a) \(\left(2x-1\right)\left(x^2-x+1\right)+x^2\left(3-2x\right)=2\)

\(\Rightarrow2x^3-2x^2+2x-x^2+x-1+3x^2-2x^3=2\)

\(\Rightarrow\left(2x^3-2x^3\right)-\left(2x^2+x^2-3x^2\right)+\left(2x+x\right)-1=2\)

\(\Rightarrow3x-1=2\)

\(\Rightarrow3x=2-1\)

\(\Rightarrow3x=1\)

\(\Rightarrow x=\dfrac{1}{3}\)

a. bằng 1/3 nha

b: =>\(3\left(x-2-4x^2+8x\right)+4\left(3x^2+7x+2\right)=24\)

=>-12x^2+27x-6+12x^2+28x+8=24

=>55x+2=24

=>55x=22

=>x=2/5

c: \(\Leftrightarrow\left(x^2+3x+2\right)\left(x+3\right)-x^3-8x^2-16x-6=0\)

=>\(x^3+6x^2+11x+6-x^3-8x^2-16x-6=0\)

=>\(-2x^2-5x=0\)

=>x=0 hoặc x=-5/2

d: \(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+8=0\)

=>6x+12=0

=>x=-2

19

Từ pt đầu ta có:

\(x^2-xy-2xy+2y^2=0\)

\(\Leftrightarrow x\left(x-y\right)-2y\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x-2y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\x=2y\end{matrix}\right.\)

TH1: \(x=y\) thế xuống pt dưới:

\(y^2-y-y^2=1\Rightarrow y=-1\Rightarrow x=-1\)

TH2: \(x=2y\) thế xuống pt dưới:

\(\left(2y\right)^2-2y-y^2=1\Leftrightarrow3y^2-2y-1=0\)

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

Vậy nghiệm của hệ là: \(\left(x;y\right)=\left(-1;-1\right);\left(1;2\right);\left(-\dfrac{1}{3};-\dfrac{2}{3}\right)\)

21.

Từ pt đầu:

\(xy+2=2x+y\Leftrightarrow xy-y+2-2x=0\)

\(\Leftrightarrow y\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

TH1: \(x=1\) thế xuống pt dưới:

\(2y+y^2+3y=6\Leftrightarrow y^2+5y-6=0\)

\(\Rightarrow\left[{}\begin{matrix}y=1\\y=-6\end{matrix}\right.\)

TH2: \(y=2\) thế xuông pt dưới

\(4x+4+6=6\Rightarrow x=-1\)

Vậy nghiệm của pt là: \(\left(x;y\right)=\left(1;1\right);\left(1;-6\right);\left(-1;2\right)\)

22.

\(\Leftrightarrow\left\{{}\begin{matrix}9x^2-6xy+y^2+6xy=10\\\left(3x-y\right)\left(10-6xy\right)=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-y\right)^2=10-6xy\\\left(3x-y\right)\left(10-6xy\right)=8\end{matrix}\right.\)

Thế \(10-6xy\) từ pt trên xuống dưới ta được 

\(\left(3x-y\right)\left(3x-y\right)^2=8\)

\(\Leftrightarrow\left(3x-y\right)^3=2^3\)

\(\Leftrightarrow3x-y=2\)

\(\Leftrightarrow y=3x-2\)

Thế vào pt đầu:

\(9x^2+\left(3x-2\right)^2=10\)

\(\Leftrightarrow18x^2-12x-6=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-\dfrac{1}{3}\Rightarrow y=-3\end{matrix}\right.\)

a: Ta có: \(\left(\dfrac{3}{5}-\dfrac{7}{4}\right):\dfrac{23}{10}\)

\(=\dfrac{12-35}{20}\cdot\dfrac{10}{23}\)

\(=\dfrac{-23\cdot10}{23\cdot20}=-\dfrac{1}{2}\)

b: Ta có: \(\left(-\dfrac{2}{3}+\dfrac{3}{4}\right)^2\cdot\dfrac{12}{5}-\dfrac{11}{5}\)

\(=\dfrac{1}{144}\cdot\dfrac{12}{5}-\dfrac{11}{5}\)

\(=\dfrac{1}{60}-\dfrac{11}{5}=\dfrac{1}{60}-\dfrac{132}{60}=\dfrac{-131}{60}\)