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

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

\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)   

\(=0\)   

Vậy giá trị biểu thức không phụ thuộc vào biến x 

( x - 1 )³ - ( x - 1 )( x² + x + 1 ) - 3( 1 - x )x 

= x³ - 3x² + 3x - 1 - ( x³ - 1 ) - 3x + 3x²

= x³ - 3x² + 3x - 1 - x³ + 1 - 3x + 3x²

= 0

Vậy biểu thức không phụ thuộc vào biến ( đpcm )

\(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)( ĐKXĐ : \(x\ne\pm2\))

\(=\frac{5\left(x+2\right)}{2\left(2x-4\right)}\cdot\frac{-\left(2x-4\right)}{x+2}\)

\(=\frac{-5\left(x+2\right)\left(2x-4\right)}{2\left(2x-4\right)\left(x+2\right)}\)

\(=-\frac{5}{2}\)

\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}\)( ĐKXĐ : \(x\ne-5;x\ne6\))

\(=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\frac{3}{-\left(x-6\right)}\)

\(=\frac{3\left(x-6\right)\left(x+6\right)}{-2\left(x+5\right)\left(x-6\right)}\)

\(=\frac{3\left(x+6\right)}{-2\left(x+5\right)}=\frac{3x+18}{-2x-10}=-\frac{3x+18}{2x+10}\)

a) 

Điều kiện : \(\hept{\begin{cases}4x-8\ne0\\x+2\ne0\end{cases}}\)    

\(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)    

\(=\frac{5\left(x+2\right)}{-2\left(4-2x\right)}\cdot\frac{4-2x}{x+2}\)    

\(=\frac{-5}{2}\)    

b) 

Điều kiện : \(\hept{\begin{cases}2x+10\ne0\\6-x\ne0\end{cases}}\)    

\(\hept{\begin{cases}x\ne-5\\x\ne6\end{cases}}\)     

\(=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}\)   

\(=\frac{-6\left(x+6\right)\cdot3}{2x+10}\)   

\(=\frac{-9\left(x+6\right)}{x+5}\)  

\(=\frac{-9x-54}{x+5}\)  

\(=\frac{-9\left(x+5\right)-9}{x+5}\) 

\(=-9-\frac{9}{x+5}\)

a) ( x - 1 )³ + 3x( x - 1 )² + 3x²( x - 1 ) + x³

= [ ( x - 1 ) + x ) ]³ ( HĐT số 4 )

= [ x - 1 + x ]³

= [ 2x - 1 ]³ 

=> đpcm

b) ( x² - 2xy )³ + 3( x² - 2xy )y² + 3( x² - 2xy )y⁴ + y⁶

= [ ( x² - 2xy ) + y² ]³ ( HĐT số 4 )

= [ x² - 2xy + y² ]³

= [ ( x - y )² ]³

= ( x - y )⁶

=> đpcm

Áp dụng công thức : (A + B)³ = A³ + 3A²B + 3AB² + B³

(A - B)³ = A³ - 3A²B + 3AB² -B³

a) (3x + 1)³ = (3x)³ + 3.(3x)².1 + 3.3x.1 + 1³ = 27x³ + 27x² + 9x + 1

b) \(\left(\frac{x}{3}-1\right)^3=\left(\frac{x}{3}\right)^3-3\cdot\left(\frac{x}{3}\right)^2\cdot1+3\cdot\left(\frac{x}{3}\right)\cdot1^2-1^3\)

\(=\frac{x^3}{27}-3\cdot\frac{x^2}{9}\cdot1+3\cdot\frac{x}{3}\cdot1-1\)

= \(\frac{x^3}{27}-\frac{x^2}{3}+x-1\)

c) \(\left(2x-\frac{1}{x}\right)^3=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot\frac{1}{x}+3\cdot2x\cdot\left(\frac{1}{x}\right)^2-\left(\frac{1}{x}\right)^3\)

\(=8x^3-3\cdot4x^2\cdot\frac{1}{x}+6x\cdot\frac{1}{x^2}-\frac{1}{x^3}\)

\(=8x^3-12x+\frac{6}{x}-\frac{1}{x^3}\)

d) \(\left(-y^2+3x\right)^3=\left(3x-y^2\right)^3=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot y^2+3\cdot3x\cdot y^4-y^6\)

= 27x³ - 27x²y² + 9xy⁴ - y⁶

= -y⁶ + 9xy⁴ - 27x²y² + 27x³

Tương tự câu cuối :>

a) \(8a^2xy-18b^2xy=2xy\left(4a^2-9b^2\right)=2xy\left(2a-3b\right)\left(2a+3b\right)\)

b) \(32a^2b^2-4=4\left(8a^2b^2-1\right)\)

c) \(x^2-49z^2-4xy+4y^2=\left(x^2-4xy+4y^2\right)-49z^2\)

\(=\left(x-2y\right)^2-\left(7z\right)^2=\left(x-2y+7z\right)\left(x-2y-7z\right)\)

d) \(3x^2+6x+3-3y^2=3\left(x^2+2x+1-y^2\right)=3.\left[\left(x+1\right)^2-y^2\right]\)

\(=3\left(x-y+1\right)\left(x+y+1\right)\)

e) \(12x^2y-12y^3+36xy+27y=3y\left(4x^2-4y^2+12x+9\right)\)

\(=3y\left[\left(4x^2+12x+9\right)-4y^2\right]=3y\left[\left(2x+3\right)^2-\left(2y\right)^2\right]\)

\(=3y\left(2x-2y+3\right)\left(2x+2y+3\right)\)

a) 8a²xy - 18b²xy 

= 2xy( 4a² - 9b² )

= 2xy( [ ( 2a )² - ( 3b )² ]

= 2xy( 2a - 3b )( 2a + 3b )

b) 32a²b² - 4

= 4( 8a²b² - 1 )

c) x² - 49z² - 4xy + 4y²

= ( x² - 4xy + 4y² ) - 49z²

= ( x - 2y )² - ( 7z )²

= ( x - 2y - 7z )( x - 2y + 7z )

d) 3x² + 6x + 3 - 3y²

= 3( x² + 2x + 1 - y² )

= 3[ ( x² + 2x + 1 ) - y² ]

= 3[ ( x + 1 )² - y² ]

= 3( x - y + 1 )( x + y + 1 )

e) 12x²y - 12y³ + 36xy + 27y

= 3y( 4x² - 4y² + 12x + 9 )

= 3y[ ( 4x² + 12x + 9 ) - 4y² ]

= 3y[ ( 2x + 3 )² - ( 2y )² ]

= 3y( 2x - 2y + 3 )( 2x + 2y + 3 )

Kéo dài AE cắt CD tại M, kéo dài BE cắt CD tại N

=> ^BAM=^AMD (góc so le trong), mà ^BAM = ^DAM (đề bài) => ^DAM = ^AMD => tg ADM cân tại D => AD=DM

Chứng minh tương tự ta cũng có tg BNC cân tại C => BC=CN

=> AD+BC = DM+CN=DN+MN+MN+CM=(DN+MN+CM)+MN=CD+MN

Mà CD=AD+BC (theo đề bài) => MN=0 hay M trùng N chính là giao của AE và BE => E trùng M trùng N => E thuộc CD

tìm nghiệm của 3x^2-2x+1

Do \(x,y,z\inℤ\)

nen tu gia thiet suy ra

\(x^2+4y^2+z^2-2xy-2y+2z\le-1\)

\(\Leftrightarrow\left(x-y\right)^2+\left(z+1\right)^2+\left(y-1\right)^2+2y^2\le1\)

mat khac

\(\hept{\begin{cases}\left(y-1\right)^2+2y^2>0\\\left(x-y\right)^2+\left(z+1\right)^2\ge0\end{cases}}\)

nen \(\left(x-y\right)^2+\left(z+1\right)^2+\left(y-1\right)^2+2y^2=1\)

den day ban lap bang cac gia tri se tim duoc \(\left(x,y,z\right)=\left(0,0,-1\right)\)

Bài làm:

1) đk: \(x\ne0;x\ne-5\)

Ta có: \(\frac{30}{x}-\frac{30}{x+5}=1\)

\(\Leftrightarrow\frac{30\left(x+5\right)-30x}{x\left(x+5\right)}=1\)

\(\Leftrightarrow x^2+5x=150\)

\(\Leftrightarrow x^2+5x-150=0\)

\(\Leftrightarrow\left(x-10\right)\left(x+15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+15=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-15\end{cases}}\)

2) đk: \(x\ne0;x\ne-2\)

Ta có: \(\frac{60}{x}-\frac{60}{x+2}=1\)

\(\Leftrightarrow\frac{60\left(x+2\right)-60x}{x\left(x+2\right)}=1\)

\(\Leftrightarrow x^2+2x=120\)

\(\Leftrightarrow x^2+2x-120=0\)

\(\Leftrightarrow\left(x-10\right)\left(x+12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-12\end{cases}}\)

\(\frac{30}{x}-\frac{30}{x+5}=1\)( ĐKXĐ : \(x\ne0;x\ne-5\))

<=> \(30\left(\frac{1}{x}-\frac{1}{x+5}\right)=1\)

<=> \(30\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{x\left(x+5\right)}\right)=1\)

<=> \(30\left(\frac{5}{x\left(x+5\right)}\right)=1\)

<=> \(\frac{5}{x\left(x+5\right)}=\frac{1}{30}\)

<=> \(5\cdot30=x\left(x+5\right)\)

<=> \(x^2+5x-150=0\)

<=> \(x^2+15x-10x-150=0\)

<=> \(x\left(x+15\right)-10\left(x+15\right)=0\)

<=> \(\left(x-10\right)\left(x+15\right)=0\)

<=> \(\orbr{\begin{cases}x-10=0\\x+15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-15\end{cases}}\)( tmđk )

Vậy S = { 10 ; -15 }

\(\frac{60}{x}-\frac{60}{x+2}=1\)( ĐKXĐ : \(x\ne0;x\ne-2\))

<=> \(60\left(\frac{1}{x}-\frac{1}{x+2}\right)=1\)

<=> \(60\left(\frac{x+2}{x\left(x+2\right)}-\frac{x}{x\left(x+2\right)}\right)=1\)

<=> \(60\left(\frac{2}{x\left(x+2\right)}\right)=1\)

<=> \(\frac{2}{x\left(x+2\right)}=\frac{1}{60}\)

<=> \(2\cdot60=x\left(x+2\right)\)

<=> \(x^2+2x-120=0\)

<=> \(x^2+12x-10x-120=0\)

<=> \(x\left(x+12\right)-10\left(x+12\right)=0\)

<=> \(\left(x-10\right)\left(x+12\right)=0\)

<=> \(\orbr{\begin{cases}x-10=0\\x+12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-12\end{cases}}\)

Vậy S = { 10 ; -12 }