a) 

<=> \(3x-12x^2+12x^2-6x=9\)

<=> \(-3x=9\)

<=> \(x=-3\)

b)

<=> \(6x-24x^2-12x+24x^2=6\)

<=> \(-6x=6\)

<=> \(x=-1\)

c) 

<=> \(6x-4-3x+6=1\)

<=> \(3x+2=1\)

<=> \(x=-\frac{1}{3}\)

d) 

<=> \(9-6x^2+6x^2-3x=9\)

<=> \(-3x=0\)

<=> \(x=0\)

e) KO HIỂU ĐỀ

f) 

<=> \(4x^2-8x+3-\left(4x^2+9x+2\right)=8\)

<=> \(-17x+1=8\)

<=> \(x=-\frac{7}{17}\)

g) 

<=> \(-6x^2+x+1+6x^2-3x=9\)

<=> \(-2x=8\)

<=> \(x=-4\)

h)

<=> \(x^2-x+2x^2+5x-3=4\)

<=> \(3x^2+4x=7\)

<=> \(\orbr{\begin{cases}x=1\\x=-\frac{7}{3}\end{cases}}\)

a. \(3x\left(1-4x\right)+6x\left(2x-1\right)=9\)

\(\Rightarrow3x-12x^2+12x^2-6x=9\)

\(\Rightarrow-3x=9\)

\(\Rightarrow x=-3\)

b. \(3x\left(2-8x\right)-12x\left(1-2x\right)=6\)

\(\Rightarrow6x-24x^2-12x+24x^2=6\)

\(\Rightarrow-6x=6\)

\(\Rightarrow x=-1\)

c. \(2\left(3x-2\right)-3\left(x-2\right)=1\)

\(\Rightarrow6x-4-3x+6=1\)

\(\Rightarrow3x+2=1\)

\(\Rightarrow3x=-1\)

\(\Rightarrow x=-\frac{1}{3}\)

\(=\left(a^2-1\right)^3-\left(a^6-1\right)\)

\(=a^6-3a^4+3a^2-1-a^6+1\)

\(=-3a^4+3a^2\)

\(=-3a^2\left(a^2-1\right)\)

\(\left(a^2-1\right)^3-\left(a^4+a^2+1\right)\left(a^2-1\right)\)

\(=a^6-3a^4+3a^2-1-\left(a^6-a^4+a^4-a^2+a^2-1\right)\)

\(=a^6-3a^4+3a^2-1-a^6+1\)

\(=-3a^4+3a^2\)

\(=-3a^2\left(a^2-1\right)\)

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

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-4\left(a^3+b^3+c^3\right)-12abc\)

\(=-3\left(a^3+b^3+c^3\right)+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-12abc\)

\(=-3\left(\left(a^3+b^3+c^3\right)-\left(a+b\right)\left(b+c\right)\left(c+a\right)+4abc\right)\)

XONG NHAAAAA :3333333

a = 11111...111(2n chứ số 1) = \(\frac{10^{2n}-1}{9}\)

b = 22222...222(n chữ số 2) = \(\frac{2\left(10^n-1\right)}{9}\)

a - b = \(\frac{10^{2n}-1}{9}-\frac{2.10^n-2}{9}=\frac{10^{2n}-1-2.10^n+2}{9}\)

\(=\frac{10^{2n}-2.10^n+1}{9}=\frac{\left(10^n-1\right)^2}{3^2}=\left(\frac{10^n-1}{3}\right)^2\)là số chính phương

=> đpcm

Ta có :

b = 22222...22222 ( n chữ số 2 ) = 2m

a = 11111...111 ( 2n chữ số 1 ) = 10ⁿ . 11111...111 ( n chữ số ) + 11...1111 ( n chữ số )

\(=\left(9m+1\right)m+m=9m^2+2m\) 

Lấy vế a trừ vế b ta được  \(9m^2+2m-2m=9m^2=\left(3a\right)^2\) là SCP 

=> Đpcm

Câu b) khó hiểu, ko có dấu "=" nhaaaa !!!!!!

\(B=3\left(\frac{a}{b}+\frac{b}{a}\right)+\frac{a}{b}+\frac{b}{a+b}\ge3.2+\frac{a}{b}+\frac{b}{a+b}=6+\frac{a}{b}+\frac{b}{a+b}>6+0+0=6\)

Vậy ta thấy \(B>6\)

=> Ko có dấu "=" xảy ra.

a) ( x - 3 )( x + 3 )( x + 2 ) - ( x - 1 )( x² - 3 ) - 5x( x + 4 )² - ( x - 5 )²

= ( x² - 3² )( x + 2 ) - ( x³ - x² - 3x + 3 ) - 5x( x² + 8x + 16 ) - ( x² - 10x + 25 )

= x³ + 2x² - 9x - 18 - x³ + x² + 3x - 3 - 5x³ - 40x² - 80x - x² + 10x - 25

= ( x³ - x³ - 5x³ ) + ( 2x² + x² - 40x² - x² ) + ( -9x + 3x + 10x - 80x ) + ( -18 - 3 - 25 )

= -5x³ - 38x² - 76x - 46 

b) 2x( x - 4 )² - ( x + 5 )( x - 2 )( x + 2 ) + 2( x + 5 )² - ( x - 1 )²

= 2x( x² - 8x + 16 ) - ( x + 5 )( x² - 4 ) + 2( x² + 10x + 25 ) - ( x² - 2x + 1 )

= 2x³ - 16x² + 32x - ( x³ + 5x² - 4x - 20 ) + 2x² + 20x + 50 - x² + 2x - 1

= 2x³ - 16x² + 32x - x³ - 5x² + 4x + 20 + 2x² + 20x + 50 - x² + 2x - 1

= ( 2x³ - x³ ) + ( -16x² - 5x² + 2x² - x² ) + ( 32x + 4x + 20x + 2x ) + ( 20 + 50 - 1 )

= x³ - 20x² + 58x + 69

c) ( x + 5 )² - 4x( 2x + 3 )² - ( 2x - 1 )( x + 3 )( x - 3 )

= x² + 10x + 25 - 4x( 4x² + 12x + 9 ) - ( 2x - 1 )( x² - 9 )

= x² + 10x + 25 - 16x³ - 48x² - 36x - ( 2x³ - x² - 18x + 9 )

= x² + 10x + 25 - 16x³ - 48x² - 36x - 2x³ + x² + 18x - 9

= ( -16x³ - 2x³ ) + ( x² - 48x² + x² ) + ( 10x - 36x + 18x ) + ( 25 - 9 )

= -18x³ - 46x² - 8x + 16

d) -2x( 3x + 2 )( 3x - 2 ) + 5( x + 2 )² - ( x - 1 )( 2x + 1 )( 2x - 1 )

= -2x( 9x² - 4 ) + 5( x² + 4x + 4 ) - ( x - 1 )( 4x² - 1 )

= -18x³ + 8x + 5x² + 20x + 20 - ( 4x³ - 4x² - x + 1 )

= -18x³ + 8x + 5x² + 20x + 20 - 4x³ + 4x² + x - 1

= ( -18x³ - 4x³ ) + ( 5x² + 4x² ) + ( 8x + 20x + x ) + ( 20 - 1 )

= -22x³ + 9x² + 29x + 19

Ta có:63^2-47^2=3969-2209=1760

215^2-105^2=46225-11025=35200

ks nhé!Học tốt!