dễ vậy cũng hỏi

đkxđ \(x\ne\pm2;-1\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}-\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{1}{\left(x+2\right)\left(x-2\right)}\)

\(\Leftrightarrow x-2-3x-6=x+1\)

\(\Leftrightarrow x-2-3x-6-x-1=0\)

\(\Leftrightarrow-5x-9=0\)

\(\Leftrightarrow x=-\frac{9}{5}\left(tmđk\right)\)

Ta có : 1/6= 2/12

Số thứ 1 là: 51:( 12+5).5=15

Số thứ 2 là: 51-15= 36

Vậy....

Đặt \(n^2+n+17=a^2\left(a\inℕ^∗\right)\)

\(\Leftrightarrow\left(2n\right)^2+4n+68=\left(2a\right)^2\)

\(\Leftrightarrow\left(2n+1\right)^2+67=\left(2a\right)^2\)

\(\Leftrightarrow\left(2a\right)^2-\left(2n+1\right)^2=67\)

\(\Leftrightarrow\left(2a-2n-1\right)\left(2a+2n+1\right)=67\)

Ta thấy : \(a,n\inℕ^∗\) \(\Rightarrow\hept{\begin{cases}2a-2n-1,2a+2n+1\inℕ^∗\\2a+2n+1>2a-2n-1\end{cases}}\)

Do đó ta xét TH sau :

\(\hept{\begin{cases}2a-2n-1=1\\2a+2n+1=67\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}n=32\\a=33\end{cases}}\) ( thỏa mãn )

Vậy : \(n=32\) thỏa mãn đề.

a) Ta có :

\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

Ta thấy : \(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\forall x,y,z\)

Do đó : \(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(3x-3\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\) ( thỏa mãn )

Vậy : \(\left(x,y,z\right)=\left(1,3,-1\right)\)

Câu (b) nữa Vinh ơi

\(\frac{3}{x}=\frac{4}{y}\Rightarrow x=\frac{3y}{4}\)

\(C=\frac{2x+3y}{3x+4y}=\frac{2\cdot\frac{3}{4y}+3y}{3\cdot\frac{3y}{4}\cdot4y}\)

\(=\frac{2\cdot\frac{3}{4}+3}{3\cdot\frac{3}{4}+4}=\frac{\frac{9}{2}}{\frac{25}{4}}\)

\(=\frac{9}{2}\cdot\frac{4}{25}=\frac{18}{25}\)

\(\frac{3}{x}=\frac{4}{y}\Rightarrow x=\frac{3y}{4}\)

\(\Rightarrow C=\frac{\frac{3y}{2}+3y}{\frac{9y}{4}+4y}=\frac{\frac{9y}{2}}{\frac{25y}{4}}=\frac{18}{25}\)

\(5x^3+6x^2+12x+8=0\)

\(\Leftrightarrow4x^3+\left(x^2+6x+12x+8\right)=0\)

\(\Leftrightarrow4x^3+\left(x+2\right)^3=0\)

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

\(\Leftrightarrow x+2-\sqrt[3]{4x}=0\)

\(\Leftrightarrow x\left(1-\sqrt[3]{4}\right)=0\)

\(\Leftrightarrow x=-\frac{2}{1-\sqrt[3]{4}}=\frac{2}{\sqrt[3]{4}-1}\)

bài này chắc là xét mấy TH ta ??? nhưng tìm 1 KQ dễ lắm bn 

\(5x^3+6x^2+12x+8=0\)

\(5.x.x.x+6x.x+12x=-8\)

\(23x=-8\)

\(x=-\frac{8}{23}\)

giúp mk với 

làm ơn đi mà

\(^{2012^{2013}:7=287.571428571426}\)

(dư 571428571426)

\(ĐKXĐ:x\ne1\)

Phương trình đã có 1 nghiệm bằng 2. Ta cần giải phương trình:

\(2x+\frac{1}{x-1}=0\)

\(\Leftrightarrow\frac{2x\left(x-1\right)+1}{x-1}=0\)

\(\Leftrightarrow2x^2-2x+1=0\)

Ta có \(\Delta=2^2-4.2.1=-4< 0\)(vô nghiệm)

Vậy nghiệm duy nhất là 2

Giải :

\(\left(x-2\right)\left(2x+\frac{1}{x-1}\right)=0\)

\(\Leftrightarrow x-2=0\text{ hoặc }2x+\frac{1}{x-1}=0\)

* Trường hợp 1 :

\(x-2=0\Leftrightarrow x=2\)

* Trường hợp 2 :

\(2x+\frac{1}{x-1}=0\) \(\left(\text{ĐKXĐ : }x-1\ne0\Leftrightarrow x\ne1\right)\)

\(\Leftrightarrow\frac{2x\left(x-1\right)}{x-1}+\frac{1}{x-1}=0\)

\(\text{Khử mẫu : }2x\left(x-1\right)+1=0\)

\(\Leftrightarrow2x^2-2x+1=0\)

\(\Leftrightarrow x^2-x+\frac{1}{2}=0\)

\(\Leftrightarrow x^2-x+\frac{1}{4}+\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{-1}{4}\)

\(\Leftrightarrow x\in\varnothing(\text{vì }\left(x-\frac{1}{2}\right)^2\ge0)\)

Vậy \(S=\left\{2\right\}\).