Biến đổi mỗi đa thức theo hướng làm xuất hiện thừa số x+y-2 \(M=x^3+x^2y-2x^2-xy-y^2+3y+x-1\)

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

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

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

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

\(M=x^2.0+y.0+0+1\)

\(M=1\)

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

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

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

\(N=\left(x^2x+x^2y-2x^2\right)-\left(xyx+xyy-2xy\right)+\left(2x+2y-4\right)+2\)

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

\(N=x^2.0-xy.0+2.0+2\)

\(N=2\)

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

\(P=\left(x^4+x^3y-2x^3\right)+\left(x^3y+x^2y^2-2x^2y\right)-\left(x^2+xy-2x\right)+3\)\(P=\left(x^3x+x^3y-2x^3\right)+\left(x^2y.x+x^2yy-2x^2y\right)-\left(xx+xy-2x\right)+3\)

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

\(P=x^3.0+x^2y.0-x.0+3\)

\(P=3\)

Tích mình nha!

Mà bài này hình như học ở lớp 7 rồi!

a, \(\left(\frac{1}{2}+x\right)^2=\left(\frac{1}{2}\right)^2+2.\frac{1}{2}.x+x^2=\frac{1}{4}+x+x^2\)

\(\left(2x+1\right)^2=\left(2x\right)^2+2.2x.1+1^2=4x^2+4x+1\)

b, \(\left(2x+3y\right)^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=4x^2+12xy+9y^2\)

\(\left(0,01+xy\right)^2=\frac{1}{10000}+\frac{1}{50}xy+x^2y^2\)

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

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

\(56.64=\left(60-4\right)\left(60+4\right)=60^2-4^2\)

MINH CUNG VAY

2) Ta có:

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

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

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

Do \(x+y-2=0\Rightarrow x+y=2\)

\(\Rightarrow B=\left(x^4+x^3y-2x^3\right)+\left(x^3y+x^2y^2-2x^2y\right)-\left[2x-2x\right]+3\)

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

\(=0+0+3\)

\(=3\)

Vậy \(B=3\)

1) Ta có:

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

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

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

\(=0+0+0+1\)

\(=1\)

Vậy \(A=1\)

a) P = \(x^2+3x+y^2-3y-2xy+90\)

= \(\left(x-y\right)^2+3\left(x-y\right)+90\)

= \(5^2+3.5+90=130\)

b) P = \(4x^2+9y^2-12xy-12x+24xy-18y+118\)

= \(4x^2+9y^2+12xy-12x-18y+118\)

= \(\left(2x+3y\right)^2-6\left(2x+3y\right)+118\)

= \(\left(-7\right)^2-6.\left(-7\right)+118=209\)

Các bạn ơi cho tui hỏi câu này : noise in / kept / night / the / awake / city / at / the / him / .

Giúp mình với , cảm ơn.

a) \(P=x\left(x+3\right)+y\left(y-3\right)-2xy+90\\ P=x^2+3x+y^2-3y-2xy+90\\ P=\left(x^2+y^2-2xy\right)+\left(3x-3y\right)+90\\ P=\left(x-y\right)^2+3\left(x-y\right)+90\\ P=5^2+3.5+90\\ P=130\)

b) \(P=\left(2x-3y\right)^2-12x\left(1-2y\right)-18y+118\\ P=4x^2-12xy+9y^2-12x+24xy-18y+118\\ P=\left(4x^2+12xy+9y^2\right)-6\left(2x+3y\right)+118\\ P=\left(2x+3y\right)^2-6\left(2x+3y\right)+118\\ P=\left(-7\right)^2-6.\left(-7\right)+118\\ P=49+42+118=209\)

Lời giải:

 $\frac{x}{y}=\frac{2}{3}\Rightarrow \frac{x}{2}=\frac{y}{3}$. Đặt $\frac{x}{2}=\frac{y}{3}=k$ thì:

$x=2k; y=3k$

Khi đó: $3x-2y=3.2k-3.2k=0$. Mẫu số không thể bằng $0$ nên $A$ không xác định. Bạn xem lại.

$B=\frac{2(2k)^2-2k.3k+3(3k)^2}{3(2k)^2+2.2k.3k+(3k)^2}=\frac{29k^2}{33k^2}=\frac{29}{33}$