Question

tìm x

\(\left(11-x^2-2x\right)\cdot\left(11+x^2+2x\right)=121\)

Answer 1

Đặt \(x^2+2x=t\left(đk:t\ge-1\right).\)

\(\Rightarrow\left(11-t\right)\left(11+t\right)=121\Leftrightarrow121-t^2=121\Leftrightarrow t=0\)

\(\Rightarrow x^2+2x=0\Leftrightarrow\orbr{\begin{cases}-2\\0\end{cases}}\)

Answer 2

Thực ra bài này đặt cho dễ nhìn, ta hoàn toàn có thể ssuy ra mà ko cần đặt ha!

Answer 3

= -2 hoặc 0 

~ Hok tốt ~

#JH

Answer 4

\(\left(11-x^2-2x\right)\left(11+x^2+2x\right)=121\) 

\(\Leftrightarrow-x^4-4x^3-4x^2+121=121\)

\(\Leftrightarrow-x^4-4x^3-4x^2=121-121\)

\(\Leftrightarrow-x^4-4x^3-4x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

Answer 5

Ta đặt \(x^2+2x=m\)Theo bài ra, ta có :

\(\left(11-x^2-2x\right).\left(11+x^2+2x\right)=121\Leftrightarrow\left[11-\left(x^2+2x\right)\right].\left(11+x^2+2x\right)=121\)

\(\Leftrightarrow\left(11-m\right).\left(11+m\right)=121\Leftrightarrow11.\left(11+m\right)-m.\left(11+m\right)=121\)

\(\Leftrightarrow121+11m-11m-m^2=121\Leftrightarrow121-m^2=121\Leftrightarrow m^2=0\Leftrightarrow m=0\)

\(\Rightarrow x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x+2=0\Leftrightarrow x=-2\end{cases}}\Rightarrow x\in\left\{-2;0\right\}.\)

Answer 6

\(\Rightarrow11^2-\left(x^2+2x\right)^2=121\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

Answer 7

\(\left(11-x^2-2x\right).\left(11+x^2+2x\right)=121\)

\(\Rightarrow[11-\left(x^2+2x\right)].[11+\left(x^2+2x\right)]=121\)

\(\Rightarrow11^2-\left(x^2-2x\right)=121\)

\(\Rightarrow x^2-2x=0\)

\(\Rightarrow x\left(x-2\right)=0\)\(\Rightarrow x=0\)hoặc \(x=2\)

Vậy \(x=0\)hoặc \(x=2\)