a) x=-3/20

b) x=-5/7

a)  \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{8}{12}\)

\(\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11-8}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{5}{20}-\dfrac{8}{20}\)

\(\Leftrightarrow x=\dfrac{-3}{20}\)

Vậy x= \(\dfrac{-3}{20}\)

b)  \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{8-15}{20}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(\Leftrightarrow x=\dfrac{1}{4}.\dfrac{-20}{7}\)

\(\Leftrightarrow x=\dfrac{-5}{7}\)

Vậy x= \(\dfrac{-5}{7}\)

a) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow x+\dfrac{2}{5}=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{11}{12}-\dfrac{8}{12}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}-\dfrac{2}{5}=\dfrac{5}{20}-\dfrac{8}{20}=\dfrac{-3}{20}\)

b) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{-7}{20}\)

hay \(x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{1}{4}\cdot\dfrac{-20}{7}=\dfrac{-5}{7}\)

\(a,\dfrac{11}{12}x+\dfrac{3}{4}=-\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{11}{12}x=-\dfrac{1}{6}-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{11}{12}x=-\dfrac{11}{12}\)

\(\Leftrightarrow x=-\dfrac{11}{12}:\dfrac{11}{12}\)

\(\Leftrightarrow x=-\dfrac{11}{12}.\dfrac{12}{11}\)

\(\Leftrightarrow x=-1\)

\(b,3-\left(\dfrac{1}{6}-x\right).\dfrac{2}{3}=\dfrac{2}{3}\)

\(\Leftrightarrow3-\dfrac{2}{3}.\left(\dfrac{1}{6}-x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow3-\dfrac{1}{9}+\dfrac{2}{3}x=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{2}{3}-3+\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{2}{3}x=-\dfrac{20}{9}\)

\(\Leftrightarrow x=-\dfrac{20}{9}:\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{10}{3}\)

a) Ta có : \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\\ x+2=0\Rightarrow x=-2\)

Lập bảng xét dấu:

TH : Xét x < -2

Ta có : - ( x+ 2) - (x - \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)

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

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

-2x = 2\(\dfrac{1}{4}\)

=> x = \(-1\dfrac{1}{8}\) ( loại )

TH 2: \(-2\le x< \dfrac{1}{2}\)

Ta có : x + 2 + ( -x + \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)

=> \(2,5=\dfrac{3}{4}\) ( loại )

TH3 : \(x\ge\dfrac{1}{2}\)

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

2x + 1,5 = \(\dfrac{3}{4}\)

x = -0,375( loại )

vậy ....

b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{3}-2x=-\dfrac{3}{4}\\ \Rightarrow2x=1\dfrac{5}{12}\\ \Rightarrow x=\dfrac{17}{24}\)

c) \(\left|x-1\right|+2.\left(x+4\right)=10\\ \Rightarrow\left|x-1\right|=10-2x-8\\ \Rightarrow\left|x-1\right|=2-2x\)

TH1 : \(x-1\ge0\) \(\Rightarrow x\ge1\)

\(\Rightarrow x-1=2-2x\\ \Rightarrow3x=3\\ \Rightarrow x=1\left(TM\right)\)

TH2 : \(x-1< 0\Rightarrow x< 1\)

=> \(x-1=-2+2x\\ \Rightarrow-x=-1\Rightarrow x=1\)(loại)

Vậy x = 1

b. \(\left(\dfrac{2}{3}-2x\right)\cdot1\dfrac{1}{2}=\dfrac{3}{4}\Rightarrow\dfrac{2}{3}-2x=\dfrac{3}{4}:\dfrac{3}{2}=\dfrac{1}{2}\Rightarrow-2x=\dfrac{1}{2}-\dfrac{2}{3}=-\dfrac{1}{6}\Rightarrow x=-\dfrac{1}{6}:\left(-2\right)=\dfrac{1}{12}\)

Vậy \(x=\dfrac{1}{12}\)

c. \(\left|x-1\right|+2\left(x+4\right)=10\Rightarrow\left|x-1\right|+2x+8=10\Rightarrow\left|x-1\right|+2x=10-8=2\)

\(\Rightarrow\left[{}\begin{matrix}x-1+2x=2;x-1\ge0\\-\left(x-1\right)+2x=2;x-1< 0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+2x=2+1+3;x\ge1\\-x+1+2x=2;x< 1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=3;x\ge1\\-x+2x=2-1=1;x< 1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3:3=1;x\ge1\\x=1;x< 1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x\in\varnothing\end{matrix}\right.\)

Vậy x = 1

d. \(\dfrac{11}{12}+\dfrac{11}{12\cdot23}+...+\dfrac{11}{89\cdot100}+x=1\dfrac{2}{3}\)

\(\Rightarrow\dfrac{11}{12}+\dfrac{11}{276}+...+\dfrac{11}{8900}+x=\dfrac{5}{3}\)

\(\Rightarrow\dfrac{22}{23}+...+\dfrac{11}{8900}+x=\dfrac{5}{3}\)

\(\Rightarrow\dfrac{99}{100}+x=\dfrac{5}{3}\Rightarrow x=\dfrac{5}{3}-\dfrac{99}{100}=\dfrac{203}{300}\)

Vậy \(x=\dfrac{203}{300}\)

e. \(\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)-x+4\dfrac{221}{231}=2\dfrac{1}{3}\)

\(\Rightarrow\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)-x=\dfrac{7}{3}-4\dfrac{221}{231}\)

\(\Rightarrow x=\dfrac{\dfrac{7}{3}-\dfrac{1145}{231}}{\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}}=\dfrac{-\dfrac{202}{77}}{\dfrac{2}{143}+\dfrac{2}{195}+...+\dfrac{2}{399}}=\dfrac{-\dfrac{202}{77}}{\dfrac{10}{231}}=\dfrac{-202}{77}\cdot\dfrac{231}{10}=\dfrac{-303}{5}\)

Vậy \(x=-\dfrac{303}{5}\)

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

Lời giải:

a) Hiển nhiên vế trái $\geq 0$ do tính chất của trị tuyệt đối.

$\Rightarrow 4x\geq 0\Rightarrow x\geq 0$. Đến đây ta có thể phá bỏ dấu trị tuyệt đối

$|x+\frac{11}{17}|+|x+\frac{2}{17}|+|x+\frac{4}{17}|=4x$

$x+\frac{11}{17}+x+\frac{2}{17}+x+\frac{4}{17}=4x$

$3x+1=4x$

$x=1$

b) Hiển nhiên vế trái $\geq 0$ nên $11x\geq 0\Rightarrow x\geq 0$

Khi đó:

$|x+\frac{1}{2}|+|x+\frac{1}{6}|+|x+\frac{1}{12}|+...+|x+\frac{1}{110}|=x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}$

$=10x+(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110})$

$=10x+(1-\frac{1}{11})=10x+\frac{10}{11}=11x$

$\Rightarrow x=\frac{10}{11}$

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a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)

a: =>11(x-3)=6(x-5)

=>11x-33=6x-30

=>5x=3

=>x=3/5

b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3

=>-9/4x+11/12=8/3

=>-9/4x=32/12-11/12=21/12=7/4

=>x=-7/9

c: =>1/2x-1/3-2/3x-1=x

=>-1/6x-4/3=x

=>-7/6x=4/3

=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7

d: =>1-2x-3x+1=7/2

=>-5x=3/2

=>x=-3/10