a: \(M=\left(x+y\right)^3+2\left(x^2+2xy+y^2\right)\)

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

\(=7^3+2\cdot49=441\)

b: \(A=x^2+2x+y^2-2y-2xy+37\)

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

\(=7^2+2\cdot7+37\)

\(=49+14+37=100\)

b: \(x^2+y^2=\left(x+y\right)^2-2xy=25-12=13\)

c: \(\left(x-y\right)^2=\left(x+y\right)^2-4xy=5^2-4\cdot6=1\)

=>x-y=1 hoặc x-y=-1

a) \(x^2+10x+26+y^2+2y\)

= \(x^2+10x+25+y^2+2y+1\)

= \(\left(x+5\right)^2+\left(y+1\right)^2\)

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

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

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

c) \(z^2-6z+5-t^2-4t\)

= \(z^2-6z+9-\left(t^2+4t+4\right)\)

= \(\left(z-3\right)^2-\left(t+2\right)^2\)

d) \(4x^2-12x-y^2+2y+1\)

Hình như câu này sai đề -_-

a, \(x^2+10x+26+y^2+2y\)

\(=\left(x^2+2.x.5+5^2\right)+\left(1^2+2.1.y+y^2\right)\)

\(=\left(x+5\right)^2+\left(y+1\right)^2\)

b, \(x^2-2xy+2y^2+2y+1\)

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

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

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

c,\(z^2 -6z+5-t^2-4t\)

\(=-\left(t^2+4t-z^2+6z-5\right)\)

\(=-\left(t^2+2.t.2+2^2-z^2+2.z.3-3^2\right)\)

\(=-\left(\left(t^2+2.t.2+2^2\right)-\left(z^2-2.z.3+3^2\right)\right)\)

\(=-\left(\left(t+2\right)^2-\left(z-3\right)^2\right)\)

\(=\left(z-3\right)^2-\left(t+2\right)^2\)

d, Không biết làm hihi :)

* là nhân❌

1, (x+y+4). (x+y-4)=(x+y)²-4²=(x+y)²-16

2, (x-y+6). (x+y-6)=(x+y)²-6²=(x+y)²-36

3, (x+2y+3z). (2y+3z-x)=(2y+3z)²-x²

\(1.\left[\left(x+y\right)-4\right]\left[\left(x+y\right)+4\right]=\left(x+y\right)^2-4^2\)

\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{\left(1+1+1\right)^2}{x^2+y^2+z^2+2xy+2yz+2xz}=\frac{9}{\left(x+y+z\right)^2}=9\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y+z=1\\x=y=z\end{cases}\Leftrightarrow x=y=z=\frac{1}{3}}\)

ban chung minh gium mik bdt nha

sorry minh chua hoc bat dang thuc nay