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

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

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

\(\Leftrightarrow4\cdot6x=5x+38\)

\(\Leftrightarrow24x-5x=38\)

\(\Leftrightarrow19x=38\Leftrightarrow x=\dfrac{38}{19}=2\)

Vậy \(S=\left\{2\right\}\)

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

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

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

\(\Leftrightarrow x^3-2x^2+x+x^2-2x+1-2x-2x^2+2=0\)

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

PT vô nghiệm , không tìm được x 

Vậy \(S=\varnothing\)

c) \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)

\(\Leftrightarrow3\left(x^2-2x+4\right)+9\left(x-1\right)=3\left(x^2+x-3\right)\)

\(\Leftrightarrow3x^2-6x+12+9x-9=3x^2+3x-9\)

\(\Leftrightarrow3x^2-6x+12+9x-9-3x^2-3x+9=0\)

\(\Leftrightarrow0x+12=0\)

PT vô nghiệm 

Vậy \(S=\varnothing\)

Câu cuối tương tự 

Nhiều câu quá >.<

a/ \(2x\left(x+5\right)=\left(x+3\right)^2+\left(x-1\right)^2+20.\)

\(2x^2+10x=x^2+6x+9+x^2-2x+1+20.\)

\(10x=4x+30\)

\(6x=30\Rightarrow x=5\)

các câu còn lại tương tự

\(a,2x\left(x+5\right)=\left(x+3\right)^2+\left(x-1\right)^2+20\)

\(\Leftrightarrow2x^2+10x=x^2+6x+9+x^2-2x+1+20\)

\(\Leftrightarrow2x^2+10x=2x^2+4x+30\)

\(\Leftrightarrow2x^2+10x-2x^2-4x=30\)

\(\Leftrightarrow6x=30\)

\(\Leftrightarrow x=5\)

Vậy ...........

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

\(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3x^2+15x-6x-30\)

\(\Leftrightarrow4x^2-8x+4=4x^2+11x-29\)

\(\Leftrightarrow4x^2-8x-4x^2-11x=-29-4\)

\(\Leftrightarrow-19x=-33\)

\(\Leftrightarrow x=\frac{33}{19}\)

Vậy...........

\(c,\left(x-1\right)^2+\left(x+3\right)^2=2\left(x-2\right)\left(x+1\right)+38\)

\(\Leftrightarrow x^2-2x+1+x^2+6x+9=2x^2+2x-4x-4+38\)

\(\Leftrightarrow2x^2+4x+10=2x^2-2x+34\)

\(\Leftrightarrow2x^2+4x-2x^2+2x=34-10\)

\(\Leftrightarrow6x=24\)

\(\Leftrightarrow x=4\)

Vậy.............

\(d,\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-18\)

\(\Leftrightarrow x^3+6x+12x+8-\left(x^3-6x+12x-8\right)=12x^2-12x-8\)

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

\(\Leftrightarrow12x=-24\)

\(\Leftrightarrow x=-2\)

Vậy............

a, (x-1)^2 + (x+3)^2 = 2(x-2)(x+1) + 38

<=> x^2 -2x +1 + x^2 + 6x +9 = 2x^2 +2x -4x -4 +38

<=> x^2 -2x +x^2 +6x -2x^2 -2x +4x= -4 +38 -10

<=> 6x= 24

<=> x = 4

=> S={4}

b, 5(2x-3)-4(5x-7)= 19 -2(x+11)

<=> 10x -15 -20x +28 = 19-2x-22

<=> 10x -20x +2x = 19 -22 +15 -28

<=> -8x = -16

<=> x = 2

=> S={2}

a: =>9x^2+12x+4-9x^2+12x-4=5x+38

=>24x=5x+38

=>19x=38

=>x=2

e: =>x^3+1-2x=x^3-x

=>-2x+1=-x

=>-x=-1

=>x=1

f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1

=>12x-9=3x+1

=>9x=10

=>x=10/9

b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)

=>-3x+3=3x-9

=>-6x=-12

=>x=2