Tìm x:
\(5x\left(x-1\right)=x-1\)
\(5x\left(x-1\right)-\left(x-1\right)=0\)
\(\left(5x-1\right)\left(x-1\right)=0\)
\(\Rightarrow\)\(\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\)\(\Rightarrow\)\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)
Vậy x=\(\dfrac{1}{5}\)hoặc x=1
\(2\left(x+5\right)-x^2-5x=0\)
\(2\left(x+5\right)-x\left(x+5\right)=0\)
\(\left(2-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy...
A)\(x^2+5x-6=x^2-x+6x-6\\ =\left(x-1\right)\left(x+6\right)\)
B)\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)\\ =\left(x+y\right)\left(5x-1\right)\)
C)\(7x-6x^2-2=-6x^2+3x+4x-2\\ =-3x\left(2x-1\right)+2\left(2x-1\right)=\left(2x-1\right)\left(2-3x\right)\)
D)\(x^2+4x+3=x^2+x+3x+3=\left(x+1\right)\left(x+3\right)\)
E)\(2x+3x-5=5x-5=5\left(x-1\right)\)
F)\(16x-5x^3=x\left(16-5x^2\right)\)
Lời giải:
Từ \(a+b+c=0\) ta có:
\(\left\{{}\begin{matrix}a+b+c=0\\\left(a+b+c\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\end{matrix}\right.\)
Nên \(a^3+b^3+c^3=\left(a+b+c\right)^3+3abc=\left(a+b+c\right)\left(a^2+b^2+c^2+ab+bc+ac\right)+3abc=0+3abc=3abc\)Nên \(a^3+b^3+c^3=3abc\)
Phân tích đa thức thành nhân tử
a)\(x^2+5x-6\)
\(=x^2+6x-x-6\)
\(=x\left(x+6\right)-\left(x+6\right)\)
\(=\left(x-1\right)\left(x+6\right)\)
b)\(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(5x-1\right)\left(x+y\right)\)
c)\(7x-6x^2-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(2-3x\right)\)
d)\(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
e)\(2x^2+3x-5\)
\(=2x^2-2x+5x-5\)
\(=2x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(2x+5\right)\)
Còn bài f) không ra, chúc bạn học tốt!^^
2.
a. \(5x\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-x+1=0\Leftrightarrow5\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=\dfrac{1}{5}\)
b. \(2\left(x+5\right)-x^{^2}-5x=0\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy x=-5 hoặc x=2
c. thay \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\) ta có:
\(a^{^3}+b^{^3}+c^{^3}=3abc^{^3}\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left[\left(a+b\right)^3+c^{^3}\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^{^2}-c\left(a+b\right)+c^{^2}\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^{^2}+2ab+b^{^2}-ac-bc+c^{^2}-3ab\right)...\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^{^2}+b^{^2}+c^{^2}-ab-ac-bc\right)=0\) luôn đúng vì a+b+c=0
\(\Rightarrow a^{^3}+b^{^3}+c^{^3}=3abc\left(đpcm\right)\)
