ae ơi đề bài lại như này nhé chứng minh a 1 + a2 +....+a99 <1

\(a_k=\frac{2k+1}{k^2\left(k+1\right)^2}=\frac{k^2+2k+1-k^2}{k^2\left(k+1\right)^2}=\frac{\left(k+1\right)^2}{k^2\left(k+1\right)^2}-\frac{k^2}{k^2\left(k+1\right)^2}=\frac{1}{k^2}-\frac{1}{\left(k+1\right)^2}\)

\(S=\frac{1}{1^2}-\frac{1}{\left(1+1\right)^2}+\frac{1}{2^2}-\frac{1}{\left(2+1\right)^2}+\frac{1}{3^2}-\frac{1}{\left(3+1\right)^2}+...+\frac{1}{99^2}-\frac{1}{\left(99+1\right)^2}\)

\(S=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=1-\frac{1}{100^2}< 1\) ( đpcm ) 

... 

M = x⁴ - 6x³ + 10x² - 6x + 9

M = (x² - 6x + 9) + x⁴ - 6x³ + 9x²

M = (x - 3)² + x²(x² - 6x + 9)

M = (x - 3)².(1 + x²)

Ta có:\(\left(x-3\right)^2\ge0;\left(1+x^2\right)\ge1\)

\(\Rightarrow M\ge1\)

Dấu 'x' xảy ra khi:

\(\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy Mmin = 1 khi x = 3

Chúc bạn học tốt!!!

Mình giải lại từ dòng số 6 nhé!!!

=> M = 0 

Dấu '=' xảy ra khi:

(x - 3)² = 0 => x - 3 = 0

=> x = 3

Vậy Mmin = 0 khi x = 3

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

    \(=\left[\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}\right]^2\)

     \(=\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}\)

Dấu "="\(\Leftrightarrow x=-\frac{1}{2}\)

Vậy ............

nhanh lên hộ t đi :>>>>>>>>>>

\(4xy-3\left(x+y\right)=59\Rightarrow16xy-12x-12y=59.4\)

\(\Rightarrow4x\left(4y-3\right)-3\left(4y-3\right)=59.4+9\)

\(\Rightarrow\left(4x-3\right)\left(4y-3\right)=245\)

Ta có bảng sau: 

Mà x,y nguyên dương

Vậy \(\left(x;y\right)\in\left\{\left(1;62\right),\left(2;13\right),\left(13;2\right),\left(62;1\right)\right\}\)

 

\(A=2x^2+9y^2-6xy-6x-12y+2015\)

\(A=\left(x^2-6xy+9y^2\right)+x^2-6x-12y+2015\)

\(A=\left(x-3y\right)^2+4.\left(x-3y\right)-10x+x^2+2015\)

\(A=\left(x-3y\right)^2+4.\left(x-3y\right)+4+\left(x^2-10x+25\right)+1986\)

\(A=\left(x-3y+2\right)^2+\left(x-5\right)^2+1986\)

Vì \(\left(x-3y+2\right)^2\ge0;\left(x-5\right)^2\ge0\)

\(\Rightarrow A\ge1986\)

Dấu '=' xảy ra khi:

\(\Rightarrow\hept{\begin{cases}x-3y+2=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{7}{3}\\x=5\end{cases}}}\)

Vậy A_min= 1986 khi x = 5, y = 7/3

Chúc bạn học tốt!!!