\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)

\(\Leftrightarrow\left[x\left(x+1\right)\right]^2+4x\left(x+1\right)=12\)

\(\Leftrightarrow x^2\left(x+1\right)^2+4x\left(x+1\right)=12\)

\(\Leftrightarrow x^4+2x^3+x^2+4x^2+5x=12\)

\(\Leftrightarrow x^4+2x^3+5x^2+5x=12\)

\(\Leftrightarrow x^4+2x^3+5x^2+5x-12=0\)

\(\Leftrightarrow\left(x^3+3x^2+8x+12\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)=0\)

vi \(x^2+x+6\ne0\)nen:

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)

sai dau sua ho toi

Đặt \(x^2+x=u\)

Phương trình trở thành \(u^2+4u=12\)

\(\Leftrightarrow u^2+4u-12=0\)

Ta có \(\Delta=4^2+4.12=64,\sqrt{\Delta}=8\)

\(\Rightarrow\orbr{\begin{cases}u=\frac{-4+8}{2}=2\\u=\frac{-4-8}{2}=-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2+x=2\\x^2+x=-6\end{cases}}\)

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

Ta có \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+3}{2}=1\\x=\frac{-1-3}{2}=-2\end{cases}}\)

+) \(x^2+x=-6\Leftrightarrow x^2+x+6=0\)

Mà \(x^2+x+6=\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0\)

Vậy phương trình chỉ có 2 nghiệm là 1 và -2

\(ĐKXĐ:x\ne\pm1\)

\(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)

\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4\left(x+1\right)}-\frac{12x-1}{4\left(x-1\right)}=0\)

\(\Leftrightarrow\frac{24+20\left(x^2-1\right)-\left(8x-1\right)\left(x-1\right)-\left(12x-1\right)\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow24+20x^2-20-8x^2+9x-1-12x^2-11x+1=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow x=2\)

Vậy tập nghiệm của phương trình là \(S=\left\{2\right\}\)

ĐKXĐ: \(x\ne\pm1\)

\(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)

\(\Leftrightarrow\frac{6}{\left(x+1\right)\left(x-1\right)}+5=\frac{8x-1}{4\left(x+1\right)}-\frac{12x-1}{4\left(1-x\right)}\)

\(\Leftrightarrow24\left(1-x\right)+20\left(x+1\right)\left(x-1\right)\left(1-x\right)=\left(8x-1\right)\left(x-1\right)\left(1-x\right)\)\(-\left(12x-1\right)\left(x+1\right)\left(1-x\right)\)

\(\Leftrightarrow4-4x+20x^2-20x^3=18x^2-20x^3+2x\)

\(\Leftrightarrow4-4x+20x^2=18x^2+2x\)

\(\Leftrightarrow4-4x+20x^2-18x^2-2x=0\)

1. He is used to------------ up early in the morning.
A. get B. getting C. have gotten D. got
2. The boy was kept indoors----------- naughty.
A. to be B. have been C. for being D. when he
3. I shall never forget-----------with you to Paris last year.
A. staying B. to staying C. to stay D. stayed
4. I am looking forward to -----------you.
A. having seen B. seeing C. to see D. all are wrong
5. I am always remember------------ off the lights before I leave my house.
A. turning B. to turn C. turned D. being turned

1, \(=\left[\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}-x\right]:\frac{1-x^2}{\left(1-x\right)-x^2\left(1-x\right)}\)

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

2, để B<0 <=> (x²+1)(1-x)<0

vì x^2+1 > 0 với mọi x

=> \(\hept{\begin{cases}x^2+1>0\\1-x< 0\end{cases}\Leftrightarrow x>1}\)

3, \(\left|x-4\right|=5\Leftrightarrow\orbr{\begin{cases}x=9\\x=-1\left(loại\right)\end{cases}}\)

Thay x=9 vào B ta có: B=(9²+1)(1-9)=82.(-8)=-656