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(x2 - x + 1)2 - 5x(x2 - x + 1) + 4x2
Đặt x2 - x + 1 = a
<=> a2 - 5xa + 4x2 = x2 - 4xa - xa + 4x2
= a(a - 4x) - x(a - 4x) = (a - x)(a - 4x)
= (x2 - x + 1 - x)(x2 - x + 1 - 4x)
= (x2 - 2x + 1)(x2 - 5x + 1) = (x - 1)2(x2 - 5x + 1)
Đặt x2 - x + 1 = y
đthức <=> y2 - 5xy + 4x2
= y2 - xy - 4xy + 4x2
= y( y - x ) - 4x( y - x )
= ( y - x )( y - 4x )
= ( x2 - x + 1 - x )( x2 - x + 1 - 4x )
= ( x2 - 2x + 1 )( x2 - 5x + 1 )
= ( x - 1 )2( x2 - 5x + 1 )
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
ta có : \(\frac{4a-3b}{a}=\frac{4bk-3b}{bk}=\frac{b\left(4k-3\right)}{bk}=\frac{4k-3}{k}\)
\(\frac{4c-3d}{c}=\frac{4dk-3d}{dk}=\frac{d\left(4k-3\right)}{dk}=\frac{4k-3}{k}\)
\(\Rightarrow\frac{4a-3b}{a}=\frac{4c-3d}{c}\)
\(\left(a^2+4b^2-5\right)^2-16\left(ab+1\right)^2\)
\(=\left(a^2+4b^2-5\right)^2-4^2\left(ab+1\right)^2\)
\(=\left(a^2+4b^2-5\right)^2-\left[4\left(ab+1\right)\right]^2\)
\(=\left(a^2+4b^2-5\right)^2-\left[4ab+4\right]^2\)
\(=\left(a^2+4b^2-5-4ab-4\right)\left(a^2+4b^2-5+4ab+4\right)\)
\(=\left(a^2+4b^2-4ab-9\right)\left(a^2+4b^2+4ab-1\right)\)
\(\left(a^2+4b^2-5\right)^2-16\left(ab+1\right)^2\)
= \(\left(a^2+4b^2-5\right)^2-\left[4\left(ab+1\right)\right]^2\)
= \(\left(a^2+4b^2-5\right)^2-\left(4ab+4\right)^2\)
= \(\left(a^2+4b^2-5-4ab-4\right)\left(a^2+4b^2-5+4ab+4\right)\)
= \(\left(a^2+4b^2-4ab-9\right)\left(a^2+4b^2+4ab-1\right)\)
= \(\left[\left(a-2b\right)^2-3^2\right]\left[\left(a+2b\right)^2-1^2\right]\)
= \(\left(a-2b-3\right)\left(a-2b+3\right)\left(a+2b-1\right)\left(a+2b+1\right)\)
toán 8 mà bạn
chọn đại thôi.he he..........