B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé

giúp em vớiiiiiii

vì (x+y)^2-xy=x^2+y^2+2xy-xy=x^2+y^2+xy

a)B=3x³ -2y³-6x²y²+xy

   B=(3x³-6x²y²)+(xy-2y³)

   B=3x²(x-2y²)+y(x-2y²)

    B=(x-2y²)(3x²+y)
tại x=\(\frac{2}{3}\)và y=\(\frac{1}{2}\)ta có B=(x-2y²)(3x²+y)=(\(\frac{2}{3}\)-2*^{\(\frac{1}{2}\)}^2 )(3*\(\frac{2}{3}\)^2+\(\frac{1}{2}\))=\(\frac{1}{6}\)*\(\frac{11}{6}\)=\(\frac{11}{36}\)

b)C= 2x+xy²-x²y-2y

   C=(2x-2y)+(xy²-x²y)

   C=2(x-y)-xy(x-y)

   C=(2-xy)(x-y)

tại x=\(-\frac{1}{2}\)và y=\(-\frac{1}{3}\)ta có C=(2-xy)(x-y)=(2-\(-\frac{1}{2}\)*\(-\frac{1}{3}\))(\(-\frac{1}{2}\)+\(\frac{1}{3}\))=\(\frac{-11}{36}\)

A(1/2^2022)=1/2^2022+1/2^4044+...+1/2^(2022^2021)

=>2^2022*A=1+1/2^2022+...+1/2^(2022^2020)

=>A*(2^2022-1)=1-1/2^(2022^2021)

=>\(A=\dfrac{2^{2022^{2021}}-1}{2^{2022}-1}\)

\(\Rightarrow x^3+x^2y-2x^2-xy-y^2+2y+y+x+2020\)

\(x^2.\left(x+y-2\right)-y\left(x+y-2\right)+y+x+2020\)(1)

Thay x+y-2=0 vào (1) , ta được :

\(x^2.0-y.0+y+x+2020\\ =0+y+x+2020\)

\(=x+y+2022-2\\ =\left(x+y-2\right)+2022\\ \)(2)

Thay x+y-2 vào (2), ta được

\(=0+2022=2022\)

_ Tham khảo thôi ậ, nếu sai thì mong mn thông cảm_

_# yum #_

A       = 2\(x^2\)y + \(xy\) - 3\(xy\)

Thay \(x\) = -2; y = 4 vào biểu thức A ta có: 

A      =  2\(\times\) (-2)² \(\times\) 4 + (-2) \(\times\) 4 - 3 \(\times\) (-2) \(\times\) 4

A     = 2 \(\times\) 4 \(\times\) 4 - 8 + 6 \(\times\) 4

A     = 8 \(\times\) 4 - 8 + 24

A     =  32 - 8 + 24

A    =  24 + 24

A    =    48

B = (2\(x^2\) + \(x\) - 1) - ( \(x^2+5x-1\) )

Thay \(x\) = - 2 vào biểu thức B ta có:

B = { 2\(\times\)(-2)² + (-2) - 1} - { (-2)² +5\(\times\)(-2) - 1}

B = { 2 \(\times\) 4  - 3} - { 4 - 10 - 1}

B = { 8 - 3} - { 4 - 11}

B = 5 - (-7)

B = 5 + 7

B = 12

1/ \(\left(x^2+1\right)\left(x-2\right)+2x=4.\)

\(\left(x^2+1\right)\left(x-2\right)+2x-4=0\)

\(\left(x^2+1\right)\left(x-2\right)+\left(2x-4\right)=0\)

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

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

\(\left(x-2\right)\left(x^2+3\right)=0\)

TH1:\(x-2=0\Rightarrow x=2\)

TH2: \(x^2+3=0\)

\(\Rightarrow x^2=-3\)(vô lí)

\(\Rightarrow x\in\left\{2\right\}\)

2/ \(A=a\left(b-3\right)-b\left(b-1\right)\)

đề sai f ko ạ, do mik đâu thấy C mà bạn lại cho đề c=2???

\(B=xy\left(x+y\right)-2x-2y\)

\(B=xy\left(x+y\right)-\left(2x+2y\right)\)

\(B=xy\left(x+y\right)-2\left(x+y\right)\)

\(B=\left(x+y\right)\left(xy-2\right)\)

có xy=8 ; x+y=7

\(\Rightarrow B=\left(x+y\right)\left(xy-2\right)\)

\(\Rightarrow B=8\cdot\left(8-2\right)=8\cdot6=48\)

a) \(2x=5y\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x+y}{5+2}=\dfrac{-21}{7}=-3\)

Khi đó: \(\left\{{}\begin{matrix}\dfrac{x}{5}=-3\\\dfrac{y}{2}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3.5=-15\\y=-3.2=-6\end{matrix}\right.\)

\(\Rightarrow O=x^2-xy+2y=\left(-3\right)^2-\left(-15\right).\left(-6\right)+2.\left(-6\right)=9-90-12=-93\)

b)

Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)

\(\Rightarrow2k.5k=90\\ \Leftrightarrow10k^2=90\\ \Leftrightarrow k^2=9\\ \Leftrightarrow\left[{}\begin{matrix}k=-3\\k=3\end{matrix}\right.\)

Nếu k = -3

\(\Rightarrow\left\{{}\begin{matrix}x=-3.2=-6\\y=-3.5=-15\end{matrix}\right.\)

\(\Rightarrow Q=-93\)

Nếu k = 3

\(\Rightarrow\left\{{}\begin{matrix}x=3.2=6\\y=3.5=15\end{matrix}\right.\)

\(\Rightarrow Q=6^2-6.15+2.15=-24\)