ko có yêu cầu ai làm đc

lam gium minh 

\(x^2-22.x-110=0\)

<=>\(x^2-22.x=110\)

<=> \(x^2-22.x+11^2=110+11^2\)( cộng cả hai vế với \(11^2\)để được hằng đẳng thức)

<=>\(\left(x-11\right)^2=231\)

<=>\(\hept{\begin{cases}x-11=\sqrt{231}\\x-11=-\sqrt{231}\end{cases}}\)

<=>\(\hept{\begin{cases}x=11+\sqrt{231}\\x=11-\sqrt{231}\end{cases}}\)

vậy phương trình có hai nghiệm \(x_1=11+\sqrt{231};x_2=11-\sqrt{231}\)

Bài1:

\(a,\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\\ \Leftrightarrow\left(2x+3\right)^2-4x^2+1=22\\ \Leftrightarrow\left(2x+3-2x\right)\left(2x+3+2x\right)=21\\ \Leftrightarrow3\left(4x+3\right)=21\\ \Leftrightarrow4x+3=7\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\\ Vậy....\\ b,\left(2x-1\right)^3-4x^2\left(2x-3\right)=5\\ \Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\\ \Leftrightarrow6x=6\\ \Leftrightarrow x=1\\ Vậy...\)

Các câu sau cũng như thế

Bài2:

\(A=x^2+20x+9\\ =\left(x^2+20x+100\right)-91\\ =\left(x+10\right)^2-91\)

Với mọi x thì \(\left(x+10\right)^2\ge0\\ \Rightarrow\left(x+10\right)^2-91\ge-91\)

Hay \(A\ge-91\)

Để A=-91 thì

\(\left(x+10\right)^2=0\\ \Leftrightarrow x+10=0\\ \Leftrightarrow x=-10\)

Vậy...

\(B=4x^2+5x+7\\ =\left(4x^2+5x+\dfrac{25}{16}\right)+5,4375\\ =\left(2x+\dfrac{5}{4}\right)^2+5,4375\)

Với mọi x;y thì \(\left(2x+\dfrac{5}{4}\right)^2+5,4375\ge5,4375\)

Hay \(A\ge5,4375\)

Để \(A=5,4375\) thì \(\left(2x+\dfrac{5}{4}\right)^2=0\\ \Leftrightarrow2x+\dfrac{5}{4}=0\\ \Leftrightarrow x=\dfrac{-5}{8}\)

Vậy....

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

b) \(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)

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

c) \(2x^3-12x^2+17x-2=2x^3-4x^2-8x^2+16x+x-2=2x^2\left(x-2\right)-8x\left(x-2\right)+\left(x-2\right).\)

\(=\left(x-2\right)\left(2x^2-8x+1\right)\)

d) \(2x^4+x^3-22x^2+15x+36=2x^4+2x^3-x^3-x^2-21x^2-21x+36x+36.\)

\(=2x^3\left(x+1\right)-x^2\left(x+1\right)-21x\left(x+1\right)+36\left(x+1\right)\)

\(=\left(x+1\right)\left(2x^3-x^2-21x+36\right)\)

\(x^3+x^2+9x-10x^2-10x+25x+25\)

\(=x^2\left(x+1\right)-10x\left(x+1\right)+25\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-10x+25\right)=\left(x+1\right)\left(x-5\right)^2\)

tao dep trai lam

c) 49x²+14x+1=0

=>(7x+1)²=0

= > 7x+1=0

=> 7x=-1

=> x=-\(\dfrac{1}{7}\)

c)

\(x^2-x-12=0\\ \Leftrightarrow x^2+3x-4x-12=0\\ \Leftrightarrow x\cdot\left(x+3\right)-4\cdot\left(x+3\right)=0\\ \Leftrightarrow\left(x-4\right)\cdot\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

e)

\(\frac{6x+22}{x+2}-\frac{2x+7}{x+3}=\frac{x+4}{x^2+5x+6}\\ \Leftrightarrow\frac{6x^2+40x+66}{x^2+5x+6}-\frac{2x^2+11x+14}{x^2+5x+6}-\frac{x+4}{x^2+5x+6}=0\\ \Leftrightarrow6x^2+40x+66-2x^2-11x-14-x-4=0\\ \Leftrightarrow4x^2+28x+48=0\\ \Leftrightarrow4\cdot\left(x^2+7x+12\right)=0\\ \Leftrightarrow4\cdot\left(x^4+4x+3x+12\right)=0\\ \Leftrightarrow4\cdot\left[x\cdot\left(x+4\right)+3\cdot\left(x+4\right)\right]=0\\ \Leftrightarrow4\cdot\left(x+4\right)\cdot\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+4=0\\x+3=0\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=-4\\x=-3\end{matrix}\right.\)

b)

\(\frac{4x}{x^2+4x+3}-1=6\cdot\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\\ \Leftrightarrow\frac{4x}{\left(x+1\right)\cdot\left(x+3\right)}-1=6\cdot\left(\frac{1}{x+3}-\frac{1}{2\cdot\left(x+1\right)}\right)\\ \Leftrightarrow4x-\left(x+1\right)\cdot\left(x+3\right)=6\cdot\left(\frac{1}{x+3}-\frac{1}{2\cdot\left(x+2\right)}\right)\cdot\left(x+1\right)\cdot\left(x+3\right)\\ \Leftrightarrow-x^2-3=\frac{6x^2}{x+3}+\frac{24x}{x+3}+\frac{18}{x+3}-\frac{3x^2}{x+1}-\frac{12x}{x+1}-\frac{9}{x++1}\\ \Leftrightarrow-x^2\cdot\left(x+3\right)\cdot\left(x+1\right)-3\cdot\left(x+3\right)\cdot\left(x+1\right)=6x^2\cdot\left(x+1\right)+24x\cdot\left(x+1\right)+18\cdot\left(x+1\right)-3x^2\cdot\left(x+3\right)-12x\cdot\left(x+3\right)-9\cdot\left(x+3\right)\\ \Leftrightarrow-x^4-4x^3-6x^2-12x-9=3x^3+9x^2-3x-9\\ \Leftrightarrow-x^4-4x^3-6x^2-12x=3x^3+9x^2-3x\\ \Leftrightarrow x^4+4x^3+6x^2+12x+3x^3+9x^2-3x=0\\ \Leftrightarrow x^4+7x^3+15x^2+9x=0\\ \Leftrightarrow x\cdot\left(x^3+7x^2+15x+9\right)=0\\ \Leftrightarrow x\cdot\left(x^2+6x+9\right)\cdot\left(x+1\right)=0\\ \Leftrightarrow x\cdot\left(x+3\right)^2\cdot\left(x+1\right)=0\)

\(\Rightarrow x=\left[{}\begin{matrix}0\\-3\\-1\end{matrix}\right.\)

a) 3x - 2 = 2x-3

<=> 3x-2 -2x +3 = 0

<=> x +1 = 0

<=> x = -1

c) 3 - 4y+24+6y=y+27+3y

<=> 3 - 4y+24+6y - y - 27 - 3y = 0

<=> -2y =0

<=> y = 0

b,7-2x = 22 - 3x

<=> 7-2x -22 +3x = 0

<=> -15 +x = 0

<=> x = 15

d) x-12+4x = 25+2x-1

<=> x-12+4x -25-2x+1=0

<=> 3x -36 = 0

<=> 3x = 36

<=> x = 12

còn câu e bạn tự làm nha

\(a,3x-2=2x-3\)

\(3x-2x=-3+2\)

\(x=-1\)

Vậy pt cs nghiệm là  { -1 }

\(b,7-2x=22-3x\)

\(-2x+3x=22-7\)

\(x=15\)

Vậy pt cs nghiệm là { 15 }

bn lm nốt nha ... 

a/ \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

<=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)

<=> \(\left(2x+3\right)^2-4x^2+1=22\)

<=> \(\left(2x+3-2x\right)\left(2x+3+2x\right)=21\)

<=> \(3\left(4x+3\right)=21\)

<=> \(4x+3=7\)

<=> \(4x=4\)

<=> \(x=1\)

......................?

mik ko biết

mong bn thông cảm 

nha ................