a) \(\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x-22\)

\(\Leftrightarrow-5x^2+2x-1=-5x^2-x-22\)

\(\Leftrightarrow3x+21=0\)

\(\Leftrightarrow x=-7\)

Vậy tập nghiệm của phương trình là \(S=\left\{-7\right\}\)

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

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

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

\(\Leftrightarrow x=1\)

Vậy tập nghiệm của phương trình là \(S=\left\{1\right\}\)

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

\(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)

\(\Leftrightarrow x^3+3x^2+12x-9=x^3+3x^2+3x+1\)

\(\Leftrightarrow9x-10=0\)

\(\Leftrightarrow x=\frac{10}{9}\)

Vậy tập nghiệm của phương trình là \(S=\left\{\frac{10}{9}\right\}\)

\(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)

\(\Leftrightarrow9x^2+12x+4-9x^2+12x-4=5x+38\)

\(\Leftrightarrow24x-5x-38=0\)

\(\Leftrightarrow19x-38=0\)

\(\Leftrightarrow19\left(x-2\right)=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

VẬY ..

Đáp án:

\(S=\left\{2\right\}\)

Lời giải:

a) \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)

\(\Leftrightarrow\left[\left(3x+2\right)-\left(3x-2\right)\right].\left(3x+2+3x-2\right)=5x+38\)

\(\Leftrightarrow\left(3x+2-3x+2\right).6x=5x+38\)

\(\Leftrightarrow24x=5x+38\)

\(\Leftrightarrow24x-5x=38\)

\(\Leftrightarrow19x=38\)

\(\Leftrightarrow x=2\)

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

Ta có : \(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4+2x^3+5x^2+10x-6x-12=0\)

\(\Leftrightarrow x^3\left(x+2\right)+5x\left(x+2\right)-6\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^3+5x-6\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^3-x^2+x^2-x+6x-6\right)=0\)

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

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

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

hoặc   \(x-1=0\)

hoặc   \(x^2+x+6=0\)

\(\Leftrightarrow\) \(x=-2\)(tm)

hoặc    \(x=1\)(tm)

hoặc   \(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\)(ktm)

Vậy tập nghiệm của phương trình là \(S=\left\{-2;1\right\}\)

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

Ta có \(\Delta=5^2+4.2.3=49,\sqrt{\Delta}=7\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-5+7}{4}=\frac{1}{2}\\x=\frac{-5-7}{4}=-3\end{cases}}\)

a) \(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)

\(=\left(\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{x^2-4}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=\left(\frac{2x+4}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{x^2-4}-\frac{x+2}{x^2-4}\right)\)

\(=\frac{2x}{\left(x+2\right)^2}:\frac{-x}{x^2-4}=\frac{2x}{\left(x+2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{-x}\)

\(=\frac{-2\left(x-2\right)}{\left(x+2\right)}=\frac{-2x+4}{x+2}\)

b) \(x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

+) x = 0 \(\Rightarrow A=\frac{-2.0+4}{0+2}=\frac{4}{2}=2\)

+) x = 3 \(\Rightarrow A=\frac{-2.3+4}{3+2}=\frac{-2}{5}\)

bạn giúp mk câu c vs

các bạn giúp mình với a. Mình cảm ơn trước

\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+7}{93}+\frac{x+9}{91}+\frac{x+11}{89}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+3}{97}+1+\frac{x+5}{95}+1\)\(=\frac{x+7}{93}+1+\frac{x+9}{91}+1+\frac{x+11}{89}+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}\)\(=\frac{x+100}{93}+\frac{x+100}{91}+\frac{x+100}{89}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}\)\(-\frac{x+100}{93}-\frac{x+100}{91}-\frac{x+100}{89}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{89}\right)=0\)

Mà \(\left(\frac{1}{99}< \frac{1}{97}< \frac{1}{95}< \frac{1}{93}< \frac{1}{91}< \frac{1}{89}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{89}\right)< 0\)

\(\Rightarrow x+100=0\Leftrightarrow x=-100\)

Vậy x = -100

b, \(\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)

     \(\frac{x+200}{99}+\frac{x+200}{98}=\frac{x+200}{97}+\frac{x+200}{96}\)

   \(\frac{x+200}{99}+\frac{x+200}{98}-\frac{x+200}{97}-\frac{x+200}{96}=0\)

\(\left(x+200\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

mà\(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\ne0\)

==> x+200=0

<=>x=-200

Vậy nghiệm của phương trình là x=-200

c,  \(\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)

      \(\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)

\(\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)

mà  \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)

==>200-x=0

<=>x=200

vậy nghiệm của pt là x=200

Ta có : \(x^3-5x^2+1997x-14077=0\)

\(\Leftrightarrow x^3-7x^2+2x^2-14x+2011x-14077=0\)

\(\Leftrightarrow x^2\left(x-7\right)+2x\left(x-7\right)+2011\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x^2+2x+2011\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x^2+2x+2011=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\left(tm\right)\\\left(x+1\right)^2+2010=0\left(ktm\right)\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{7\right\}\)