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\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)
1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)
2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)=\left(3x-2y\right)\left(3x+2y+2\right)\)
3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)
1) Ta có: \(x^3+2x^2-6x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)+2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
2: Ta có: \(9x^2+6x-4y^2-4y\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(3x+2y+2\right)\)
a: \(x^2+12x+36=0\)
=>\(x^2+2\cdot x\cdot6+6^2=0\)
=>\(\left(x+6\right)^2=0\)
=>x+6=0
=>x=-6
b: \(4x^2-4x+1=0\)
=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)
=>\(\left(2x-1\right)^2=0\)
=>2x-1=0
=>2x=1
=>x=1/2
c: \(x^3+6x^2+12x+8=0\)
=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)
=>\(\left(x+2\right)^3=0\)
=>x+2=0
=>x=-2
a) \(x^2-xz-9y^2+3yz\)
\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
c) \(x^3+2x^2-6x-27\)
\(=\left(x^3-27\right)+\left(2x^2-6x\right)\)
\(=\left(x-3\right)\left(x^2-3x+9\right)+2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-3x+9+2x\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
Bài 1.
\(a\Big) 9(4x+3)^2=16(3x-5)^2\\\Leftrightarrow 9[(4x)^2+2\cdot 4x\cdot3+3^2]=16[(3x)^2-2\cdot3x\cdot5+5^2]\\\Leftrightarrow9(16x^2+24x+9)=16(9x^2-30x+25)\\\Leftrightarrow 144x^2+216x+81=144x^2-480x+400\\\Leftrightarrow (144x^2-144x^2)+(216x+480x)=400-81\\\Leftrightarrow 696x=319\\\Leftrightarrow x=\dfrac{11}{24}\\Vậy:x=\dfrac{11}{24}\\---\)
\(b\Big)(x-3)^2=4x^2-20x+25\\\Leftrightarrow(x-3)^2=(2x)^2-2\cdot2x\cdot5+5^2\\\Leftrightarrow(x-3)^2=(2x-5)^2\\\Leftrightarrow (x-3)^2-(2x-5)^2=0\\\Leftrightarrow (x-3-2x+5)(x-3+2x-5)=0\\\Leftrightarrow (-x+2)(3x-8)=0\\\Leftrightarrow \left[\begin{array}{} -x+2=0\\ 3x-8=0 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} -x=-2\\ 3x=8 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} x=2\\ x=\dfrac{8}{3} \end{array} \right.\\Vậy:...\)
a) \(4x^2+20x+25=\left(2x+5\right)^2\)
b) \(x^2-6x+9=\left(x-3\right)^2\)
c) \(4x^2+12x+9=\left(2x+3\right)^2\)