a)(a-b)2012x+2012y-ax-ay

=(a-b)[x(2012-a)+y(2012-a0]

=(a-b)(2012-a)(x+y)

b)(a-b)144y²-x²+2x-1

=(a-b)[144y²-(x²-2x+1)]

=(a-b)[(12y)²-(x-1)²]

=(a-b)(12y+x-1)(12y-x+1)

Bạn ấy làm chuẩn rồi

k cho mk nha

Ta có :

\(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)

\(=\left(a^2+ab+bc+ca\right)\left(b^2+ab+bc+ca\right)\left(c^2+ab+bc+ca\right)\)

\(=\left[\left(a^2+ab\right)+\left(bc+ca\right)\right]\left[\left(b^2+ab\right)+\left(bc+ca\right)\right]\left[\left(c^2+bc\right)+\left(ab+ca\right)\right]\)

\(=\left(a+c\right)\left(a+b\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)\left(b+c\right)\)

\(=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)

Vậy ...

Bạn tham khảo bài làm ở đường link phía dưới nhé

Câu hỏi của Nguyễn Desmond - Toán lớp 8 - Học toán với OnlineMath

a) 4x² - 3x - 7 = 4x² + 4x - 7x - 7 = 4x(x + 1) - 7(x + 1) = (4x - 7)(x + 1) = 0\(\Rightarrow\orbr{\begin{cases}4x-7=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=-1\end{cases}}}\)

b) x² - 5x - 6 = x² + x - 6x - 6 = x(x + 1) - 6(x + 1) = (x - 6)(x + 1) = 0\(\Rightarrow\orbr{\begin{cases}x-6=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=6\\x=-1\end{cases}}}\)

c) 2x² + 3x - 5 = 2x² - 2x + 5x - 5 = 2x(x - 1) + 5(x - 1) = (2x + 5)(x - 1) = 0\(\Rightarrow\orbr{\begin{cases}2x+5=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2,5\\x=1\end{cases}}}\)

d) (3x - 1)² - (x + 2)² = [(3x - 1) - (x + 2)].[(3x - 1) + (x + 2)] = (2x - 3)(4x + 1) = 0\(\Rightarrow\orbr{\begin{cases}2x-3=0\\4x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1,5\\x=-0,25\end{cases}}}\)

a) 4x² - 3x - 7 = 0

<=> 4x² + 4x - 7x - 7 = 0

<=> 4x ( x + 1 ) - 7 ( x + 1 ) = 0

<=> ( x + 1 ) ( 4x - 7 ) = 0

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\4\text{x}-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{7}{4}\end{cases}}}\)

Vậy x = { -1 ; \(\frac{7}{4}\)}

b) x² - 5x - 6 = 0

<=> x² +  x - 6x  - 6 = 0

<=> x ( x + 1 ) - 6 ( x + 1 ) = 0

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

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

Vậy ______

c) 2x² + 3x - 5 = 0

<=> 2x² -  2x + 5x - 5 = 0

<=> 2x ( x - 1 ) + 5 ( x - 1 ) = 0

<=> ( x - 1 ) ( 2x + 5 ) = 0

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

Vậy _____

d) ( 3x - 1 )² - ( x + 2 )² = 0

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

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

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

Vậy _______