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,(3x-1) mũ 2=9/16
<=> (3x-1)^2 = ( ±3/4)^2
<=> l3x-1l  = 3/4
Hoặc 3x-1 = 3/4
 <=> 3x= 3/4 + 1
<=> x = 7/4 : 3
<=> x= 7/1

Bài 1: 

a) Ta có: \(\left(15x^2\cdot y^2\cdot z\right):3xyz\)

\(=\dfrac{15x^2y^2z}{3xyz}\)

\(=5xy\)

b) Ta có: \(3x^2\cdot\left(5x^2-4x+3\right)\)

\(=3x^2\cdot5x^2-3x^2\cdot4x+3x^2\cdot3\)

\(=15x^4-12x^3+9x^2\)

c) Ta có: \(\left(2x^2-3x\right):\left(x-4\right)\)

\(=\dfrac{2x^2-8x+5x-20+20}{x-4}\)

\(=\dfrac{2x\left(x-4\right)+5\left(x-4\right)+20}{x-4}\)

\(=2x+5+\dfrac{20}{x-4}\)

d) Ta có: \(-5xy\cdot\left(3x^2y-5xy+y^2\right)\)

\(=-5xy\cdot3x^2y+5xy\cdot5xy-5xy\cdot y^2\)

\(=-15x^3y^2+25x^2y^2-5xy^3\)

1) x² + x²y - y - 1

= x²( 1 + y ) - ( 1 + y )

= ( 1 + y )( x² - 1 )

= ( 1 + y )( x - 1 )( x + 1 )

2) x² + y² - 2xy - 25

= ( x² - 2xy + y² ) - 25

= ( x - y )² - 5²

= ( x - y - 5 )( x - y + 5 )

3) ( 2x - 1 )( x² + 2x - 1 ) - ( 1 - 2x )( x - 3 )

= ( 2x - 1 )( x² + 2x - 1 ) + ( 2x - 1 )( x - 3 )

= ( 2x - 1 )( x² + 2x - 1 + x - 3 )

= ( 2x - 1 )( x² + 3x - 4 )

= ( 2x - 1 )( x² - x + 4x - 4 )

= ( 2x - 1 )[ x( x - 1 ) + 4( x - 1 ) ]

= ( 2x - 1 )( x - 1 )( x + 4 )

4) a² + x² - 16 + 2ax

= ( a² + 2ax + x² ) - 16

= ( a + x )² - 4²

= ( a + x - 4 )( a + x + 4 )

Trả lời:

a, 5x² + 10xy + 5y² = 5 ( x² + 2xy + y² ) = 5 ( x + y )² 

b, x² + 3x - y² + 3y = ( x² - y² ) + ( 3x + 3y ) = ( x - y )( x + y ) + 3 ( x + y ) = ( x + y )( x - y + 3 )

c, x² + 5x - y² + 5y = ( x² - y² ) + ( 5x + 5y ) = ( x - y )( x + y ) + 5 ( x + y ) = ( x + y )( x - y + 5 )

d, 3x² - 3y² - 2 ( x - y )² = 3 ( x² - y² ) - 2 ( x - y )² = 3 ( x - y )( x + y ) - 2 ( x - y )² = ( x - y )[ 3 ( x + y ) - 2 ] = ( x - y )( 3x + 3y - 2 )

e, x² - 2x - 4y² - 4y = ( x² - 4y² ) - ( 2x + 4y ) = ( x - 2y )( x + 2y ) - 2 ( x + 2y ) = ( x + 2y )( x - 2y - 2 )

a) 5x²+10xy+5y²

=5(x²+2xy+y²)

=5(x+y)²

b) x²+3x-y²+3y

=(x²-y²)+(3x+3y)

=(x-y)(x+y)+3(x+y)

=(x+y)(x-y+3)

c) x²+5x-y²+5y

=(x²-y²)+(5x+5y)

=(x-y)(x+y)+5(x+y)

=(x+y)(x-y+5)

d) 3x²-3y²-2(x-y)²

=3(x²-y²)-2(x-y)²

=3(x-y)(x+y)-2(x-y)²

=(x-y)[3(x+y)-2(x-y)]

e) x²-2x-4y²-4y

=(x²-4y²)-(2x+4y)

=(x-2y)(x+2y)-2(x+2y)

=(x+2y)(x-2y-2)

#H

Bài 1:

a) Ta có: \(2x=5y.\)

=> \(\frac{x}{y}=\frac{5}{2}\)

=> \(\frac{x}{5}=\frac{y}{2}\) và \(x.y=90.\)

Đặt \(\frac{x}{5}=\frac{y}{2}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)

Có: \(x.y=90\)

=> \(5k.2k=90\)

=> \(10k^2=90\)

=> \(k^2=90:10\)

=> \(k^2=9\)

=> \(k=\pm3.\)

TH1: \(k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.5=15\\y=3.2=6\end{matrix}\right.\)

TH2: \(k=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).5=-15\\y=\left(-3\right).2=-6\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(15;6\right),\left(-15;-6\right).\)

e) Ta có: \(\frac{x}{y}=\frac{4}{5}.\)

=> \(\frac{x}{4}=\frac{y}{5}\) và \(x.y=20.\)

Đặt \(\frac{x}{4}=\frac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)

Có: \(x.y=20\)

=> \(4k.5k=20\)

=> \(20k^2=20\)

=> \(k^2=20:20\)

=> \(k^2=1\)

=> \(k=\pm1.\)

TH1: \(k=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.4=4\\y=1.5=5\end{matrix}\right.\)

TH2: \(k=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).4=-4\\y=\left(-1\right).5=-5\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(4;5\right),\left(-4;-5\right).\)

Chúc bạn học tốt!