Question

Tìm x biết:

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

b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\)

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

d) \(\dfrac{11}{12}+\dfrac{11}{12.23}+...+\dfrac{11}{89.100}+x=1\dfrac{2}{3}\)

e) \(\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{19.21}\right)-x+4\dfrac{221}{231}=2\dfrac{1}{3}\)

Answer 1

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

Lập bảng xét dấu:

x		-2		\(\dfrac{1}{2}\)
x + 2	-	0	+		+
x - \(\dfrac{1}{2}\)	-		-	0	+

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 ....

Answer 2

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

Answer 3

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 \(x=-\dfrac{303}{5}\)