1) Phân tích đa thức thành nhân tử bằng phương pháp dùng hằng đẳng thức:
a)x\(^4\)+y\(^2\)-2x\(^2\)y
b)(2a+b)\(^2\)-(2b+a)\(^2\)
c)(x\(^2\)+1)\(^2\)-4x\(^2\)
d)a\(^3\)+b\(^3\)+c\(^3\)-3abc
2)Chứng minh:
a)(7n-2)\(^2\)-(2n-7)\(^2\)\(⋮\)7 với n \(\in\)Z
b)n\(^3\)-n\(⋮\)6 với n\(\in\)Z
3) Tìm x pk
a)4x\(^3\)-36x=0
b)x\(^2\)-x+\(\frac{1}{4}\)=0
c)x\(^3\)-0,25x=0
d)x\(^2\)-10x=-25
Dạng 1:
a) \(x^4+y^2-2x^2y=\left(x^2-y\right)^2\)
b) \(\left(2a+b\right)^2-\left(2b+a\right)^2\)
\(=\left(2a+b-2b-a\right)\left(2a+b+2b+a\right)\)
\(=\left(a-b\right)\left(3a+3b\right)\)
\(=3\left(a-b\right)\left(a+b\right)\)
c) \(\left(x^2+1\right)^2-4x^2\)
\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)^2\)
d) \(a^3+b^3+c^3-3abc\)
\(=a^3+3a^2b+3ab^2+b^3+c^3-3abc-3a^2b-3ab^2\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ca-bc-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Dạng 2:
a) \(\left(7n-2\right)^2-\left(2n-7\right)^2\)
\(=\left(7n-2-2n+7\right)\left(7n-2+2n-7\right)\)
\(=\left(5n+5\right)\left(9n-9\right)\)
\(=45\cdot\left(n+1\right)\cdot\left(n-1\right)⋮3;5;9\) chứ không chia hết cho 7
Bạn xem lại đề.
b) \(n^3-n=n\left(n^2-1\right)=n\left(n-1\right)\left(n+1\right)\)
Vì \(n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên tích đó chia hết cho 2 và 3.
Mặt khác \(\left(2;3\right)=1\)
Do đó \(n\left(n-1\right)\left(n+1\right)⋮2.3=6\) ( đpcm
Dạng 3:
a) \(4x^3-36x=0\)
\(\Leftrightarrow4x\left(x^2-9\right)=0\)
\(\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy...
b) \(x^2-x+\frac{1}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy...
c) \(x^3-0,25x=0\)
\(\Leftrightarrow x\left(x^2-0,25\right)=0\)
\(\Leftrightarrow x\left(x-0,5\right)\left(x+0,5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\)
Vậy...
d) \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x=5\)
Vậy...
3) Tìm x :
a) \(4x^3-36x=0\)
\(\Leftrightarrow4.\left(x^3-9x\right)=0\)
\(\Leftrightarrow x.\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9=3^2=\left(-3\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy : ....
b) \(x^2-x+\frac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow=\frac{1}{2}\)
Vậy : ....
