\)

th2: \(5-2x=-12\)

\(2x=5+12\)

\(2x=17\)

\(x=17:2\)

\(x=8,5\)

c) \(-3+\left|x\right|=-1\)

\(\left|x\right|=-1+3\)

\(\left|x\right|=2\)

th1: \(x=-2\)

th2 : \(x=2\)

d)\(\left|2\dfrac{1}{3}-x\right|=\dfrac{1}{6}\)

\(\left|\dfrac{7}{3}-x\right|=\dfrac{1}{6}\)

th1 :\(\dfrac{7}{3}-x=\dfrac{1}{6}\)

\(x=\dfrac{7}{3}-\dfrac{1}{2}\)

\(x=\dfrac{11}{6}\)

th2: \(\dfrac{7}{3}-x=\dfrac{-1}{6}\)

\(x=\dfrac{7}{3}+\dfrac{1}{6}\)

\(x=\dfrac{-5}{2}\)

e) \(\dfrac{5}{7}-\left|x+1\right|=\dfrac{1}{14}\)

\(\left|x+1\right|=\dfrac{5}{7}-\dfrac{1}{14}\)

\(\left|x+1\right|=\dfrac{9}{14}\)

th1 :\(x+1=\dfrac{9}{14}\)

\(x=\dfrac{9}{14}-1\)

\(x=\dfrac{-5}{14}\)

th2 : \(x+1=\dfrac{-9}{14}\)

\(x=\dfrac{-9}{14}-1\)

\(x=\dfrac{-5}{14}\)

a) \(\left(\frac{2^2}{5}\right)+5\frac{1}{2}.\left(4,5-2,5\right)+\frac{2^3}{-4}\)

\(=\frac{4}{5}+\frac{11}{2}.2+\frac{-8}{4}\)

\(=\frac{4}{5}+11-2\)

\(=\frac{4}{5}+9\)

\(=\frac{49}{9}\)

b) \(\left(-2^3\right)+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)

\(=-8+4-5+64\)

= 55

c) \(\frac{\sqrt{3^2+\sqrt{39}^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}\)

\(=\frac{\sqrt{9+39}}{91-\sqrt{49}}\)

\(=\frac{\sqrt{48}}{91-7}\)

\(=\frac{4\sqrt{3}}{84}\)

\(=\frac{\sqrt{3}}{41}\)

d) Xem lại đề nhé em!

e) \(\sqrt{25}-3\sqrt{\frac{4}{9}}\)

\(=5-3.\frac{2}{3}\)

= 5 - 2

= 3

h) \(\left(-3^2\right).\frac{1}{3}-\sqrt{49}+\left(5^3\right):\sqrt{25}\)

\(=-9.\frac{1}{3}-7+125:5\)

\(=-3-7+25\)

= 15