x³ - 9x - 8x - 24 = 0 

<=> x(x² - 9) - 8(x + 3) = 0

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

<=> (x + 3)(x² - 3x - 8) = 0 

<=> \(\left(x+3\right)\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{41}{4}\right)=0\)

<=> \(\left(x+3\right)\left[\left(x-\frac{3}{2}\right)^2-\left(\frac{\sqrt{41}}{2}\right)^2\right]=0\)

<=> \(\left(x+3\right)\left(x-\frac{3}{2}+\frac{\sqrt{41}}{2}\right)\left(x-\frac{3}{2}-\frac{\sqrt{41}}{2}\right)=0\)

<=> \(\orbr{\begin{cases}x=-3\\x=\frac{3\pm\sqrt{41}}{2}\end{cases}}\)

(x+2y)^2=x^2+2.x.2y+(2y)^2=x^2+4xy+4y^2

thank, đáp án đứng r

a) A = ( x - 2y )³ + ( x + 2y )³ - 2x ( x² + y )=

       = x³ - 6x²y + 12xy² - 8y³ + x³ + 6x²y + 12xy² + 8y³ - 2x³ - 2xy

       =                       24xy² - 2xy

b) B = ( x - 1 )( x + y ) ( x - y )  - x²( x - 1 )= 

       =  ( x -1 )( x² - y² ) - x² ( x - 1 )

       = ( x - 1 )( x² - y² - x² )

       = -y² ( x - 1 ) 

c ) C =  ( x + 2)² - 2( x + 2 )( x - 8 ) + ( x - 8 ) ² = 

        = ( x + 2 - x + 8 ) ² 

        = 10²

          = 100

HOk tốt!!!!!!!!!! 

Bài 2 : a) \(2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}}\)

b) \(2\left(2x-1\right)+6x\left(2x-1\right)=0\Leftrightarrow\left(2x-1\right)\left(2+6x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\2+6x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\6x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{2}{6}=-\frac{1}{3}\end{cases}}}\)

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

\(\Leftrightarrow\left(-x-9\right)\left(3x+3\right)=0\Leftrightarrow\orbr{\begin{cases}-x-9=0\\3x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-9\\x=-1\end{cases}}}\)

Tự KL cho các phần 

cái đầu tiên là x²+2y² nha

a)

\(x+2y=5\Leftrightarrow x=5-2y\)

Thay vào ta được

\(M=\left(5-2y\right)^2+2y^2=25-20y+4y^2+y^2=6y^2-20y+25=6\left(y^2-\frac{10}{3}y+\frac{25}{9}\right)+\frac{25}{3}=6\left(y-\frac{5}{3}\right)^2+\frac{25}{3}\)

Mà \(6\left(y-\frac{5}{3}\right)^2\ge0\forall y\Leftrightarrow6\left(y-\frac{5}{3}\right)^2+\frac{25}{3}\ge\frac{25}{3}\)

Dấu '' = '' xảy ra \(\Leftrightarrow y=\frac{5}{3}\)

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

\(\Rightarrow MinM=\frac{25}{3}\Leftrightarrow x=y=\frac{5}{3}\)

A = -x² + 2x + 5 = -( x² - 2x + 1 ) + 6 = -( x - 1 )² + 6 ≤ 6 ∀ x

Dấu "=" xảy ra <=> x = 1 . => MaxA = 6

B = 3x² + x + 1 = 3( x² + 1/3x + 1/36 ) + 11/12 = 3( x + 1/6 )² + 11/12 ≥ 11/12 ∀ x

Dấu "=" xảy ra <=> x = -1/6 . => MinB = 11/12

C = -4x² - x + 1 = -4( x² + 1/4x + 1/64 ) + 17/16 = -4( x + 1/8 )² + 17/16 ≤ 17/16 ∀ x

Dấu "=" xảy ra <=> x = -1/8 . => MaxC = 17/16

D = 9x² - x + 3 = 9( x² - 1/9x + 1/324 ) + 107/36 = 9( x - 1/18 )² + 107/36 ≥ 107/36 ∀ x

Dấu "=" xảy ra <=> x = 1/18 . => MinD = 107/36

Trả lời:

Bài 10:

b, \(x^2-x=-2x^2+2x\)

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

\(\Leftrightarrow3x^2-3x=0\)

\(\Leftrightarrow3x\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Vậy x = 0; x = 1 là nghiệm của pt.

c, \(2x^2\left(x-1\right)+x^2=x\)

\(\Leftrightarrow2x^2\left(x-1\right)+x^2-x=0\)

\(\Leftrightarrow2x^2\left(x-1\right)+x\left(x-1\right)=0\)

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

\(\Leftrightarrow x=0;x=1;x=-\frac{1}{2}\)

Vậy x = 0; x = 1; x = - 1/2 là nghiệm của pt.

d, \(\left(x-2\right)\left(x^2+4\right)=x^2-2x\)

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

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

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

\(\Leftrightarrow x-2=0\) (vì \(x^2-x+4=\left(x-\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\forall x\) )

\(\Leftrightarrow x=2\)

Vậy x = 2 là nghiệm của pt.

x² + x = -2x² + 2x

<=> 3x² - x = 0

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

<=> x = 0 hoặc x = 1/3