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

Câu 1:

c: \(\frac19+\frac28+\frac37+\cdots+\frac91\)

\(=\left(\frac19+1\right)+\left(\frac28+1\right)+\cdots+\left(\frac82+1\right)+1\)

\(=\frac{10}{2}+\frac{10}{3}+\cdots+\frac{10}{10}=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)

Ta có: \(\left(\frac12+\frac13+\frac14+\cdots+\frac{1}{10}\right)\cdot x=\frac19+\frac28+\frac37+\cdots+\frac91\)

=>\(x\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)

=>x=10

Câu 2:

d: \(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023\cdot2024}\)

\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023}-\frac{1}{2022\cdot2023\cdot2024}\right)\)

\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2022\cdot2023\cdot2024}\right)\)

Hẹ hẹ

a) \(x+2x+3x+...+100x=-213\)

\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)

\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)

\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)

d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)

                                                                    \(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)

c) \(3\left(x-2\right)+2\left(x-1\right)=10\)

\(\Rightarrow3x-6+2x-2=10\)

\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)

a) \(x+2x+3x+4x+...+100x=-213\)

\(x.\left(1+2+3+4+...+100\right)=-213\)

\(x.5050=-213\)

\(x=-\frac{213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)

\(\frac{1}{2}x=-\frac{43}{12}\)

\(x=\frac{-43}{6}\)

\(\frac14-2x=5\)

\(2x=\frac14-5\)

\(2x=\frac{-19}{4}\)

\(x=-\frac{19}{4}:2\)

\(x=\frac{-19}{8}\)

\(\frac12x-\frac13=25\%\)

\(\frac12x-\frac13=\frac14\)

\(\frac12x=\frac14+\frac13\)

\(\frac12x=\frac{7}{12}\)

\(x=\frac{7}{12}:\frac12\)

\(x=\frac76\)

b: 2x^3-1=15

=>2x^3=16

=>x=2

\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)

=>y-25=32; z+9=50

=>y=57; z=41

d: 3/5x=2/3y

=>9x=10y

=>x/10=y/9=k

=>x=10k; y=9k

x^2-y^2=38

=>100k^2-81k^2=38

=>19k^2=38

=>k^2=2

TH1: k=căn 2

=>\(x=10\sqrt{2};y=9\sqrt{2}\)

TH2: k=-căn 2

=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)

\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)

\(\Leftrightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)

\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)

\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)

\(\Leftrightarrow x+1=0\)( \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\)) 

\(\Leftrightarrow x=-1\)

Vậy x=-1

mỗi phân số + 1 thì sẽ có tử chung là x + 1

chuyển vế có \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\))  =0

mà tổng các phân số kia khác 0 nên x+1 bằng 0 

=> x=-1

a: =>x/3=-5/2

hay x=-15/2

b: \(\Leftrightarrow\dfrac{7}{3}:x=\dfrac{1}{5}-\dfrac{4}{9}=\dfrac{9-20}{45}=\dfrac{-11}{45}\)

\(\Leftrightarrow x=\dfrac{7}{3}:\dfrac{-11}{45}=\dfrac{7}{3}\cdot\dfrac{-45}{11}=\dfrac{-105}{11}\)

c: \(\Leftrightarrow x=\dfrac{-7}{2}\cdot2=-7\)

d: =>x/27=-1/3+2/9=2/9-3/9=-1/9=-3/27

=>x=-3

Hay quá, mk cũng đang tìm câu này nè

a) ⇒ \(\dfrac{5}{3}x\) \(=\) \(\dfrac{5}{6}+\dfrac{1}{4}\)

⇒ \(\dfrac{5}{3}x=\dfrac{13}{12}\)

⇒ \(x=\dfrac{13}{12}:\dfrac{5}{3}\)

⇒\(x=\dfrac{13}{20}\)

a) \(32< 2^x< 128\)

=> \(2^5< 2^x< 2^7\)

=> x = 6

b) \(2^{x-1}+4\cdot2^x=9\cdot2^5\)

=> \(2^{x-1}+2^2\cdot2^x=9\cdot2^5\)

=> \(2^{x-1}+2^{2+x}=9\cdot2^5\)

=> 9.2^x-1 = 9.2⁵

=> 2^x-1 = \(\frac{9\cdot2^5}{9}=2^5\)

=> x - 1 = 5 => x = 6

c) \(9\cdot27\le3^x\le243\)

=> \(243\le3^x\le243\)

=> x = 5

d) Giống câu b)

e) \(3^{x-1}+5\cdot3^{x-2}=216\)

=> 8.3^x-2 = 216

=> 3^x-2 = 27

=> 3^x-2 = 3³

=> x - 2 = 3 => x = 5

f) 27^x-3 = 9^x+3 

=> 27^x-3 = 9^x+3

=> (3³)^x-3 = (3²)^x+3

=> 3^3x-9 = 3^{2x + 6}

=> không thỏa mãn x vì x là phân số mà theo đề bài là số nguyên

g) x²⁰¹⁹ = x => x²⁰¹⁹ - x = 0 => x(x²⁰¹⁸ - 1) = 0 => x = 0 hoặc x = 1

a) 

\(2^5< 2^x< 2^7\) 

\(5< x< 7\) 

\(x=6\) 

b) 

\(2^{x-1}+2^2\cdot2^x=9\cdot2^5\) 

\(2^{x-1}+2^{2+x}=9\cdot2^5\) 

\(2^{x-1}\left(1+2^3\right)=9\cdot2^5\) 

\(2^{x-1}\cdot9=9\cdot2^5\) 

\(2^{x-1}=2^5\) 

\(x-1=5\) 

\(x=6\)