a: \(\dfrac{3x+2}{4}-\dfrac{3x+1}{3}=\dfrac{5}{6}\)

=>3(3x+2)-4(3x+1)=10

=>9x+6-12x-4=10

=>-3x+2=10

=>-3x=8

=>x=-8/3

b: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{9x-10}{4-x^2}\)

=>(x-1)(x-2)-x(x+2)=-9x+10

=>x^2-3x+2-x^2-2x=-9x+10

=>-5x+2=-9x+10

=>x=2(loại)

\(\frac{22}{9}-\left(x+\frac{1}{2}\right)^2=\frac{7}{3}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{22}{9}-\frac{7}{3}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{1}{9}\)

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

\(\Rightarrow x+\frac{1}{2}=\frac{1}{3}\)

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

\(x=-\frac{1}{6}\)

\(\frac{22}{9}-\left(x+\frac{1}{2}\right)^2=\frac{7}{3}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{22}{9}=\frac{7}{3}\)\

\(\left(x+\frac{1}{2}\right)^2=\frac{22}{9}-\frac{21}{9}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{1}{9}\)

\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{3}\)

TH1:\(x+\frac{1}{2}=\frac{1}{3}\)

       \(x=\frac{1}{3}-\frac{1}{2}\)

        \(x=-\frac{1}{6}\)

TH2:\(x+\frac{1}{2}=-\frac{1}{3}\)

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

       \(x=-\frac{5}{6}\)

Vậy \(x\in\left\{-\frac{1}{6};-\frac{5}{6}\right\}\)

x  +(22/9 -1/2) =7/3

x   +(44/18-9/18)=7/3 + 35/18

x                         =42/18+35/18

x                          =77/18

k nha bạn                    

`A=(x^2-2)(x^2+x-1)-x(x^3+x^2-3x-2)`

`=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x`

`=(x^4-x^4)+(x^3-x^3)+(3x^2-x^2-2x^2)+(2x-2x)+2`

`=2`

\(\frac{2}{3}-\frac{1}{5}.\left(\frac{3.x}{2}-\frac{1}{4}\right)=\frac{11}{2}-\frac{1}{4}\)

\(\Leftrightarrow\frac{2}{3}-\frac{1}{5}.\left(\frac{3.x}{2}-\frac{1}{4}\right)=\frac{21}{4}\)

\(\Leftrightarrow\frac{1}{5}.\left(\frac{3.x}{2}-\frac{1}{4}\right)=\frac{2}{3}-\frac{21}{4}\)

\(\Leftrightarrow\frac{1}{5}.\left(\frac{3.x}{2}-\frac{1}{4}\right)=\frac{-55}{12}\)

\(\Leftrightarrow\frac{3.x}{2}-\frac{1}{4}=\frac{-55}{12}:\frac{1}{5}\)

\(\Leftrightarrow\frac{3.x}{2}-\frac{1}{4}=\frac{-275}{12}\)

\(\Leftrightarrow\frac{3.x}{2}=\frac{-275}{12}+\frac{1}{4}\)

\(\Leftrightarrow\frac{3.x}{2}=\frac{-68}{3}\)

\(\Leftrightarrow\left(3.x\right).3=-136\)

\(\Leftrightarrow3.x=-136:3\)

\(\Leftrightarrow3.x=\frac{-136}{3}\)

\(\Leftrightarrow x=\frac{-136}{3}:3\)

\(\Leftrightarrow x=\frac{-136}{9}\)

Tích trước đi

mình k rồi đó bạn trả lời đi

a/ \(C=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(x+y-1\right)\)

\(C=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-1\right)=x+y-1\) (do x+y-2=0)

Mà x+y-2=0 => x+y-1=1 => C=1

b/  Với x=2; y=2 Ta nhận thấy \(x^3-2y^2=2^3-2.2^2=2^3-2^3=0\) => D=0

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

rút gọn à banj

đúng rồi á

(x²+4xy+4y²)-(2x+4y)+10=(x+2y)²-2(x+2y)+10=5²-10+10=25 :333

:333 là biểu cảm nhé

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