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a) M=\(\frac{-1}{9}\)x4y3(2xy2)2=\(\frac{-1}{9}\)x4y3(4x2y4)=\(\frac{-1}{9}\)x6y7
b) y=\(\frac{-x}{3}\)=> x=-3y
mà x+y=2
=>-3y+y=2 <=> -2y=2 => y=-1 => x=-3y=-3*-1=3
Thay x=3; y=-1 vào M...=>M=\(\frac{-1}{9}\)(36)(-17)=81
Bài 1 :
A + B = 4x2 - 5xy + 3y2 + 3x2 + 2xy - y2
= ( 4x2 + 3x2 ) - ( 5xy - 2xy ) + ( 3y2 - y2 )
= 7x2 - 3xy + 2y2
A - B = 4x2 - 5xy + 3y2 - ( 3x2 + 2xy - y2 )
= 4x2 - 5xy + 3y2 - 3x2 - 2xy + y2
= ( 4x2 - 3x2 ) - ( 5xy + 2xy ) + ( 3y2 + y2 )
= x2 - 7xy + 4y2
Bài 2 :
a) M + (5x2 - 2xy) = 6x2 + 9xy - y2
M = 6x2 + 9xy - y2 - (5x2 - 2xy)
M = 6x2 + 9xy - y2 - 5x2 + 2xy
M = ( 6x2 - 5x2 ) + ( 9xy + 2xy ) - y2
M = x2 + 11xy - y2
Vậy M = x2 + 11xy - y2
b) (3xy - 4y2) - N = x2 - 7xy + 8y2
N = 3xy - 4y2 - x2 - 7xy + 8y2
N = ( 3xy - 7xy ) - ( 4y2 - 8y2 ) - x2
N = -4xy + 4y2 - x2
Vậy N = -4xy + 4y2 - x2
3, Cho đa thức
A(x)+B(x) = (3x4-\(\dfrac{3}{4}\)x3+2x2-3)+(8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\))
= 3x4-\(\dfrac{3}{4}\)x3+2x2-3+8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\)
= (3x4+8x4)+(-3/4x3+1/5x3)+(-3+2/5)+2x2-9x
= 11x4 -0.55x3-2.6+2x2-9x
A(x)-B(x)=(3x4-\(\dfrac{3}{4}\)x3+2x2-3)-(8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\))
= 3x4-\(\dfrac{3}{4}\)x3+2x2-3-8x4-\(\dfrac{1}{5}\)x3+9x-\(\dfrac{2}{5}\)
= (3x4-8x4)+(-3/4x3-1/5x3)+(-3-2/5)+2x2+9x
= -5x4-0.95x3-3.4+2x2+9x
B(x)-A(x)=(8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\))-(3x4-\(\dfrac{3}{4}\)x3+2x2-3)
=8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\)-3x4+\(\dfrac{3}{4}\)x3-2x2+3
=(8x4-3x4)+(1/5x3+3/4x3)+(2/5+3)-9x-2x2
= 5x4+0.95x3+2.6-9x-2x2
a) A + B
\(=4x^5-7y^2+2xy-x-5y-\frac{1}{4}+6x^5-2y^2+3x-\frac{1}{6}y+6\)
\(=\left(4x^5+6x^5\right)-\left(7y^2+2y^2\right)+2xy+\left(3x-x\right)-\left(5y+\frac{1}{6}y\right)+\left(6-\frac{1}{4}\right)\)
\(=10x^5-9y^2+2xy+2x-\frac{31}{6}y+\frac{23}{4}\)
A - B
\(=4x^5-7y^2+2xy-x-5y-\frac{1}{4}-6x^5+2y^2-3x+\frac{1}{6}y-6\)
\(=\left(4x^5-6x^5\right)+\left(2y^2-7y^2\right)+2xy-\left(x+3x\right)+\left(\frac{1}{6}y-5y\right)-\left(\frac{1}{4}+6\right)\)
\(=-2x^5-5y^2+2xy-2x-\frac{29}{6}y-\frac{25}{4}\)
A = 3x2 - 2x + 1
| x | = 1/2 => x = ±1/2
Với x = 1/2 => A = 3.(1/2)2 - 2.1/2 + 1
= 3.1/4 - 1 + 1
= 3/4
Với x = -1/2 => A = 3(-1/2)2 - 2.(-1/2) + 1
= 3.1/4 + 1 + 1
= 3/4 + 2 = 11/4
B = 2x + 2xy - y
| x | = 2, 5 => x = ±5/2
Với x = 5/2 ; y = -3/4
=> B = 2.5/2 + 2.5/2.(-3/4) - (-3/4)
= 5 - 15/4 + 3/4
= 2
Với x = -5/2 ; y = -3/4
=> B = 2.(-5/2) + 2.(-5/2).(-3/4) - (-3/4)
= -5 + 15/4 + 3/4
= -1/2
a) \(\left|x\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-1}{2}\end{cases}}\)
+) Thay \(x=\frac{1}{2}\)vào biểu thức A ta có :
\(A=3.\left(\frac{1}{2}\right)^2-2.\frac{1}{2}+1=\frac{3}{4}-1+1=\frac{3}{4}\)
+) Thay \(x=\frac{-1}{2}\)vào biểu thức A ta có :
\(A=3.\left(\frac{-1}{2}\right)^2-2.\left(\frac{-1}{2}\right)+1=\frac{3}{4}+1+1=\frac{11}{4}\)
vậy .............
b) Ta có : \(\left|x\right|=2,5\Leftrightarrow\orbr{\begin{cases}x=2,5\\x=-2,5\end{cases}}\)
+) Thay x = 2,5 vào biểu thức B , ta có :
\(B=2.2,5+2.2,5.\frac{-3}{4}-\frac{-3}{4}=2\)
+) Thay x = -2,5 vào biểu thức B , ta có :
\(B=2.\left(-2,5\right)+2.\left(-2,5\right).\frac{-3}{4}-\frac{-3}{4}=-\frac{1}{2}\)
Vậy ...............