a, Ta có 

A= x(x+2)+y(y-2)-2xy +37

=x²+2x+y²-2y-2xy+37

=x²-2xy+y²+2(x-y)+37

=(x-y)²+2(x-y)+37

Vì x-y=7

=>(x-y)²+2(x-y)+37=7²+14+37=100

KL

b, Ta có B=x²+4y²-2x+10+4xy-4y

=x²+4xy+4y²-2x-4y+10

=(x+2y)²-2(x+2y)+10

Vì x+2y=5 

=>(x+2y)²-2(x+2y)+10=5²-10+10=25

KL

1: a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=7^2+2.7+37\) (Vì \(x-y=7\))

\(=100\)

Vậy \(A=100\)

b) Ta có: \(B=x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)

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

\(=5^2-2.5+10\)

\(=25\)

Vậy \(B=25\)

c) Ta có : \(C=\left(x-y\right)^2\)

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

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

\(=26-2.5\) (Vì \(x^2+y^2=26\) ; \(xy=5\))

\(=16\)

Vậy \(C=16\)

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

\(=x^2+2xy\)

\(=x\left(x+2y\right)\) \(\left(dpcm\right)\)

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

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

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

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

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

Chúc bn học tốt ✔✔✔

bài 2: a bạn có thể thêm bớt y^2 vào vế bên phải

bài 2 c thì bạn có thể mở ngoặc ở vế phải rồi tính sau đó áp dụng hđt

\(a.\)

\(x\left(x+z\right)+y\left(y-z\right)-2xy+37\)

\(=x^2+xz+y^2-yz-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+z\left(x-y\right)+37\)

\(=\left(x-y\right)^2+z\left(x-y\right)+37\)

\(=7^2+x.7^2+37\)

\(=86+49x\)

\(b.\)

\(x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)

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

\(=5^2-2.5+10\)

\(=25\)

a) Ta có:

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

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

Thay x + y = 3 vào A

\(A=3^2-4.3+1\)

\(A=9-12+1\)

\(A=-2\)

b) Sửa đề:

\(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(B=x^2+2x+y^2-2y-2xy+37\)

\(B=\left(x^2+y^2+1+2x-2y-2xy\right)+36\)

\(B=\left(x-y+1\right)^2+36\)

Thay x - y = 7 vào B

\(B=\left(7+1\right)^2+36\)

\(B=100\)

c) Ta có:

\(C=x^2+4y^2-2x+10+4xy-4y\)

\(C=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)

\(C=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

Thay x + 2y = 5 vào C

\(C=5^2-2.5+10\)

\(C=25-10+10\)

\(C=25\)

Câu 2: 

\(B=x^2+2x+y^2-2x-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=7^2+2\cdot7+37=49+37+14=100\)

Câu 3: 

\(C=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)

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

\(=5^2-2\cdot5+10=25\)

a.

\(\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)

Nhân vế:

\(-4\left(x^3-y^3\right)=\left(16x-4y\right)\left(5x^2-y^2\right)\)

\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)

\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4y}{7}\\y=-3x\end{matrix}\right.\)

Thế vào \(y^2=5x^2+4...\)

b. Đề bài không hợp lý ở \(4x^2\)

c.

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)

Trừ vế:

\(x^3-y^3-3x^2-6y^2=9-3x+12y\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)

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

\(\Leftrightarrow x-1=y+2\)

\(\Leftrightarrow y=x-3\)

Thế vào \(x^2=2y^2=x-4y\) ...

b.

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2+y^4-4xy^3=1\\4x^2+2y^2-4xy=2\end{matrix}\right.\)

\(\Rightarrow y^4-2y^2-4xy^3+4xy=-1\)

\(\Leftrightarrow\left(y^2-1\right)^2-4xy\left(y^2-1\right)=0\)

\(\Leftrightarrow\left(y^2-1\right)\left(y^2-1-4xy\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\\x=\dfrac{y^2-1}{4y}\end{matrix}\right.\)

Thế vào \(2x^2+y^2-2xy=1\) ...

Với \(x=\dfrac{y^2-1}{4y}\) ta được:

\(2\left(\dfrac{y^2-1}{4y}\right)^2+y^2-2\left(\dfrac{y^2-1}{4y}\right)y=1\)

\(\Leftrightarrow5y^4-6y^2+1=0\)