1. <=> \(\left(3x+2\right)^3-\left(\left(3x\right)^3+2^3\right)=0\)

<=> \(\left(\left(3x\right)^3+2^3+3\left(3x+2\right).3x.2\right)-\left(\left(3x\right)^3+2^3\right)=0\)

<=>3 (3x + 2) . 3x.2 = 0 

<=> (3x + 2 ) . x = 0 

<=> x = -2/3 hoặc x = 0

2. Tương tự

1 

\(\left(3x+2\right)^3-\left[\left(3x\right)^3+2^3\right]=0\) 

\(\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot2+3\cdot3x\cdot2^2+2^3-\left(3x\right)^3-2^3=0\) 

\(54x^2+36x=0\)  

\(18x\left(3x+2\right)=0\) 

\(\orbr{\begin{cases}x=0\\3x+2=0\end{cases}}\) 

\(\orbr{\begin{cases}x=0\\x=\frac{-2}{3}\end{cases}}\) 

2 

\(\left(2x+1\right)^3-\left[\left(2x\right)^3-1^3\right]=0\) 

\(\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3-\left(2x\right)^3-1^3=0\)  

\(12x^2+6x=0\) 

\(6x\left(2x+1\right)=0\)  

\(\orbr{\begin{cases}x=0\\2x+1=0\end{cases}}\)  

\(\orbr{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)

a) x^2 - 11x + 18 = 0 

=> x^2 - 2x - 9x + 18 = 0 

=> x ( x- 2 ) - 9 ( x- 2 ) = 0 

=> ( x- 9 )( x- 2 )= 0 

=> x- 9 = 0 hoặc x - 2 = 0 

=> x= 9 hoặc x = 2 

1)2x³+3x²+2x+3=0

=> (2x³+3x²)+(2x+3)=0

=> x²(2x+3)+(2x+3)=0

=> (2x+3)(x²+1)=0

=>\(\hept{\begin{cases}2x+3=0\\x^2+1=0\end{cases}}\)=>\(\hept{\begin{cases}2x=-3\\x^2=-1\end{cases}}\)=>\(\hept{\begin{cases}x=\frac{-3}{2}\\vo.nghiem\end{cases}}\)

Vậy x=-3/2

2)x²-3x-18=0

=> (x²+3x)-(6x+18)=0

=> x(x+3)-6(x+3)=0

=> (x+3)(x-6)=0

=> \(\hept{\begin{cases}x+3=0\\x-6=0\end{cases}}\)=>\(\hept{\begin{cases}x=-3\\x=6\end{cases}}\)

Vậy x=-3 hoặc x=6

3)Sai đề rồi bạn, 30 thành 30x mới đúng

x³-11x²+30x=0

=> x(x²-11x+30)=0

=> x[(x²-5x)-(6x-30)]=0

=> x[x(x-5)-6(x-5)]=0

=> x(x-5)(x-6)=0

=>\(\hept{\begin{cases}x=0\\x-5=0\\x-6=0\end{cases}}\)=>\(\hept{\begin{cases}x=0\\x=5\\x=6\end{cases}}\)

Vậy x=0 hoặc x=5 hoặc x=6

\(2x\left(x^2-25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-25=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

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

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

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x-5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{3}\end{cases}}\)

\(9\left(3x-2\right)-x\left(2-3x\right)=0\)

\(9\left(3x-2\right)+x\left(3x-2\right)=0\)

\(\left(9+x\right)\left(3x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}9+x=0\\3x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-9\\x=\frac{2}{3}\end{cases}}\)

\(\left(2x-1\right)^2=25\)

\(\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)

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

\(\Leftrightarrow-5\left(4x+3\right)=0\)

\(\Leftrightarrow4x+3=0\)

\(\Leftrightarrow4x=-3\)

\(\Leftrightarrow x=\frac{-3}{4}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)

h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)

\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)

\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)

\(\Leftrightarrow-9x+2x-3-10x=30\)

\(\Leftrightarrow-17x-3=30\)

\(\Leftrightarrow-17x=33\)

\(\Leftrightarrow x=\frac{-33}{17}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)

a) ( x +2 )² - ( 3x - 1 ) ( x +2 ) = 0

<=> (x+2)(x+2-3x+1) = 0

<=> (x+2)(-2x+3) = 0

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\-2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{3}{2}\end{cases}}}\)

b) ( 2x - 1 )² - ( 2x + 5 ) ( 2x - 5 ) = 18

<=> 4x² -4x +1  - (4x²-25) =18

<=> 4x² -4x +1  - 4x²  + 25 = 18

<=> - 4x + 26 = 18

<=> - 4x  = 18 - 26

<=> - 4x   = -8

<=> x = 2

c) ( 2x + 3 )² - ( x - 5 )² = 0

<=> [( 2x + 3 ) - ( x - 5 )].[( 2x + 3 ) + ( x - 5 )] = 0

<=> (2x +3 -x +5) . (2x +3  + x  - 5) = 0

<=> (x +8)(3x-2) = 0

\(\Leftrightarrow\orbr{\begin{cases}x+8=0\\3x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-8\\x=\frac{2}{3}\end{cases}}}\)

d) 5x³ + 3x - 8 = 0

<=> (5x³ -5x) +(8x-8) = 0

<=> 5x(x² - 1) + 8(x-1) = 0

<=> 5x(x - 1)(x+1) + 8(x-1) = 0

<=> (x - 1)[5x(x+1) + 8] = 0

<=> (x-1)(5x²+5x +8 ) = 0

<=> (x-1).5.(x²+x+8/5) = 0

<=> 5.(x-1)(x²+x+1/4 + 27/20) = 0

\(\Leftrightarrow\left(x-1\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{27}{20}\right]\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\\left(x+\frac{1}{2}\right)^2+\frac{27}{20}=0\end{cases}\Leftrightarrow x=1}\)vỉ \(\left(x+\frac{1}{2}\right)^2+\frac{27}{20}>0\)với mọi x

Vậy x = 1

a)

pt <=>     \(x^2+4x+4+x^2-6x+9=2x^2+14x\)

<=>     \(2x^2-2x+13=2x^2+14x\)

<=>     \(16x=13\)

<=>     \(x=\frac{13}{16}\)

b)

pt <=>     \(x^3+3x^2+3x+1+x^3-3x^2+3x-1=2x^3\)

<=>   \(2x^3+6x=2x^3\)

<=>   \(6x=0\)

<=>   \(x=0\)

c)

pt <=>    \(\left(x^3-3x^2+3x-1\right)-125=0\)

<=>   \(\left(x-1\right)^3=125\)

<=>   \(x-1=5\)

<=>   \(x=6\)

d)

pt <=>   \(\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)

<=>   \(\left(x-1\right)^2+\left(y+2\right)^2=0\)     (1)

CÓ:   \(\left(x-1\right)^2;\left(y+2\right)^2\ge0\forall x;y\)

=>   \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\)       (2)

TỪ (1) VÀ (2) =>    DÁU "=" XẢY RA <=>   \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}}\)

<=>     \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

e)

pt <=>   \(2x^2+8x+8+y^2-2y+1=0\)

<=>   \(2\left(x+2\right)^2+\left(y-1\right)^2=0\)

TA LUÔN CÓ:   \(2\left(x+2\right)^2+\left(y-1\right)^2\ge0\forall x;y\)

=> DẤU "=" XẢY RA <=>   \(\hept{\begin{cases}2\left(x+2\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\) 

<=>     \(\hept{\begin{cases}x=-2\\y=1\end{cases}}\)

a) ( x + 2 )² + ( x - 3 )² = 2x( x + 7 )

<=> x² + 4x + 4 + x² - 6x + 9 = 2x² + 14x

<=> x² + 4x + x² - 6x - 2x² - 14x = -4 - 9

<=> -16x = -13

<=> x = 13/16

b) ( x + 1 )³ + ( x - 1 )³ = 2x³

<=> x³ + 3x² + 3x + 1 + x³ - 3x² + 3x - 1 = 2x³

<=> x³ + 3x² + 3x + x³ - 3x² + 3x - 2x³ = -1 + 1

<=> 6x = 0

<=> x = 0

c) x³ - 3x² + 3x - 126 = 0

<=> ( x³ - 3x² + 3x - 1 ) - 125 = 0

<=> ( x - 1 )³ = 125

<=> ( x - 1 )³ = 5³

<=> x - 1 = 5

<=> x = 6

d) x² + y² - 2x + 4y + 5 = 0

<=> ( x² - 2x + 1 ) + ( y² + 4y + 4 ) = 0

<=> ( x - 1 )² + ( y + 2 )² = 0 (*)

\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

e) 2x² + 8x + y² - 2y + 9 = 0

<=> 2( x² + 4x + 4 ) + ( y² - 2y + 1 ) = 0

<=> 2( x + 2 )² + ( y - 1 )² = 0 (*)

\(\hept{\begin{cases}2\left(x+2\right)^2\ge0\forall x\\\left(y-1\right)^2\ge0\forall y\end{cases}}\Rightarrow2\left(x+2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

Đẳng thức xảy ra ( tức xảy ra (*) ) <=> \(\hept{\begin{cases}x+2=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=1\end{cases}}\)