Ta có: 

\(\left(\frac{x}{y}+\frac{y}{x}\right)^2=\frac{x^2}{y^2}+2.\frac{x}{y}.\frac{y}{x}+\frac{y^2}{x^2}=\left(\frac{x}{y}-\frac{y}{x}\right)^2+4.\frac{x}{y}.\frac{y}{x}\)

\(=\left(\frac{x}{y}-\frac{y}{x}\right)^2+4\ge4\) với mọi x y >0

Vì x, y >0 => \(\frac{x}{y}+\frac{y}{x}>0\) mà \(\left(\frac{x}{y}+\frac{y}{x}\right)^2\ge4\)

=> \(\frac{x}{y}+\frac{y}{x}\ge2>\frac{1}{2}\)với mọi x, y >0

"=" xảy ra <=> x =y

Em kiểm tra lại đề bài nha.

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100+120+451+115x0=00000000000

100 + 120 + 451 + 115 x 0

= 100 + 120 + 451 + 0

= 220 + 451

= 671

a)

\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)

\(=16\)

\(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)

\(=\left(a-b+c+b-c\right)\left(a-b+c-b+c\right)+2ab-2ac\)

\(=a\left(a-2b+2c\right)+2ab-2ac\)

\(=a^2-2ab+2ac+2ab-2ac\)

\(=a^2\)

\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)

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

\(=\left(-4\right)^2=16\)

Sửa đề: Cho \(a^2+b^2+c^2=m\)

Tính: \(A=\left(2a+2b-c\right)^2+\left(2b+2c-a\right)^2+\left(2c+2a-b\right)^2\)

Giải: 

Ta có: \(\left(x+y-z\right)^2=\left(x+y\right)^2-2\left(x+y\right).z+z^2=x^2+y^2+z^2+2xy-2xz-2yz\)

Ứng dụng vào bài trên:

\(A=\left[\left(2a\right)^2+\left(2b\right)^2+c^2+2\left(2a\right)\left(2b\right)-2\left(2a\right)c-2\left(2b\right)c\right]\)

\(+\left[\left(2b\right)^2+\left(2c\right)^2+a^2+2\left(2b\right)\left(2c\right)-2\left(2b\right)a-2\left(2c\right)a\right]\)

\(+\left[\left(2c\right)^2+\left(2a\right)^2+b^2+2\left(2c\right)\left(2a\right)-2\left(2c\right)b-2\left(2a\right)b\right]\)

\(=4a^2+4b^2+c^2+8ab-4ac-4bc\)

\(+4b^2+4c^2+a^2+8bc-4ba-4ca\)

\(+4c^2+4a^2+b^2+8ca-4cb-4ab\)

\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)

\(=9m\).

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

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

\(=3^2-4.3+q\)

\(=q-3\)

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

\(=4\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)

\(=\left(x-y\right)\left(x+y+4\right)\)

Thank you nha !!!