Phân tích đa thức thành nhân tử:
ax – bx – a2 + 2ab – b2
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Ta có
25 – a 2 + 2 a b – b 2 = 25 – ( a 2 – 2 a b + b 2 ) = 5 2 – ( a – b ) 2
= (5 + a – b)(5 – a + b)
Đáp án cần chọn là: D
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(a\left(b^2+c^2\right)+b\left(a^2+c^2\right)+c\left(a^2+b^2\right)-2abc-a^3-b^3-c^3\)
\(=c\left(a-b\right)^2+\left[ab^2+ac^2+a^2b+bc^2-a^3-b^3-c^3\right]\)
\(=c\left(a-b\right)^2+c^2\left(a+b-c\right)+ab^2+a^2b-a^3-b^3\)
\(=c\left(a-b\right)^2+c^2\left(a+b-c\right)-\left(a^3-a^2b\right)+\left(ab^2-b^3\right)\)
\(=c\left(a-b\right)^2+c^2\left(a+b-c\right)-a^2\left(a-b\right)+b^2\left(a-b\right)\)
\(=c\left(a-b\right)^2+c^2\left(a+b-c\right)-\left(a+b\right)\left(a-b\right)^2\)
\(=-\left(a-b\right)^2\left(a+b-c\right)+c^2\left(a+b-c\right)\)
\(=\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)\)
\(a^2-b^2-2x\left(a-b\right)=\left(a-b\right)\left(a+b\right)-2x\left(a-b\right)=\left(a-b\right)\left(a+b-2x\right)\)
\(a^2-b^2-2x\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b\right)-2x\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b-2x\right)\)
\(=\left(b^2+c^2+2bc-a^2\right)\left(b^2+c^2-2bc-a^2\right)\)
\(=\left(b+c-a\right)\left(b+c+a\right)\left(b-c-a\right)\left(b-c+a\right)\)
a2 – b2 – 4a + 4
= a2 – 4a + 4 – b2
= (a – 2)2 – b2
= (a – 2 + b)(a – 2 – b)
= (a + b – 2)(a – b – 2)
Lời giải:
$x^2-y^2+a^2-b^2+2ax+2by=(x^2+a^2+2ax)-(y^2+b^2-2by)$
$=(x+a)^2-(y-b)^2=(x+a-y+b)(x+a+y-b)$
\(a^2-b^2-ac+bc=\left(a^2-b^2\right)-\left(ac-bc\right)=\left(a-b\right)\left(a+b\right)-c\left(a-b\right)=\left(a-b\right)\left(a+b-c\right)\)
ax - bx - a² + 2ab - b²
= (ax - bx) - (a² - 2ab + b²)
= x(a - b) - (a - b)²
= (a - b)(x - a + b)