\(\Leftrightarrow\left(2x-1\right)\left(-2x-1+2x-1\right)=22\)

\(\Leftrightarrow2x-1=-11\)

\(\Leftrightarrow x=-5\)

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

\(\Leftrightarrow1-4x^2+4x^2-4x+1=22\)

hay x=-5

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

\(\Leftrightarrow1-\left(2x\right)^2+\left(2x\right)^2-2.2x.1+1^2=22\)

\(\Leftrightarrow4x^2-4x^2-4x=22-1-1\)

\(\Leftrightarrow-4x=20\Leftrightarrow x=20:\left(-4\right)\Leftrightarrow x=-5\)

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

\(\Leftrightarrow\left(4x^2+12x+9\right)-\left(4x^2-1\right)=22\)

\(\Leftrightarrow4x^2+12x+9-4x^2+1=22\)

\(\Leftrightarrow12x+10=22\)

\(\Leftrightarrow12x=12\)

\(\Leftrightarrow x=1\)

2(x−1)−3(2x+2)−4(2x+3)=16

⇔2x−2−6x−6−8x−12=16

⇔−12x−20=16

⇔−12x=16+20

⇔−12x=36

⇔x=−3

Vậy x=−

P.s:-.- Ko chắc đâu ạ 

TL:

\(\left(2x+3\right)^2-4x^2+1\) =22

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

(4x+3)3+1=22

12x+9+1=22

12x+10=22

12x=12

x=1

vậy ....

hc tốt

Ta có : (2x + 3)² - (2x + 1)(2x - 1) = 22

=> 4x² + 12x + 9 - 4x² + 1 = 22

=> 12x + 10 = 22

=> 12x = 12

=> x = 1

Vậy x = 1

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

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

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

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

\(\Leftrightarrow3.\left(4x+3\right)=21\)

\(\Leftrightarrow4x+3=7\)

\(\Leftrightarrow4x=4\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

( 2x + 3 )² - ( 2x + 1 )( 2x - 1 ) = 22

<=> 4x² + 12x + 9 - [ ( 2x )² - 1 ] = 22

<=> 4x² + 12x + 9 - 4x² + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1 

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)

`=> (x-3)5 = (2x+1)3`

`=> 5x-15 = 6x+3`

`=> 5x-6x = 15+3`

`=> -x=18`

`=> x=-18`

\(\dfrac{x+1}{22}=\dfrac{6}{x}\)

`=> (x+1)x = 22*6`

`=> (x+1)x = 132`

`=> x^2 + x = 132`

`=> x^2+x-132=0`

`=> (x-11)(x+12)=0`

`=>`\(\left[{}\begin{matrix}x-11=0\\x+12=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=11\\x=-12\end{matrix}\right.\)

\(\dfrac{2x-1}{2}=\dfrac{5}{x}\)

`=> (2x-1)x = 2*5`

`=> 2x^2 - x =10`

`=> 2x^2 - x - 10 =0`

`=> 2x^2 + 4x - 5x - 10 =0`

`=> (2x^2 + 4x) - (5x+10)=0`

`=> 2x(x+2) - 5(x+2)=0`

`=> (2x-5)(x+2)=0`

`=>`\(\left[{}\begin{matrix}2x-5=0\\x+2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)

\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)

`=> (2x-1)(2x+1)=21*3`

`=> 4x^2 + 2x - 2x - 1 = 63`

`=> 4x^2 - 1=63`

`=> 4x^2 - 1 - 63=0`

`=> 4x^2 - 64 = 0`

`=> 4(x^2 - 16)=0`

`=> 4(x^2 + 4x - 4x - 16)=0`

`=> 4[(x^2+4x)-(4x+16)]=0`

`=> 4[x(x+4)-4(x+4)]=0`

`=> 4(x-4)(x+4)=0`

`=>`\(\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)

`=> (2x+1)(x+1) = 9*5`

`=> (2x+1)(x+1)=45`

`=> 2x^2 + 2x + x + 1 = 45`

`=> 2x^2 + 3x + 1 =45`

`=> 2x^2 + 3x + 1 - 45 =0`

`=> 2x^2+3x-44=0`

`=> 2x^2 + 11x - 8x - 44=0`

`=> (2x^2 +11x) - (8x+44)=0`

`=> x(2x+11) - 4(2x+11)=0`

`=> (x-4)(2x+11)=0`

`=>`\(\left[{}\begin{matrix}x-4=0\\2x+11=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)

\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \left(x-3\right)\cdot5=\left(2x+1\right)\cdot3\\ x5-15=6x+3\\ x5-6x=3+15\\ -x=18\\ \Rightarrow x=-18\)

\(\dfrac{x+1}{22}=\dfrac{6}{x}\\ \left(x+1\right)\cdot x=6\cdot22\\ \left(x+1\right)\cdot x=2\cdot3\cdot2\cdot11\\ \left(x+1\right)\cdot x=12\cdot11\\ \Rightarrow x=11\)

\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\\ \left(2x-1\right)\cdot\left(2x+1\right)=21\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot3\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot9\\ \Rightarrow2x+1=9\\ 2x=8\\ x=4\)

b) Ta có: \(\left(x^2-7\right)\left(x+2\right)-\left(2x-1\right)\left(x-14\right)+x\left(x^2-2x-22\right)+35\)

\(=x^3+2x^2-7x-14-\left(2x^2-28x-x+14\right)+x^3-2x^2-22x+35\)

\(=2x^3-29x+21-2x^2+29x-14\)

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

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

\(\Rightarrow\left(1+2x\right)\left(1-2x\right)+\left[\left(2x\right)^2-2.2x+1^2\right]=22\)

\(\Rightarrow1^2-\left(2x\right)^2+\left(4x^2-4x+1\right)=22\)

\(\Rightarrow1-4x^2+4x^2-4x+1=22\)

\(\Rightarrow2-4x=22\)

\(\Rightarrow-4x=22-2=20\)

\(\Rightarrow x=20:\left(-4\right)=-5\)

b/ \(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2.\left(x+1\right)^2=0\)

\(\Rightarrow\left(x^2-2.x.5+5^2\right)+\left(x^2-3^2\right)+2.\left(x^2+2.x.1+1^2\right)=0\)

\(\Rightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)

\(\Rightarrow x^2-10x+25+x^2-9-2x^2-4x-2=0\)

\(\Rightarrow-14x+14=0\)

\(\Rightarrow-14x=0-14=-14\)

\(\Rightarrow x=\left(-14\right):\left(-14\right)=1\)

b/\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)

\(\Leftrightarrow x^2-10x+25+x^2-3^2-2\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow x^2-10x+25+x^2-9-2x^2-4x-2=0\)

\(\Leftrightarrow14x=14\Leftrightarrow x=1\)

c/\(\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(4x^2-9\right)=0\)

\(\Leftrightarrow4x^2+12x+9+4x^2-12x+9-8x^2+18=0\)

\(\Leftrightarrow0x=-36\Leftrightarrow x=0\)

a/\(\left(2x+1\right).\left(1-2x\right)+\left(2x-1\right)^2=22\Leftrightarrow2x-4x^2+1-2x+4x^2-4x+1=22\Leftrightarrow-4x=20\Leftrightarrow x=-5\)

1) \(\left(x+1\right)^2=x^2+2x+1\)

2) \(\left(2x+1\right)^2=4x^2+4x+1\)

3) \(\left(2x+y\right)^2=4x^2+4xy+y^2\)

4) \(\left(2x+3\right)^2=4x^2+12x+9\)

5) \(\left(3x+2y\right)^2=9x^2+12xy+4y^2\)

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

7) \(\left(x^3+1\right)^2=x^6+2x^3+1\)

8) \(\left(x^2+y^3\right)^2=x^4+2x^2y^3+y^6\)

9) \(\left(x^2+2y^2\right)^2=x^4+4x^2y^2+4y^4\)

10) \(\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}xy+\dfrac{1}{9}y^2\)