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

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

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

\(\Leftrightarrow\left(2x+3\right)\left(4x^2+12x+9-x^2-3x-3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(3x^2+9x+6\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-1\\x=-2\end{matrix}\right.\)

ảnh lỗi r ạ

Lỗi rùi

lưu ảnh rồi gửi vào đây nhé

      \(\left(2x-3\right)^2-2x+3=0\)

\(\Rightarrow\left(2x-3\right)^2-\left(2x-3\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(2x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\2x-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{3}{2}\\x=2\end{cases}}\)

\(m>1\Rightarrow ac=-m-3< 0\Rightarrow\) pt luôn có 2 nghiệm trái dấu

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-m-3\end{matrix}\right.\)

\(A=\dfrac{2\left(x_1+x_2\right)^2-6x_1x_2}{x_1+x_2}=\dfrac{2.4\left(m-1\right)^2+6\left(m+3\right)}{2\left(m-1\right)}\)

\(=\dfrac{4\left(m-1\right)^2+3\left(m-1\right)+12}{m-1}=4\left(m-1\right)+\dfrac{12}{m-1}+3\)

\(A\ge2\sqrt{4\left(m-1\right).\dfrac{12}{m-1}}+3=3+8\sqrt{3}\)

Dấu "=" xảy ra khi \(4\left(m-1\right)=\dfrac{12}{m-1}\Rightarrow m=1+\sqrt{3}\)

bấn MT

cần gấp

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 

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

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

\(12x^2+6x=0\) 

\(6x\left(2x+1\right)=0\)  

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

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

1. ( 3x + 2 )³ - ( 3x + 2 ) ( 9x² - 6x + 4 ) = 0

<=> ( 3x + 2 ) [ ( 3x + 2 )² - ( 9x² - 6x + 4 ) ] = 0

<=> ( 3x + 2 ) ( 9x² + 12x + 4 - 9x² + 6x - 4 ) = 0

<=> ( 3x + 2 ) 18x = 0

<=> \(\orbr{\begin{cases}3x+2=0\\18x=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=-\frac{2}{3}\\x=0\end{cases}}\)

2. ( 2x + 1 )³ - ( 2x + 1 ) ( 4x² - 2x + 1 ) = 0

<=> ( 2x + 1 ) [ ( 2x + 1 )² - ( 4x² - 2x + 1 ) ] = 0

<=> ( 2x + 1 ) [ 4x² + 4x + 1 - 4x² + 2x - 1 ) ] = 0

<=> ( 2x + 1 ) 6x = 0

<=> \(\orbr{\begin{cases}2x+1=0\\6x=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}\)

5)

để \(\frac{5x-3}{x+1}\)là số nguyên

\(5x-3⋮x+1\)

\(x+1⋮x+1\)

\(\Rightarrow5\left(x+1\right)⋮x+1\)

\(5x-3-\left(5x-5\right)⋮x+1\)

\(-2⋮x+1\)

\(\Rightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Vậy \(x\in\left\{0;-2;1;-3\right\}\)

Ta có : x² - 2x - (x + 3)² = 6

<=> x² - 2x - x² - 6x - 9 = 6

<=> -8x - 9 = 6 

=> -8x = 15

=> x = \(\frac{15}{-8}\)

\(\left(2x-3\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\) (Thêm KL cuối dòng: Vậy \(x\in\left\{\dfrac{3}{2};\dfrac{1}{2}\right\}\))

(2x-3)x(x-1/2)=0

Đặt từng nhân tử bằng không và giải cho x:

2x - 3 = 0

2x = 3

x = 3/2

x = 0

x - 1/2 = 0

x = 1/2