\(12+13=30-x\)

\(25=30-x\)

\(\Rightarrow x=5\)

\(12+13=30-x\)

\(25=30-x\)

\(x=30-25\)

\(x=5\)

theo đề ta có: 7x = 2y => \(\frac{x}{2}=\frac{y}{7}\) và x - y = 16

áp dụng t/c DTSBN ta có:

\(\frac{x}{2}=\frac{y}{7}=\frac{x-y}{2-7}=\frac{16}{-5}=-3,2\)

=> \(\hept{\begin{cases}\frac{x}{2}=-3,2=>x=-6,4\\\frac{y}{7}=-3,2=>y=-22,4\end{cases}}\)

vậy x = -6,4 ; y = -22,4

ok nha!! 564564575676876978083452345346456456756765768768

x - y = 16 => x = 16 + y

Thay vào 7x = 2y ta được:

7(16 + y) - 2y = 0 

=> 112 + 7y - 2y = 0

=> 5y = -112

=> y = -112/5

=> x = 16 + y = 16 + (-112/5) = -32/5

\(2\left(x+1\right)^2=32\)

\(\Leftrightarrow\left(x+1\right)^2=32:2=16\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=4\\x+1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}}\)

(x+1)² = 32/2 =16 = 4²

th1: x+1 = 4

x = 3

th2; x+1 = -4

      x = -5

PT<=>\(2x^2-5x+3-3x\left(\sqrt{x+3}-2\right)\)

<=>\(\left(x-1\right)\left(2x-3\right)-3x\left(\frac{x-1}{\sqrt{x+3}+2}\right)\)

<=>\(\left(x-1\right)\left(2x-3-\frac{3x}{\sqrt{x+3}+2}\right)\)

<=> \(\hept{\begin{cases}x=1\\x=\frac{1+\sqrt{13}}{2}\end{cases}}\)

Tất nhiên là các biểu thức trên bằng 0. Mình quên ghi.

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=\frac{9999}{10000}\)

 \(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=\frac{9999}{10000}\)

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{4}\right|=4x.\)

Điều kiện \(4x\ge0\)nên 

\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}=4x\)

\(\Leftrightarrow3x+\frac{13}{12}=4x\)

\(\Leftrightarrow4x-3x=\frac{13}{12}\)

\(\Leftrightarrow x=\frac{13}{12}\)

2²+4²+6²+8²+.....+20²

=2².(1+2²+3²+4²+...+10²)

=4.385

=1540

= 10 ,12,14,16,18

them so 2 o tren nua nhe nguyen ngoc son