\(\left(a+b+c\right)^2+\left(a+b-c\right)^2-\left(2c\right)^2\)

\(=\left(a+b+c\right)^2+\left(a+b-c-2c\right)\left(a+b-c+2c\right)\)

\(=\left(a+b+c\right)^2+\left(a+b-3c\right)\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a+b+c+a+b-3c\right)\)

\(=2\left(a+b+c\right)\left(a+b-c\right)\)

VT = (a+b+c)^2

= [(a+b) + c]^2

= (a+b)^2 + 2(a+b)c + c^2

= a^2 + 2ab + b^2 + 2ac + 2bc + c^2

= a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = VP

Vậy ...

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VT= (a+b+c)^2 + a^2 + b^2 + c^2

= [(a+b) + c]^2 + a^2 + b^2 + c^2

= (a+b)^2 + 2(a+b)c + c^2 + a^2 + b^2 + c^2

= a^2 + 2ab + b^2 + 2ac + 2bc + c^2 + a^2 + b^2 + c^2

= (a^2 + 2ab + b^2) + (b^2 + 2bc + c^2) + (c^2 + 2ca + a^2)

= (a+b)^2 + (b+c)^2 + (c+a)^2 = VP

Vậy...

( a + b + c ) ² = a ( a + b + c ) + b ( a + b + c ) + c ( a + b + c ) 
= a² + ab + ac + ab + b² + bc + ac + bc + c²
= a² + b² + c² + 2ab + 2ac + 2bc
 

x³ +9x² +26x +24=x³+7x²+12x+2x²+14x+24

=x(x²+7x+12)+2(x²+7x+12)

=(x+2)(x²+7x+12)

=(x+2)[x²+4x+3x+12]

=(x+2)[x(x+4)+3(x+4)]

=(x+2)(x+3)(x+4)

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

\(3x=5\)

\(x=\frac{5}{3}\)

b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)

\(3x-8=2x-7\)

\(x=1\)

c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)

\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)

\(4x^2-3x-18=4x^2+3x\)

\(6x=-18\)

\(x=-3\)

d) Sai đề

e) ko bt

2x³ - 12x² + 17x -2

=2x³-8x²-4x²+x+16x-2

=(2x³-8x²+x)-(4x²-16x-2)

=x(2x²-8x+1)-2(2x²-8x+1)

=(x-2)(2x²-8x+1)

chắc đề thế này @@ (a+3)(9a-8) - (2+a)(9a-1) 

=9a²-8a+27a-24-9a²-17a+2

=(9a²-9a²)+(-8a+27a-17a)-24+2

=2a-22.Thay a=-3,5 vào được:2*(-3,5)-22

=-7-22=-29.Đpcm

(1) \(2-x+x-2=4x\Leftrightarrow0=4x\Leftrightarrow x=0\)

(2)\(-2-x-x+2=4x\Leftrightarrow-2x=4x\Leftrightarrow-6x=0\Leftrightarrow x=0\)

(3)\(-2-x+x-2=4x\Leftrightarrow-4=4x\Leftrightarrow x=-1\)

(4)\(2-x-x+2=4x\Leftrightarrow4-2x=4x\Leftrightarrow-6x=-4\Leftrightarrow x=\frac{2}{3}\)

\(S=\left\{-1;0;\frac{2}{3}\right\}\)