2x+1 . 3y = 12x
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a ) ( 12x - 5 )( 4x - 1 ) + 3x( 1 - 16x )
= ( 12x - 5 )( 4x - 1 ) - 3x( 16x - 1 )
= ( 12x - 5 )( 4x - 1 ) - 3x( 4x + 1 )( 4x - 1 )
= ( 4x - 1 ) [ 12x - 5 - 3x( 4x + 1 ) ]
= ( 4x - 1 ) ( 12x - 5 - 12x2 - 3x )
= ( 4x - 1 )( 12x2 - 9x + 5 )
b: \(x^2-6x+xy-6y\)
\(=x\left(x-6\right)+y\left(x-6\right)\)
\(=\left(x-6\right)\left(x+y\right)\)
c: \(2x^2+2xy-x-y\)
\(=2x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(2x-1\right)\)
e: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
\(a,=\dfrac{2y^4}{3x\left(2x-3y\right)}\\ b,=-\dfrac{2y\left(3x-1\right)^2}{3x^2}\\ c,=\dfrac{5\left(4x^2-9\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)\left(2x+3\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)}{2x+3}\\ d,=\dfrac{5x\left(x-2y\right)}{-2\left(x-2y\right)^3}=-\dfrac{5x}{2\left(x-2y\right)^2}\)
a) P = \(x^2+3x+y^2-3y-2xy+90\)
= \(\left(x-y\right)^2+3\left(x-y\right)+90\)
= \(5^2+3.5+90=130\)
b) P = \(4x^2+9y^2-12xy-12x+24xy-18y+118\)
= \(4x^2+9y^2+12xy-12x-18y+118\)
= \(\left(2x+3y\right)^2-6\left(2x+3y\right)+118\)
= \(\left(-7\right)^2-6.\left(-7\right)+118=209\)
ĐKXĐ: x ≠ 0 và x ≠ \(\dfrac{1}{2}\); y ≠ 0
Rút gọn:
\(\dfrac{10xy^2\left(2x-1\right)^3}{12x^3y\left(1-2x\right)}\)
= \(\dfrac{5y\left(2x-1\right)^3}{6x^2\left(1-2x\right)}\)
= \(\dfrac{\left(2x-1\right)^35y}{-\left(2x-1\right)6x^2}\)
= \(\dfrac{\left(2x-1\right)^2.5y}{-6x^2}\)
12x2 + 3 - 12x - 3y2
= 3( 4x2 + 1 - 4x - y2 )
= 3[ ( 4x2 - 4x + 1 ) - y2 ]
= 3[ ( 2x - 1 ) - y2 ]
= 3( 2x - y - 1 )( 2x + y - 1 )
Bài 2:
a: Ta có: \(2^{x+1}\cdot3^y=12^x\)
\(\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
\(2^{x+1}.3^y=12^x\Leftrightarrow2^{x+1}.3^y=\left(2^2.3\right)^x\)
\(\Leftrightarrow2^{x+1}.3^y=2^{2x}.3^x\)(1)
Vì 2 và 3 là các số nguyên tố nên (1) \(\Leftrightarrow\hept{\begin{cases}x+1=2x\\y=x\end{cases}\Leftrightarrow x=y=1}\)