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a: =>3M+2x^4y^4=x^4y^4
=>3M=-x^4y^4
=>M=-1/3*x^4y^4
b: x^2-2M=3x^2
=>2M=-2x^2
=>M=-x^2
c: =>M=-x^2y^3-3x^2y^3=-4x^2y^3
d: =>M=7x^2y^2-3x^2y^2=4x^2y^2
\(1)A=2x\left(x-y\right)-y\left(y-2x\right)\)
\(=2x^2-2xy-y^2+2xy\)
\(=2x^2-y^2=2.\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{1}{3}\right)^2\)
\(=\dfrac{8}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
\(2)B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(=5x^2-20xy-4y^2+20xy\)
\(=5x^2-4y^2=5.\left(-\dfrac{1}{5}\right)^2-4.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
\(3)C=\text{x.(x^2-y^2)-x^2(x+y)+y(x^2-x)}\)
\(=x^3-xy^2-x^3-x^2y+x^2y-xy\)
\(=-xy\left(x+1\right)\)
a: =-2/5x^5y^7
Hệ số: -2/5
bậc: 12
b: =3/4*x^2y^3*12/5x^4=9/5x^6y^3
Hệ số: 9/5
bậc: 9
c: =4/9x^6y^6
hệ số: 4/9
bậc: 12
d: =2/5x^6y^6
hệ số: 2/5
bậc: 12
`a, A= 4xy -xy-2xy`
`= (4-1-2)xy`
`= xy`
Thay `x=2;y=3`
Ta có : `xy=2*3=6`
`b, B= x^2 y -7x^2y-4x^2y`
`=(1-7-4)x^2y`
`= -10x^2y`
Thay `x=2;y=3`
Ta có : `-10x^2y=-10*2^2 *3= -10*4*3=-40*3=-120`
`c, C=10x^2y -x^2y-7x^2y`
`=(10-1-7)x^2y`
`= 2x^2y`
Thay `x=2;y=3`
Ta có : `2x^2y=2*2^2 *3= 2*4*3=8*3=24`
`d,D=5x^2y^2-12x^2y^2+8x^2y^2`
`= (5-12+8)x^2y^2`
`=x^2y^2`
Thay `x=2;y=3`
ta có : `x^2y^2=2^2 *3^2= 4* 9=36`
có chỗ nào bn đọc ko rõ thì ns mik nha, để mik gõ ra cho bn rõ hơn
3x2+4y2=7xy
<=> 3x2-3xy+4y2-4xy=0
<=> 3x(x-y)-4y(x-y)=0
<=> (3x-4y)(x-y)=0
<=> 3x-4y=0 hoặc x-y=0
<=> 3x=4y hoặc x=y
<=> y = \(\frac{3}{4}\)x hoặc x=y
+) y = \(\frac{3}{4}\)x, ta có:
F = \(\frac{4.\frac{3}{4}x+2x}{5.\frac{3}{4}x-7x}+\)\(\frac{3x-2.\frac{3}{4}x}{10.\frac{3}{4}x-4x}\)
F = \(\frac{5x}{-\frac{13}{4}x}+\frac{\frac{3}{2}x}{\frac{7}{2}x}\)
F = \(-\frac{20}{13}+\frac{3}{7}=-\frac{101}{91}\)
+) x = y, ta có:
F = \(\frac{4x+2x}{5x-7x}+\frac{3x-2x}{10x-4x}\)
F = \(\frac{6x}{-2x}+\frac{1x}{6x}=-3+\frac{1}{6}=-\frac{17}{6}\)
Từ \(3x^2+4y^2=7xy\Rightarrow3x^2+4y^2-7xy=0\)
\(\Rightarrow3x^2-4xy-3xy+4y^2=0\)
\(\Rightarrow x\left(3x-4y\right)-y\left(3x-4y\right)=0\)
\(\Rightarrow\left(x-y\right)\left(3x-4y\right)=0\)\(\Rightarrow\left[\begin{matrix}x=y\\x=\frac{4y}{3}\end{matrix}\right.\)
*)Xét \(x=y\) ta có \(F=\frac{4y+2y}{5y-7y}+\frac{3y-2y}{10y-4y}=\frac{6y}{-2y}+\frac{y}{6y}=-3+\frac{1}{6}=-\frac{17}{6}\)
*)Xét \(x=\frac{4y}{3}\) ta có \(F=\frac{4y+2\cdot\frac{4y}{3}}{5y-7\cdot\frac{4y}{3}}+\frac{3\cdot\frac{4y}{3}-2y}{10y-4\cdot\frac{4y}{3}}=\frac{4y+\frac{8y}{3}}{5y-\frac{28y}{3}}+\frac{4y-2y}{10y-\frac{16y}{3}}=\frac{-20}{13}+\frac{3}{7}=\frac{-101}{91}\)
( 7x + 2y )2 + ( 7x - 2y )2 - 2( 49x2 - 4y2 )
= ( 7x + 2y )2 + ( 7x - 2y )2 - 2( 7x - 2y )( 7x + 2y )
= [ ( 7x + 2y ) - ( 7x - 2y ) ]2
= ( 7x + 2y - 7x + 2y )2
= ( 4y )2 = 16y2
\(=\left(7x+2y\right)^2+\left(7x-2y\right)^2-2\left(7x-2y\right)\left(7x+2y\right)\)
\(=\left(7x+2y-7x+2y\right)^2\)
\(=\left(4y\right)^2=16y^2\)