c: \(=\left(x+y-5\right)\left(x+y+5\right)\)

\(A=4x^2-5xy+3y^2\\\Rightarrow 2A=2\cdot(4x^2-5xy+3y^2)\\\Rightarrow2A=8x^2-10xy+6y^2\\B=3x^2+2xy+y^2\\\Rightarrow3B=3\cdot(3x^2+2xy+y^2)\\\Rightarrow3B=9x^2+6xy+3y^2\\C=-x^2+3xy+2y^2\)

Khi đó: $2A-3B-C$

$=(8x^2-10xy+6y^2)-(9x^2+6xy+3y^2)-(-x^2+3xy+2y^2)$

$=8x^2-10xy+6y^2-9x^2-6xy-3y^2+x^2-3xy-2y^2$

$=(8x^2-9x^2+x^2)+(-10xy-6xy-3xy)+(6y^2-3y^2-2y^2)$

$=-19xy+y^2$

2A-3B-C

\(=2\left(4x^2-5xy+3y^2\right)-3\left(3x^2+2xy+y^2\right)+x^2-3xy-2y^2\)

\(=8x^2-10xy+6y^2-9x^2-6xy-3y^2+x^2-3xy-2y^2\)

\(=-19xy+y^2\)

`Answer:`

\(a)\left(-3x^2y-2xy^2+6\right)+\left(-x^2y+5xy^2-1\right)\)

\(=-3x^2y-2xy^2+6+-x^2y+5xy^2-1\)

\(=\left(-3x^2y-x^2y\right)+\left(-2xy^2+5xy^2\right)+\left(6-1\right)\)

\(=-4x^2y+3xy^2+5\)

\(b)\left(1,6x^3-3,8x^2y\right)+\left(-2,2x^2y-1,6x^3+0,5xy^2\right)\)

\(=1,6x^3-3,8x^2y+-2,2x^2y-1,6x^3+0,5xy^2\)

\(=\left(1,6x^3-1,6x^3\right)+\left(-3,8x^2y+-2,2x^2y\right)+0,5xy^2\)

\(=-6x^2y+0,5xy^2\)

\(c)\left(6,7xy^2-2,7xy+5y^2\right)-\left(1,3xy-3,3xy^2+5y^2\right)\)

\(=6,7xy^2-2,7xy+5y^2-1,3xy+3,3xy^2-5y^2\)

\(=\left(6,7xy^2+3,3xy^2\right)+\left(-2,7xy-1,3xy\right)+\left(5y^2-5y^2\right)\)

\(=10xy^2+-4xy\)

\(=10xy^2-4xy\)

\(d)\left(3x^2-2xy+y^2\right)+\left(x^2-xy+2y^2\right)-\left(4x^2-y^2\right)\)

\(=3x^2-2xy+y^2+x^2-xy+2y^2-4x^2+y^2\)

\(=\left(3x^2+x^2-4x^2\right)+\left(-2xy-xy\right)+\left(y^2+2y^2+y^2\right)\)

\(=-3xy+4y^2\)

\(e)\left(x^2+y^2-2xy\right)-\left(x^2+y^2+2xy\right)+\left(4xy-1\right)\)

\(=x^2+y^2-2xy-x^2-y^2-2xy+4xy-1\)

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

\(=-1\)

a) Ta có: \(A=\left(x^3-x^2y+xy^2-y^3\right)\left(x+y\right)\)

\(=x^4+x^3y-x^3y-x^2y^2+x^2y^2+xy^3-xy^3-y^4\)

\(=x^4-y^4\)

Thay x=2 và \(y=-\frac{1}{2}\) vào biểu thức \(A=x^4-y^4\), ta được:

\(A=2^4-\left(-\frac{1}{2}\right)^4\)

\(=16-\frac{1}{16}\)

\(=\frac{255}{16}\)

Vậy: \(\frac{255}{16}\) là giá trị của biểu thức \(A=\left(x^3-x^2y+xy^2-y^3\right)\left(x+y\right)\) tại x=2 và \(y=-\frac{1}{2}\)

b) Ta có: \(B=\left(a-b\right)\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)\)

\(=a^5+a^4b+a^3b^2+a^2b^3+ab^4-a^4b-a^3b^2-a^2b^3-ab^4-b^5\)

\(=a^5-b^5\)

Thay a=3 và b=-2 vào biểu thức \(B=a^5-b^5\), ta được:

\(B=3^5-\left(-2\right)^5\)

\(=243-\left(-32\right)\)

\(=243+32=275\)

Vậy: 275 là giá trị của biểu thức \(B=\left(a-b\right)\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)\) tại a=3 và b=-2

c) Ta có: \(C=\left(x^2-2xy+2y^2\right)\left(x^2+y^2\right)+2x^3-3x^2y^2+2xy^3\)

\(=x^4+x^2y^2-2x^3y-2xy^3+2x^2y^2+2y^4+2x^3-3x^2y^2+2xy^3\)

\(=x^4-2x^3y+2y^4+2x^3\)

Thay \(x=y=\frac{-1}{2}\) vào biểu thức \(C=x^4-2x^3y+2y^4+2x^3\), ta được:

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

\(=\frac{1}{16}-2\cdot\frac{-1}{8}\cdot\frac{-1}{2}+2\cdot\frac{1}{16}+2\cdot\frac{-1}{8}\)

\(=\frac{1}{16}-\frac{1}{8}+\frac{1}{8}-\frac{1}{4}\)

\(=\frac{1}{16}-\frac{1}{4}=\frac{1}{16}-\frac{4}{16}=\frac{-3}{16}\)

Vậy: \(-\frac{3}{16}\) là giá trị của biểu thức \(C=\left(x^2-2xy+2y^2\right)\left(x^2+y^2\right)+2x^3-3x^2y^2+2xy^3\) tại \(x=y=\frac{-1}{2}\)

\(a)xy+3x-2y=11\)

\(\Leftrightarrow xy+3x-2y-6=5\)

\(\Leftrightarrow x\left(y+3\right)-2\left(y+3\right)=5\)

\(\Leftrightarrow\left(y+3\right)\left(x-2\right)=5\)

\(\Leftrightarrow\hept{\begin{cases}y+3=-1\\x-2=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-4\\x=-3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+3=1\\x-2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-2\\x=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+3=-5\\x-2=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-8\\x=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+3=5\\x-2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=2\\x=3\end{cases}}\)

\(b)2x^2-2xy+x-y=12\)

\(\Leftrightarrow2x\left(x-y\right)+\left(x-y\right)=12\)

\(\Leftrightarrow\left(x-y\right)\left(2x+1\right)=12\)

\(\Rightarrow\left(x-y\right);\left(2x+1\right)\inƯ\left(12\right)\)

\(\RightarrowƯ\left(12\right)\in\left\{-1;1;-2;2;-3;3;-4;4;-6;6;-12;12\right\}\)

Vì 2x+1 luôn lẻ

\(\Rightarrow2x+1\in\left\{-1;1;-3;3\right\}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=-1\\x-y=-12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=11\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=1\\x-y=12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-12\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=-3\\x-y=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=3\\x-y=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

\(c)2xy-10y-x=13\)

\(\Leftrightarrow x\left(2y-1\right)-2y.5+5=18\)

\(\Leftrightarrow x\left(2y-1\right)-5\left(2y-1\right)=18\)

\(\Leftrightarrow\left(2y-1\right)\left(x-5\right)=18\)

\(\Leftrightarrow2y-1;x-5\inƯ\left(18\right)\)

\(\RightarrowƯ\left(18\right)\in\left\{-1;1;-2;2;-3;3;-6;6;-9;9;-18;18\right\}\)

Vì 2y-1  luôn lẻ

=>2y-1 thuộc {-1;1;-3;3;-9;9}

=> Làm  tương tự nhé

\(e)xy-2y^2+8y-3x=13\)

\(\Leftrightarrow xy-2y^2+2y+6y-3x-6=7\)

\(\Leftrightarrow y\left(x-2y+2\right)+3\left(-x+2y-2\right)=7\)

\(\Leftrightarrow y\left(x-2y+2\right)-3\left(x-2y+2\right)=7\)

\(\Leftrightarrow\left(x-2y+2\right)\left(y-3\right)=7\)

Tự khai triển như các câu trên.

Mình đg bận nên ko lm đc hết câu.

a) \(\left(3x^2-2xy+y^2\right)+\left(x^2-xy+2y^2\right)-\left(4x^2-y^2\right)\)

\(=3x^2-2xy+y^2+x^2-xy+2y^2-4x^2+y^2\)

\(=4y^2-3xy\)

b) \(\left(x^2-y^2+2xy\right)-\left(x^2+xy+2y^2\right)+\left(4xy-1\right)\)

\(=x^2-y^2+2xy-x^2-xy-2y^2+4xy-1\)

\(=-3y^2+5xy-1\)

a/ \(\left(3x^2-2xy+y^2\right)+\left(x^2-xy+2y^2\right)-\left(4x^2-y^2\right)\)

\(=3x^2-2xy+y^2+x^2-xy+2y^2-4x^2+y^2\)

\(=-3xy+4y^2\)

b/ \(\left(x^2-y^2+2xy\right)-\left(x^2+xy+2y^2\right)+\left(4xy-1\right)\)

\(=x^2-y^2+2xy-x^2-xy-2y^2+4xy-1\)

\(=-3y^2+5xy-1\)

\(Q=x^2+2xy+\left(-3x^3+3x^3\right)+\left(2y^3-y^3\right)=x^2+2xy+y^3\)

\(P=\left(\dfrac{1}{3}x^2y-\dfrac{1}{3}x^2y\right)+\left(xy^2+\dfrac{1}{2}xy^2\right)-\left(xy+5xy\right)=\dfrac{3}{2}xy^2-6xy\)