Ta có : \(M=\left(7x+7y\right)+\left(x+\frac{4}{x}\right)+\left(2y+\frac{50}{y}\right)\)

\(\ge7\left(x+y\right)+2\sqrt{x.\frac{4}{x}}+2\sqrt{2y.\frac{50}{y}}\)

\(\ge7.7+4+20=73\)

Dấu "=" xảy ra khi x = 2; y = 5

đề tuyển sinh Hà Nội. có mà ko tìm à

ĐK : \(0\le x\le1\)

\(\Rightarrow0\le1-x\le1\)\(\Rightarrow\sqrt{1-x}\ge1-x\)

Mà \(2\sqrt{x}\ge2x;\sqrt{1+x}\ge1\)

\(\Rightarrow P\ge1-x+2x+1=x+2\ge2\) 

\(\Rightarrow MinP=2\Leftrightarrow x=0\)

em đang ôn hsg lớp 9 nên không biết ạ 

số học sinh nữ là

320-(320*60/100)=128

                           Đ/S:128

\(x^7+x^2+1\)

\(=x^7+x^6+x^5+x^4+x^3+x^2+x+1\)

\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)

a) \(x^7+x^2+1=\left(x^7-x\right)+\left(x^2+x+1\right)\)

\(=x\left(x^6-1\right)+\left(x^2+x+1\right)=x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)

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

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

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)

b) \(x^7+x^5+1=\left(x^7+x^6+x^5\right)-\left(x^6-1\right)\)

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

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

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

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)

mik ghi bài toán đó ko phải nha đó là lịch sử