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\(a) (x+y)+(y-x)\)

\(= x+y+y-x\)

\(=(x-x)+(y+y)\)

\(= 2y\)

\(b) (x+y)-(y-x)\)

\(= x+y-y+x\)

\(= (x+x)+(y-y)\)

\(= 2x\)

\(c) (x-y)+(y-x)\)

\(= x-y+y-x\)

\(= (x-x)+(y-y) \)

\(=0\)

\(e) (x^2+xy-1)+(3x^2+xy+1)\)

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

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

\(= 4x^2 + 2xy\)

\(g) (x^2+xy-1)-(3x^2+xy+1)\)

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

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

\(= -2x^2 - 2\)

B

Dễ ợt ak

Hướng dẫn thôi nhé:

Lời giải:

a)\(xy+x+y+1=0\)

\(\Rightarrow x\left(y+1\right)+1\left(y+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(y+1\right)=0\)

b)\(xy-x-y=0\)

\(\Rightarrow xy-x-y+1=1\)

\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=1\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=1\)

c)\(xy-x-y-1=0\)

\(\Rightarrow xy-x-y+1=2\)

\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=2\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=2\)

d) \(xy-x-y+1=0\)

\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=0\)

e)\(xy+2x+y+11=0\)

\(\Rightarrow xy+2x+y+2=-9\)

\(\Rightarrow x\left(y+2\right)+1\left(y+2\right)=-9\)

\(\Rightarrow\left(x+1\right)\left(y+2\right)=-9\)

\(\begin{array}{l}T + H = 3{x^2}y - 2x{y^2} + xy + \left( { - 2{x^2}y + 3x{y^2} + 1} \right)\\ = 3{x^2}y - 2x{y^2} + xy - 2{x^2}y + 3x{y^2} + 1\\ = \left( {3{x^2}y - 2{x^2}y} \right) + \left( { - 2x{y^2} + 3x{y^2}} \right) + xy + 1\\ = {x^2}y + x{y^2} + xy + 1\\T - H = 3{x^2}y - 2x{y^2} + xy - \left( { - 2{x^2}y + 3x{y^2} + 1} \right)\\ = 3{x^2}y - 2x{y^2} + xy + 2{x^2}y - 3x{y^2} - 1\\ = \left( {3{x^2}y + 2{x^2}y} \right) + \left( { - 2x{y^2} - 3x{y^2}} \right) + xy - 1\\ = 5{x^2}y - 5x{y^2} + xy - 1\end{array}\)

Chọn B.

g: (x+3y)(x-3y+2)

=(x+3y)(x-3y)+2(x+3y)

=x^2-9y^2+2x+6y

h: (x+2y)(x-2y+3)

=(x+2y)(x-2y)+3(x+2y)

=x^2-4y^2+3x+6y

i: (x^2-xy+y^2)(x+y)

=x^3+x^2y-x^2y-xy^2+xy^2+y^3

=x^3+y^3

j: (x+y)(x^2-xy+y^2)=x^3+y^3

k: (5x-2y)(x^2-xy-1)

=5x*x^2-5x*xy-5x-2y*x^2+2y*xy+2y

=5x^3-5x^2y-5x-2x^2y+2xy^2+2y

=5x^3-7x^2y+2xy^2-5x+2y

l: (x^2y^2-xy+y)(x-y)

=x^3y^2-x^2y^3-x^2y^2+xy^2+xy-y^2

\(A,VT=x^3+y^3+x^3-y^3=2x^3=VP\\ B,VT=\left(x-y\right)\left(x^2+xy+y^2\right)=\left(x-y\right)\left(x^2+2xy+y^2-xy\right)\\ =\left(x-y\right)\left[\left(x+y\right)^2-xy\right]=VP\)

Sửa câu b \(cm:x^3-y^3=\left(x-y\right)\left[\left(x+y\right)^2-xy\right]\)

ĐKXĐ: \(xy\ne0;x\ne\pm y\)

\(\left\{{}\begin{matrix}\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{\dfrac{1}{y}+\dfrac{1}{x}}=\dfrac{5}{2}\\\dfrac{1}{y}-\dfrac{1}{x}+\dfrac{1}{\dfrac{1}{y}-\dfrac{1}{x}}=\dfrac{10}{3}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}a+b+\dfrac{1}{a+b}=\dfrac{5}{2}\\b-a+\dfrac{1}{b-a}=\dfrac{10}{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-\dfrac{5}{2}\left(a+b\right)+1=0\\\left(b-a\right)^2-\dfrac{10}{3}\left(b-a\right)+1=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a+b=2\\a+b=\dfrac{1}{2}\end{matrix}\right.\\\left[{}\begin{matrix}b-a=3\\b-a=\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a+b=2\\b-a=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=\dfrac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=\dfrac{5}{2}\end{matrix}\right.\)

3 TH còn lại xét tương tự