Ta có : 3x² - 7xy + 4y² = 0

=> 3x² - 3xy - 4xy + 4y² = 0

=> 3x( x - y) - 4y( x - y) = 0

=> ( x - y)( 3x - 4y) = 0

=> x = y ; 3x = 4y

Thay : x = y ; 3x = 4y vào phân thức trên ta có:

\(A=\dfrac{4y+2x}{5y-7x}+\dfrac{3x-2y}{10y-4x}\)

\(A=\dfrac{3x+2x}{5x-7x}+\dfrac{4y-2y}{10x-4x}\)

\(A=\dfrac{5x}{-2x}+\dfrac{2y}{6x}=\dfrac{5}{-2}+\dfrac{1}{3}=\dfrac{-13}{6}\)

a)\(\left\{{}\begin{matrix}8x+2y=4\\8x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\4x+1=2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1\\x=\frac{1}{4}\end{matrix}\right.\)b)

\(\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=35\\4x-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\4x-5.5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)c)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)

các câu khác tương tự

a) \(xy+3x-7y-21\)
\(\Leftrightarrow\left(xy+3x\right)-\left(7y+21\right)\)
\(\Leftrightarrow x\left(y+3\right)-7\left(y+3\right)\)
\(\Leftrightarrow\left(x-7\right)\left(y+3\right)\)

b) \(2xy-15-6x+5y\)
\(\Leftrightarrow\left(2xy-6x\right)-\left(15-5y\right)\)
\(\Leftrightarrow x\left(2y-6\right)-5\left(3-y\right)\)
\(\Leftrightarrow2x\left(y-3\right)+5\left(y-3\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(y-3\right)\)

2:

a: \(=\left(2x^2-xy\right)+\left(2xz-yz\right)\)

\(=x\left(2x-y\right)+z\left(x-2y\right)=\left(x-2y\right)\left(x+z\right)\)

b: \(=\left(x^2-4y^2\right)-\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x+2y\right)-\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x+2y-1\right)\)

c: \(=\left(y^2+10y+25\right)-9z^2\)

\(=\left(y+5\right)^2-\left(3z\right)^2\)

\(=\left(y+5+3z\right)\left(y+5-3z\right)\)

d: \(=\left(x+2y\right)^3-\left(x-2y\right)\left(x+2y\right)\)

\(=\left(x+2y\right)\left[\left(x+2y\right)^2-\left(x-2y\right)\right]\)

\(=\left(x+2y\right)\left(x^2+4xy+4y^2-x+2y\right)\)

1:

a: \(x\left(3-4x\right)+5\left(3-4x\right)=\left(3-4x\right)\left(x+5\right)\)

b: \(2y\left(5y-6\right)-4\left(6-5y\right)\)

\(=2y\left(5y-6\right)+4\left(5y-6\right)\)

\(=2\left(5y-6\right)\left(y+2\right)\)

c: \(=27\left(x-2\right)^3-3x\left(x-2\right)^2\)

\(=3\left(x-2\right)^2\cdot\left[9\left(x-2\right)-x\right]\)

\(=3\left(x-2\right)^2\left(8x-18\right)=6\left(x-2\right)^2\cdot\left(4x-9\right)\)

d: \(=6y\left(x-y\right)\left(x+y\right)-8y\left(x+y\right)^2\)

\(=2y\left(x+y\right)\left[3\left(x-y\right)-4\left(x+y\right)\right]\)

\(=2y\left(x+y\right)\left(3x-3y-4x-4y\right)\)

\(=2y\left(x+y\right)\left(-x-7y\right)\)

Bài 1

a) x(3 - 4x) + 5(3 - 4x)

= (3 - 4x)(x + 5)

b) 2y(5y - 6) - 4(6- 5y)

= 2y(5y - 6) + 4(5y - 6)

= (5y - 6)(2y + 4)

= 2(5y - 6)(y + 2)

c) 27(x - 2)³ - 3x(2 - x)²

= 27(x - 2)³ - 3x(x - 2)²

= 3(x - 2)²[9(x - 2) - x]

= 3(x - 2)²(9x - 18 - x)

= 3(x - 2)²(8x - 18)

= 6(x - 2)²(4x - 9)

d) 6y(x² - y²) - 8y(x + y)²

= 6y(x - y)(x + y) - 8y(x + y)²

= 2y(x + y)[3(x - y) - 4(x + y)]

= 2y(x + y)(3x - 3y - 4x - 4y)

= 2y(x + y)(-x - 7y)

= -2y(x + y)(x + 7y)

a. \(=-4x^5y^3+4x^5y^3-3x^4y^3+x^4y^3-6xy^2\)

\(=0-2x^4y^3-6xy^2\)

\(=-2x^4y^3-6xy^2\)

Bậc của đa thức là 5

À bậc của đa thức là 4 nha

Mik gõ nhầm 

Xl bạn

a)\(-4x^5y^3-3x^4y^3+x^4y^3-6xy^2+4x^5y^3.\)

\(=-2x^4y^3-6xy^2\)

Suy ra đa thức có bậc là 7.

b)\(2x^4-4y^5-3x^2y^3z^2+2yz^3+x^2y^3z^2.\)

\(=2x^4-4y^5-2x^2y^3z^2+2yz^3.\)

Suy ra đa thức có bậc là 7.

a: \(A=-4x^5y^3-2x^2y^3z^2-2y^4\)

b: \(B=-4x^5y^3-2x^2y^3z^2-2y^4+2x^2y^3z^2-\dfrac{2}{3}y^4+\dfrac{1}{5}x^4y^3=-4x^5y^3+\dfrac{1}{5}x^4y^3-\dfrac{8}{3}y^4\)

a, \(A=-4x^5y^3+x^4y^3-3x^2y^3z^2+4x^5y^3-x^4y^3+x^2y^3z^2-2y^4\)

\(=2x^2y^3z^2-2y^4\)

Bậc của đa thức A là 7

Vậy...

b, Ta có: \(B-2x^2y^3z^2+\dfrac{2}{3}y^4-\dfrac{1}{5}x^4y^3=A\)

\(\Rightarrow B-2x^2y^3z^2+\dfrac{2}{3}y^4-\dfrac{1}{5}x^4y^3=2x^2y^3z^2-2y^4\)

\(\Rightarrow B=2x^2y^3z^2-2y^4+2x^2y^3z^2-\dfrac{2}{3}y^4+\dfrac{1}{5}x^4y^3\)

\(=4x^2y^3z^2-\dfrac{8}{3}y^4+\dfrac{1}{5}x^4y^3\)

Vậy...