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(3x-4-x-1)(3x-4+x+1)=0
(2x-5)(4x-3)=0
2x-5 = 0 hoặc 4x-3=0
2x=5 hoặc 4x=3
x=5/2 hoặc x=3/4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Bài.1:\\ a,3x-9y=3\left(x-3y\right)\\ b,x^2-5x=x\left(x-5\right)\\ c,\left(x-3\right)\left(x-5\right)-\left(2x+1\right)\left(3-x\right)=\left(x-3\right)\left(x-5\right)+\left(x-3\right)\left(2x+1\right)\\ =\left(x-3\right)\left(x-5+2x+1\right)=\left(x-3\right)\left(3x-4\right)\\ d,3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\\ e,3\left(x+5\right)-x^2-5x=3\left(x+5\right)-x\left(x+5\right)\\ =\left(x+5\right)\left(3-x\right)\)
\(Bài.2:\\ a,x^3-9x=0\\ \Leftrightarrow x.\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=3\end{matrix}\right.\\ b,5x\left(x+2\right)-3\left(x+2\right)=0\\ \Leftrightarrow\left(5x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-2\end{matrix}\right.\\ c,x^2-7x=0\\ \Leftrightarrow x\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:
\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)
2) Bạn xem lại đề!
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Mình sẽ làm theo đề bài của mình nếu đúng thì ... nha
Biến đổi vế phải ta có :
( x + y) [ ( x - y)^2 + xy ] = ( x + y)( x^2 - 2xy + y^2 + xy)
= ( x+ y)( x^2 - xy+ y^2)
= x^3 + y^3
VẬy VT = VP đẳng thức được CM
![](https://rs.olm.vn/images/avt/0.png?1311)
bạn đăng tách ra nhé
Bài 3 :
Ta có :\(1+\dfrac{1}{2+x}=\dfrac{12}{x^3+8}\)
đk : x khác -2
\(\Rightarrow x^3+8+x^2-2x+4=12\Leftrightarrow x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(x^2+x-2\right)=0\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\Leftrightarrow x=0;x=1;x=-2\left(ktm\right)\)
Bài 2:
a,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)
\(\dfrac{1}{x}+\dfrac{2}{x-2}=0\\ \Leftrightarrow\dfrac{x-2}{x\left(x-2\right)}+\dfrac{2x}{x\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x-2+2x}{x\left(x-2\right)}=0\\ \Rightarrow3x-2=0\\ \Leftrightarrow x=\dfrac{2}{3}\left(tm\right)\)
b, ĐKXĐ:\(x\ne\pm2\)
\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2-5x}{x^2-4}\\ \Leftrightarrow\dfrac{x^2-3x+2-x^2-2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2-5x}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{-5x+2-2+5x}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow0=0\left(tm\right)\)
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\(\dfrac{4x+2}{4x-2}+\dfrac{3-6x}{6x-6}\left(dkxd:x\ne\dfrac{1}{2};x\ne1\right)\)
\(=\dfrac{2\left(2x+1\right)}{2\left(2x-1\right)}+\dfrac{3\left(1-2x\right)}{6\left(x-1\right)}\)
\(=\dfrac{2x+1}{2x-1}+\dfrac{1-2x}{2\left(x-1\right)}\)
\(=\dfrac{2x+1}{2x-1}+\dfrac{1-2x}{2x-2}\)
\(=\dfrac{\left(2x+1\right)\left(2x-2\right)}{\left(2x-1\right)\left(2x-2\right)}+\dfrac{\left(1-2x\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x-2\right)}\)
\(=\dfrac{4x^2-2x-2}{\left(2x-1\right)\left(2x-2\right)}+\dfrac{-4x^2+4x-1}{\left(2x-1\right)\left(2x-2\right)}\)
\(=\dfrac{4x^2-2x-2-4x^2+4x-1}{\left(2x-1\right)\left(2x-2\right)}\)
\(=\dfrac{2x-3}{\left(2x-1\right)\left(2x-2\right)}\)
\(=\dfrac{2x-3}{4x^2-6x+2}\)
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