Ta có 4E = 4x² - 4xy + 12y² - 8x - 40y + 80 

= 4x² - 4xy + y² - 4(2x - y) + 4 + 11y² - 44y + 44 + 32 

= (2x - y)² - 4(2x - y) + 4 + 11(y² - 4y + 4) + 32 

 = (2x - y - 2)² + (y - 2)² + 32 \(\ge32\)

=> 4E \(\ge32\)

=> E \(\ge\)8

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-y-2=0\\y-2=0\end{cases}}\Leftrightarrow x=y=2\)

Vậy Min E = 8 <=> x = y = 2 

x(a+b) - a-b

=x(a+b) - (a+b)

=(x-1)(a+b)

x( a - b ) - a + b = x( a - b ) - ( a - b ) = ( a - b )( x - 1 )

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

mấy ý kia không sửa được nữa nên nghỉ:)

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

\(< =>\left(3x-5\right)\left(3x-5-x\right)=0\)

\(< =>\orbr{\begin{cases}3x-5=0\\2x-5=0\end{cases}< =>\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{5}{2}\end{cases}}}\)

Trả lời:

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\2x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{5}{2}\end{cases}}}\)

Vậy x = 5/3; x = 5/2 là nghiệm của pt.

Trả lời:

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

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

\(\Leftrightarrow\left(-x-1\right)\left(3x-1\right)=0\)

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

Vậy x = - 1; x = 1/3 là nghiệm của pt.

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

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

\(\Leftrightarrow\left(-x-1\right)\left(3x-1\right)=0\)

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

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

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

Vậy \(S=\left\{-1;\frac{1}{3}\right\}\)

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

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

\(\Rightarrow\left(-x-1\right)\left(3x-1\right)=0\)

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

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

\(\left(x^3-x^2-5x+21\right):\left(x^2-4x+7\right)\)

\(=\left(x^3-4x^2+3x^2+7x-12x+21\right):\left(x^2-4x+7\right)\)

\(=\left[\left(x^3-4x^2+7x\right)+\left(3x^2-12x+21\right)\right]:\left(x^2-4x+7\right)\)

\(=\left[x\left(x^2-4x+7\right)+3\left(x^2-4x+7\right)\right]:\left(x^2-4x+7\right)\)

\(=\left[\left(x^2-4x+7\right)\left(x+3\right)\right]:\left(x^2-4x+7\right)\)

\(=x+3\)

đầy đủ giúp em nhé

\(5x\left(x-2\right)=x-2\)

\(\Leftrightarrow5x\left(x-2\right)-x+2=0\)

\(\Leftrightarrow5x^2-10x-x+2=0\)

\(\Leftrightarrow\left(5x^2-x\right)-\left(10x-2\right)=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2\end{cases}}\)

Vậy \(S=\left\{\frac{1}{5};2\right\}\)

tìm x biết: 5x(x-2)=x-2

x=1/5,

x=2

nha bạn 

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