đề sai

a) \(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|\left(-3,2\right)+\dfrac{2}{5}\right|\)

\(\Rightarrow\left|x-\dfrac{1}{3}\right|+0,8=\left|-3,2+0,4\right|\)

\(\Rightarrow\left|x-\dfrac{1}{3}\right|+0,8=2,8\)

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

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=2\\x-\dfrac{1}{3}=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{-5}{3}\end{matrix}\right.\)

3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

Bài 1:

a.

$|x+\frac{7}{4}|=\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)

b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$

$|2x+1|=\frac{11}{15}$

\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)

c.

$3x(x+\frac{2}{3})=0$

\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)

d.

$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$

$\Leftrightarrow x=\frac{2}{5}$

Bài 2:

$\frac{1}{100}-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}$

$=\frac{99}{100}$

$\Rightarrow A=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=\frac{-49}{50}$

Bài 1: 

a) Ta có: \(\left|x+\dfrac{7}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{7}{4}=\dfrac{1}{2}\\x+\dfrac{7}{4}=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)

b) Ta có: \(\left|2x+1\right|-\dfrac{2}{5}=\dfrac{1}{3}\)

\(\Leftrightarrow\left|2x+1\right|=\dfrac{11}{15}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{11}{15}\\2x+1=\dfrac{-11}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-4}{15}\\2x=\dfrac{-26}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{15}\\x=\dfrac{-13}{15}\end{matrix}\right.\)

c) Ta có: \(3x\left(x+\dfrac{2}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8

a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3+2x^2-x=5x\left(2-x\right)-11\left(x+2\right)\)

=>-x^2+2x-1=10x-5x^2-11x-22

=>-x^2+2x-1=-5x^2-x-22

=>4x^2+3x+21=0

=>PTVN

b: \(\Leftrightarrow\left(x+10\right)\left(x+4\right)+3\left(x+4\right)\left(x-2\right)=4\left(x+10\right)\left(x-2\right)\)

=>x^2+14x+40+3(x^2+2x-8)=4(x^2+8x-20)

=>x^2+14x+40+3x^2+6x-24=4x^2+32x-80

=>20x+16=32x-80

=>-12x=-96

=>x=8

c: \(\Leftrightarrow6\left(x-3\right)+7\left(x-5\right)=13x+4\)

=>6x-18+7x-35=13x+4

=>-53=4(loại)

d: =>3(2x-1)-5(x-2)=3(x+7)

=>6x-3-5x+10=3x+21

=>3x+21=x+7

=>x=-7

e: =>x^3-6x^2+12x-8-x^3-3x^2-3x-1=-9x^2+1

=>-9x^2+9x-9=-9x^2+1

=>9x=10

=>x=10/9

a: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

b: 

c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)

\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)