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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 6:
c: \(9x^2+6x+1=\left(3x+1\right)^2\)
d: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)
e: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2:
a: \(9x^2-1=\left(3x\right)^2-1=\left(3x-1\right)\left(3x+1\right)\)
b: \(2\left(x-1\right)+x^2-x\)
\(=2\left(x-1\right)+x\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
c: \(3x^2+14x-5\)
\(=3x^2+15x-x-5\)
\(=3x\left(x+5\right)-\left(x+5\right)=\left(x+5\right)\left(3x-1\right)\)
3:
a: \(2x\left(x-1\right)-2x^2=4\)
=>\(2x^2-2x-2x^2=4\)
=>-2x=4
=>x=-2
b: \(x\left(x-3\right)-\left(x+2\right)\left(x-1\right)=5\)
=>\(x^2-3x-\left(x^2+x-2\right)=5\)
=>\(x^2-3x-x^2-x+2=5\)
=>-4x=3
=>x=-3/4
c: \(4x^2-25+\left(2x+5\right)^2=0\)
=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)
=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)
=>4x(2x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+4}{x-3}\)
b.
\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)
\(\Rightarrow x=10\) (thỏa mãn)
2.
\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
a: \(\Leftrightarrow3x^2+2x-1=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
b: \(\Leftrightarrow x^3-4x-x^3-8=4\)
hay x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+1\right)^2-2\left(2x+1\right)\left(3-x\right)+\left(x-3\right)^2\)
\(=\left(2x+1\right)^2+2\left(2x-1\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(2x+1+x-3\right)^2\)
\(=\left(3x-2\right)^2\)
------------------------------------
\(a^3+3a^2-6a-8\)
\(=a^3+4a^2-a^2-4a-2a-8\)
\(=\left(a^3+4a^2\right)-\left(a^2+4a\right)-\left(2a+8\right)\)
\(=a^2\left(a+4\right)-a\left(a+4\right)-2\left(a+4\right)\)
\(=\left(a+4\right)\left(a^2-a-2\right)\)
\(=\left(a+4\right)\left(a^2-2a+a-2\right)\)
\(=\left(a+4\right)\left[\left(a^2-2a\right)+\left(a-2\right)\right]\)
\(=\left(a+4\right)\left[a\left(a-2\right)+\left(a-2\right)\right]\)
\(=\left(a+4\right)\left(a-2\right)\left(a+1\right)\)
---------------------------------
\(2x^2-5x+2\)
\(=2x^2-4x-x+2\)
\(=\left(2x^2-4x\right)-\left(x-2\right)\)
\(=2x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(2x-1\right)\)
-----------------------------------------
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x-4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y-2\right)\)
-------------------------------------
\(a^2-1+4b-4b^2\)
\(=a^2-\left(1-4b+4b^2\right)\)
\(=a^2-\left(1-2b\right)^2\)
\(=\left(a-1+2b\right)\left(a+1-2b\right)\)
----------------------------------------
\(a^4+6a^2b+9b^2-1\)
\(=\left(a^4+6a^2b+9b^2\right)-1\)
\(=\left(a^2+3b\right)^2-1\)
\(=\left(a^2+3b-1\right)\left(a^2+3b+1\right)\)
---------------------------------
\(2x^3+16y^3\)
\(=2\left(x^3+8y^3\right)\)
\(=2\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
Lần sau ghi đề tách riêng từng câu ra nhé em. Ghi dính chùm vậy khó nhìn lắm. Sẽ ít ai giải cho em
1: \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)
\(=\left(2x+3-2x-5\right)^2\)
=4