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

\(\Leftrightarrow2x-2+3=x+2\)

\(\Leftrightarrow2x-x=2+2-3\)

\(\Leftrightarrow x=1\)

Vậy \(S=\left\{1\right\}\)

\(b,\left(3x-7\right)\left(x+5\right)=\left(5+x\right)\left(3-2x\right)\)

\(\Leftrightarrow\left(3x-7\right)\left(x+5\right)-\left(5+x\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-7-3+2x\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(5x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{-5;2\right\}\)

Ta có: (x + 2)2 – 4 ≥ (x + 3)(x + 5) – x

⇔ x2 + 4x + 4 – 4 ≥ x2 + 5x + 3x + 15 – x

⇔ –3x ≥ 15 ⇔ x ≤ –5

Tập nghiệm: S = {x | x ≤ –5}.

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

2. \(\left(2x-5y\right)^2=4x^2-20xy+25y^2\)

3. \(\left(x+8\right)\left(x-8\right)=x^2-64\)

4. \(\left(x+4\right)^3=x^3+12x^2+48x+64\)

5. \(\left(2x-1\right)^3=8x^3-12x^2+6x-1\)

1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)

\(=x^2+4x+4+x^2-6x+9\)

\(=2x^2-2x+13\)

2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=-2x+7\)

3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)

\(=x^2-25-x^2-10x-25\)

=-10x-50

4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)

\(=x^2-6x+9-x^2+16\)

=-6x+25

5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)

\(=y^2-6y+9-y^2+6y-9\)

=0

6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)

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

=12x+18

`@` `\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`

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

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

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

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

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

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

1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)

\(=x^2+4x+4+x^2-6x+9\)

\(=2x^2-2x+13\)

2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=\left(-2x+7\right)\cdot1\)

\(=-2x+7\)

3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)

\(=x^2-25-x^2-10x-25\)

\(=-10x-50\)

a, 3y-2y=2y-3

    3y-2y-2y=3

    -y=3

     y=-3

b, 3-4x+24+6x=x+27+3x

   -4x+6x-x-3x =27-3-24

   -2x              =0

      x             =0

c, 5-(6-x)=4.(3-2x)

   5-6+x =12-8x

   x+8x  =12+6-5

  9x      =13

   x       =13/9

d, 4.(x+3)=-7x+17

   4x+12  =-7x+17

4x+7x     =17-12

11x         =5

  x          =5/11

 A)  \(3y-2y=2y-3\)     \(\Leftrightarrow y-2y=-3\) \(\Leftrightarrow y=3\)

B)  \(3-4x+24+6x=x+27-3x\)\(\Leftrightarrow3-4x+24+6x-x-27-3x=0\)

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

C)   \(5-\left(6-x\right)=4\left(3-2x\right)\)\(\Leftrightarrow5-6x+x-12+8x=0\)\(\Leftrightarrow9x-13=0\)\(\Leftrightarrow x=\frac{13}{9}\)

D)   \(4\left(x+3\right)=-7x+17\) \(\Leftrightarrow4x+12+7x-17=0\)\(\Leftrightarrow11x-5=0\)\(\Leftrightarrow x=\frac{5}{11}\)

E)  \(11x+42-2x=100-9x-22\)\(\Leftrightarrow11x+42-2x-100+9x+22=0\)

\(\Leftrightarrow18x-36=0\)\(\Leftrightarrow x=2\)

 F)  \(5x+\frac{3}{12}=1+\frac{2x}{9}\)\(\Leftrightarrow180x+9-36-8x=0\)\(\Leftrightarrow172x-27=0\)\(\Leftrightarrow x=\frac{27}{172}\)

TK nka !!! Mk giải tiếp !!