sao anh chúc ghê zậy 

Câu 1 , S = 1540

Câu 1

Đặt A=1²+2²+3²+...+10²=385

=>A.2²=2².(1²+2²+3²+...+10²)

           =1².2²+2².2²+3².2²+...+10².2²

            =2²+4²+6²+...+20²

=>A.2²=S=385.2²=1540

Vậy S =1540

a) \(\left(x+4\right)\left(x\cdot x+1\right)=0\)

\(\Rightarrow\left(x+4\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x^2+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x\notin R\end{matrix}\right.\)

\(\Rightarrow x=-4\)

Vậy \(x=-4\)

b) \(\left(\left|x\right|+2\right)\left(x\cdot x-1\right)=0\)

\(\Rightarrow\left(\left|x\right|+2\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x\right|+2=0\\x^2-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=1\\x=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Vậy \(x_1=-1;x_2=1\)

Tìm x \(\in\)Z

a) (x+4)(x.x+1)=0

(x+4)(x²+1)=0

<=>x+4 và x²+1=0

Nếu x+4=0=>x=-4

Nếu x²+1=0=>x²=-1(ko có giá tri)

Vậy...

b)(|x|+2)(x.x-1)=0

(|x|+2)(x²-1)=0

<=>|x|+2 và x²-1=0

+|x|+2=0=>|x|=-2(không tồn tại)

+x²-1=0=>x²=1=>x={1;-1}

Vậy....

a)\(\left(x+4\right)\left(x\cdot x+1\right)=0\)

\(\Rightarrow\left(x+4\right)\left(x^2+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x^2+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow x=-4\)

Vậy \(x=-4\).

b)\(\left(\left|x\right|+2\right)\left(x\cdot x-1\right)=0\)

\(\Rightarrow\left(\left|x\right|+2\right)\left(x^2-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x\right|+2=0\\x^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=1\\x=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{1;-1\right\}\).

a) \(\left(2x-1\right)\left(2x+1\right)-4x^2=3\Leftrightarrow\left(4x^2-1\right)-4x^2=3\Rightarrow-1=3\) (không đúng)

Tí làm tiếp nhé ;) h đi chơi đã

a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

\(\Leftrightarrow\left(3x-1\right)^3=1\)

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)

c) \(h\left(x\right)=\left(x+1\right)^2+\left(\dfrac{x^2+2x+2}{x+1}\right)^2=\left(x+1\right)^2+\left(x+1+\dfrac{1}{x+1}\right)^2=2\left(x+1\right)^2+\dfrac{1}{\left(x+1\right)^2}+2\ge_{AM-GM}2\sqrt{2}+2\).

Đẳng thức xảy ra khi \(2\left(x+1\right)^2=\dfrac{1}{\left(x+1\right)^2}\Leftrightarrow x=\pm\sqrt{\dfrac{1}{2}}-1\).

b) \(g\left(x\right)=\dfrac{\left(x+2\right)\left(x+3\right)}{x}=\dfrac{x^2+5x+6}{x}=\left(x+\dfrac{6}{x}\right)+5\ge_{AM-GM}2\sqrt{6}+5\).

Đẳng thức xảy ra khi x = \(\sqrt{6}\).

Câu a muốn có min thì đề bài phải là \(x\ge4\) (có dấu "=")

Còn \(x>4\) thì chắc là đề sai

a) \(x-2=\left(x-2\right)^2\)

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

\(\left(x-2\right)\left(x-2-1\right)=0\)

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

\(\Rightarrow x-2=0\) hoặc \(x-3=0\)

*) \(x-2=0\)

\(x=2\)

*) \(x-3=0\)

\(x=3\)

Vậy \(x=2;x=3\)

b) \(x+5=2\left(x+5\right)^2\)

\(2\left(x+5\right)^2-\left(x+5\right)=0\)

\(\left(x+5\right)\left[2\left(x+5\right)-1\right]=0\)

\(\left(x+5\right)\left(2x+10-1\right)=0\)

\(\left(x+5\right)\left(2x+9\right)=0\)

\(\Rightarrow x+5=0\) hoặc \(2x+9=0\)

*) \(x+5=0\)

\(x=-5\)

*) \(2x+9=0\)

\(2x=-9\)

\(x=-\dfrac{9}{2}\)

Vậy \(x=-5;x=-\dfrac{9}{2}\)

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

\(\left(x^2+1\right)\left(2x-1\right)+2x-1=0\)

\(\left(x^2+1\right)\left(2x-1\right)+\left(2x-1\right)=0\)

\(\left(2x-1\right)\left(x^2+1+1\right)=0\)

\(\left(2x-1\right)\left(x^2+2\right)=0\)

\(\Rightarrow2x-1=0\) hoặc \(x^2+2=0\)

*) \(2x-1=0\)

\(2x=1\)

\(x=\dfrac{1}{2}\)

*) \(x^2+2=0\) 

\(x^2=-2\) (vô lí)

Vậy \(x=\dfrac{1}{2}\)

d) Sửa đề:

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

\(\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\)

\(\left(x+1\right)\left(x^2+3+1\right)=0\)

\(\left(x+1\right)\left(x^2+4\right)=0\)

\(\Rightarrow x+1=0\) hoặc \(x^2+4=0\)

*) \(x+1=0\)

\(x=-1\)

*) \(x^2+4=0\)

\(x^2=-4\) (vô lí)

Vậy \(x=-1\)

a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)

=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)

=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)

=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)

=> \(x=\dfrac{-1}{11}\)

Đây toán 8 mà? :v

a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)

\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)

\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)

\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)

\(\Leftrightarrow\left(11+1\right)x=0\)

\(\Leftrightarrow11x+1=0;x=0\)

\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)

Vậy....

bai nay de\