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\(a,\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\\ b,\Leftrightarrow\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\\ \Leftrightarrow\left(2x+5\right)\left(2x-5+2x+5\right)=0\\ \Leftrightarrow4x\left(2x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
a) \(P=\dfrac{5x-2-3\left(x-2\right)+x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{x-2}\)
b) \(\left|x+3\right|=5\)\(\Rightarrow\left[{}\begin{matrix}x+3=5\\x+3=-5\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-8\left(tm\right)\end{matrix}\right.\)
\(P=\dfrac{x+2}{x-2}=\dfrac{-8+2}{-8-2}=\dfrac{3}{5}\)
c) \(P=\dfrac{x+2}{x-2}=\dfrac{x-2+4}{x-2}=1+\dfrac{4}{x-2}\in Z\)
\(\Rightarrow\left(x-2\right)\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
Kết hợp ĐKXĐ:
\(\Rightarrow x\in\left\{0;1;3;4;6\right\}\)
a: Xét tứ giác AKCI có
AK//CI
AI//KC
Do đó: AKCI là hình bình hành
\(a,=2x^3-2x-2x^3=-2x\\ b,=x^2-6x+9-x^2+4x+5=-2x+14\\ c,=8x^3+12x^2+6x+1-8x^3-12x^2+2x+3=8x+4\)
2.
a. x3 - 4x2 + 4x
= x(x2 - 4x + 4)
= x(x - 2)2
c. 3x2 + x - 4
= 3x2 + 4x - 3x - 4
= x(3x + 4) - (3x + 4)
= (x - 1)(3x + 4)
a) \(x^3-4x^2+4x=x\left(x^2-4x+4\right)=x\left(x-2\right)^2\)
b) \(x^2-y^2-6y-9=x^2-\left(y^2+6y+9\right)=x^2-\left(y+3\right)^2=\left(x-y-3\right)\left(x+y+3\right)\)
c) \(3x^2+x-4=3x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(3x+4\right)\)