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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: \(\left|\frac78x+\frac56\right|-\left|\frac12x+5\right|=0\)
=>\(\left|\frac78x+\frac56\right|=\left|\frac12x+5\right|\)
=>\(\left[\begin{array}{l}\frac78x+\frac56=\frac12x+5\\ \frac78x+\frac56=-\frac12x-5\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac78x-\frac12x=5-\frac56\\ \frac78x+\frac12x=-5-\frac56\end{array}\right.\)
=>\(\left[\begin{array}{l}\frac38x=\frac{25}{6}\\ \frac{11}{8}x=-\frac{35}{6}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{25}{6}:\frac38=\frac{25}{6}\cdot\frac83=\frac{200}{18}=\frac{100}{9}\\ x=-\frac{35}{6}:\frac{11}{8}=-\frac{35}{6}\cdot\frac{8}{11}=-\frac{140}{33}\end{array}\right.\)
\(a,x\in\left(-5379;-5378;-5377;...;5379;5380\right)\)
\(b,k+10\le k\le k+2000\)
\(\Rightarrow\hept{\begin{cases}k\ge10\\k\le2000\end{cases}}\)
\(\Rightarrow k\in\left(10;11;12;...;1999;2000\right)\)
Ta có bất đẳng thức giá trị tuyệt đối:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
Dấu \(=\)khi \(AB\ge0\).
d) \(\left|x+1\right|+\left|x+2\right|+\left|2x-3\right|\)
\(\ge\left|x+1+x+2\right|+\left|2x-3\right|\)
\(=\left|2x+3\right|+\left|3-2x\right|\)
\(\ge\left|2x+3+3-2x\right|=6\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(2x+3\right)\left(3-2x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le\frac{3}{2}\).
e) \(\left|x+1\right|+\left|x+2\right|+\left|x-3\right|+\left|x-5\right|\)
\(=\left(\left|x+1\right|+\left|3-x\right|\right)+\left(\left|x+2\right|+\left|5-x\right|\right)\)
\(\ge\left|x+1+3-x\right|+\left|x+2+5-x\right|\)
\(=4+7=11\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(3-x\right)\ge0\\\left(x+2\right)\left(5-x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le3\).
Do đó phương trình đã cho vô nghiệm.
a)(x − 12)2 = 0
=>x − 12 = 0
=> x = 12
b) (x+12)2 = 0,25
=> x + 12 = 0,5 hoặc x + 12= -0,5
=> x = -11,5 hoặc x = -12,5
c) (2x−3)3 = -8
=> 2x - 3 = -2
=> x = 0,5
d) (3x−2)5 = −243
=> 3x - 2 = -3
=> x = -1/3
e) (7x+2)-1 = 3-2
=> \(\dfrac{1}{7x+2}=\dfrac{1}{9}\)
=> 7x + 2 = 9
=> x = 1
f) (x−1)3 = −125
=> (x−1) = −5
=> x = -4
g) (2x−1)4 = 81
=> 2x - 1 = 3
=> x = 2
h) (2x−1)6 = (2x−1)8
=> 2x -1 = 0 hoặc 2x - 1 = 1 hoặc 2x - 1 = -1
=> x = 1/2 hoặc x = 1 hoặc x = 0
a/ \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
b/ \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^2\\\left(x+\dfrac{1}{2}\right)^2=\left(-\dfrac{1}{2}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{2}\\x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy ..
c/ \(\left(2x-3\right)^3=-8\)
\(\Leftrightarrow\left(2x-3\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-3=-2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
d/ \(\left(3x-2\right)^5=-243\)
\(\left(3x-2\right)^5=\left(-3\right)^5\)
\(\Leftrightarrow3x-2=-3\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy ...
e/ \(\left(x-1\right)^3=-125\)
\(\Leftrightarrow\left(x-1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x-1=-5\)
\(\Leftrightarrow x=-4\)
Vậy..
f/ \(\left(2x-1\right)^4=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=3^4\\\left(2x-1\right)^4=\left(-3\right)^4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy...
g/ \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy..
a: \(\left(x-\frac12\right)^2=0\)
=>\(x-\frac12=0\)
=>\(x=\frac12\)
b: \(\left(x-2\right)^2=1\)
=>\(\left[\begin{array}{l}x-2=1\\ x-2=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1+2=3\\ x=-1+2=1\end{array}\right.\)
c: \(\left(2x-1\right)^3=-8\)
=>\(\left(2x-1\right)^3=\left(-2\right)^3\)
=>2x-1=-2
=>2x=-1
=>\(x=-\frac12\)
Ta có :
| 2 + 3x | - | 4x - 3 | = 0
\(\Rightarrow\)| 2 + 3x | = | 4x - 3 |
\(\Rightarrow\)2 + 3x = \(\pm\)( 4x - 3 )
Ta xét 2 trường hợp :
Th 1 :
2 + 3x = 4x - 3
3x - 4x = - 3 - 2
- x = - 5
\(\Rightarrow\)x = 5
Th 2 :
2 + 3x = - ( 4x - 3 )
2 + 3x = - 4x + 3
3x + 4x = 3 - 2
7x = 1
\(\Rightarrow\)x = \(\frac{1}{7}\)
Vậy x \(\in\){ 5 ; \(\frac{1}{7}\)}
✨ 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:
✅ Kết luận:
Vậy phương trình có 2 nghiệm:
\(\boxed{x = \frac{5}{4} \text{ho}ặ\text{c} x = \frac{35}{4}}\)
Tk
7251−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...
(5−x)+21=8.5−21 \(\left(\right. 5 - x \left.\right) + \frac{1}{2} = 8\) \(5 - x = 8 - \frac{1}{2} = 7.5\) \(- x = 2.5 \textrm{ }\textrm{ } \textrm{ }\textrm{ } \Rightarrow \textrm{ }\textrm{ } \textrm{ }\textrm{ } x = - 2.5\)
thg trung su dung tri tue nhan tao kia
\(\vert5-x\vert+\frac12=8,5-\frac12\)
\(\vert5-x\vert=8,5-\frac12-\frac12\)
\(\Rightarrow\vert5-x\vert=8,5-1\)
\(\Rightarrow\vert5-x\vert=7,5\)
TH1: 5 - x = 7,5
x = 5 - 7,5
x = -2,5
TH2: 5 - x = -7,5
x = 5 - (-7,5)
x = 12,5
Vậy x ∈ {-2,5; 12,5}
\(\left\vert5-x\right\vert\) \(+\frac12\) =\(8,5-\frac12\)
\(\left\vert5-x\right\vert-\frac12=\frac{17}{2}-\frac12\)
\(\left\vert5-x\right\vert=8-\frac12\)
\(\left\vert x-5\right\vert=\frac{15}{2}\)
\(\implies x-5=\frac{15}{2}hoặcx-5=-\frac{15}{2}\)
\(TH1:x-5=\frac{15}{2}\) \(TH2:x-5=-\frac{15}{2}\)
\(\implies x=\frac{15}{2}+5\) \(\implies x=-\frac{15}{2}+5\)
\(\implies x=\frac{25}{2}\) \(\implies x=-\frac52\)
Vậy \(x\in\left\lbrace\frac{25}{2};-\frac52\right\rbrace\)