`[2-x]/[x+3] > x+1`    `ĐK: x \ne -3`

 `=>` Loại đ/á `\bb A`

Thay `x=-1` vào bất ptr có: `1,5 > 0` (Luôn đúng) `->\bb B` t/m

Thay `x=2` vào bất ptr có: `0 > 3` (Vô lí) `->\bb C` loại

Thay `x=0` vào bất ptr có: `2/3 > 1` (Vô lí) `->\bb D` loại

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      `=>` Chọn `\bb B`

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

Thay \(x=\dfrac{5}{4}\) vào D ta có:

\(D=4x+3=4.\dfrac{5}{4}+3=5+3=8\)

Thay \(x=-\dfrac{1}{4}\) vào D ta có:

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

Để \(D=\dfrac{3}{2}\)

\(\Leftrightarrow4x+3=\dfrac{3}{2}\\ \Leftrightarrow4x=-\dfrac{3}{2}\\ \Leftrightarrow x=-\dfrac{3}{8}\)

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

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

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

\(x=\dfrac{5}{4}\text{ thì }D=4.\dfrac{5}{4}+3=5+3=8\)

\(x=\dfrac{-1}{4}\text{ thì }D=4.\left(\dfrac{-1}{4}\right)+3=\left(-1\right)+3=2\)

\(D=\dfrac{-5}{2}\Leftrightarrow4x+3=\dfrac{-5}{2}\)

               \(\Leftrightarrow4x=\dfrac{-11}{2}\)

               \(\Leftrightarrow x=\dfrac{-11}{8}\)

=>1/4:(x-2/3)=2

=>(x-2/3)=1/8

=>x=1/8+2/3=3/24+16/24=19/24

13/4 - 1/4 : ( x - 2/3 )= 5/4

\(\Rightarrow\) 1/4 : ( x- 2/3 ) = 13/4 - 5/4

\(\Rightarrow\) 1/4 : ( x- 2/3)= 2

\(\Rightarrow\) x - 2/3 = 1/4 :2

\(\Rightarrow\) x- 2/3 = 1/8

\(\Rightarrow\) x= 1/8 +2/3 =19/24

Vậy x = 19/24

\(\dfrac{13}{4}-\dfrac{1}{4}\div\left(x-\dfrac{2}{3}\right)=\dfrac{5}{4}\\ \Rightarrow\dfrac{1}{4}\div\left(x-\dfrac{2}{3}\right)=\dfrac{13}{4}-\dfrac{5}{4}\\ \Rightarrow\dfrac{1}{4}\div\left(x-\dfrac{2}{3}\right)=\dfrac{8}{4}=2\\ \Rightarrow x-\dfrac{2}{3}=\dfrac{1}{4}\div2\\ \Rightarrow x-\dfrac{2}{3}=\dfrac{1}{4}\times\dfrac{1}{2}\\ \Rightarrow x-\dfrac{2}{3}=\dfrac{1}{8}\\ \Rightarrow x-\dfrac{16}{24}=\dfrac{3}{24}\\ \Rightarrow x=\dfrac{3}{24}+\dfrac{16}{24}\\ \Rightarrow x=\dfrac{19}{24}\)

Vậy \(x=\dfrac{19}{24}\) 

x-1/2=y-2/3=z-3/4 => x-1/2 = 2y-4/6 = 3z-9/12 

Theo tính chất dãy tỉ số bằng nhau ta có

x-1/2=2y-4/6=3z-9/12 =[(x-1) - (2y-4) + (3z-9)] / 2+6+12

=[(x-2y+3z)-(1-4+9)] / 20

=-10-6 /20= -16/20=-4/5

Ta có x-1/2=-4/5 => x-1=-8/5=> x=-3/5

Còn lại bạn tự làm nha (Nếu mình làm đúng thì k cho mình)

x-1/x+2=x-2/x+3

=>(x-1)(x+3)=(x-2)(x+2)

=>x(x+3)-1(x+3)=x(x+2)-2(x+2)

=>x²+3x-x-3=x²+2x-2x-4

=>x²+2x-3=x²-4=>2x-3=-4=>2x=-1

=>x=-1/2=-,5

Vậy...

Hoang Phuc: Chac k z pan???

bn có thể chia ra phần TS và MS rõ hơn đc ko

Ta có :

\(\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}=\frac{x}{3^2}+\frac{x}{3^3}+\frac{x}{3^4}\)

\(\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}-\frac{x}{3^2}-\frac{x}{3^3}-\frac{x}{3^4}=0\)

\(x^2\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}\right)=0\)

\(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}\ne0\Rightarrow x^2=0\)

\(x=0\)