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A = 5x(x - y) - y(5x - y)
A = 5x2 - 5xy - 5xy + y2
A = 5x2 - 10xy + y2 (1)
Thay x = -1; y = 3 vào (1), ta có:
5.(-1)2 - 10.(-1).3 + 32 = 44
B = 4y(x2 - 3xy + 3y2) - 2xy(2x - 6y - 3)
B = 4x2y - 12x2 + 12y3 - 4x2y + 12xy2 + 6xy
B = 12y3 + 6xy (1)
Thay x = 5; y = -1 vào (1), ta có:
12.(-1)3 + 6.5.(-1) = -42
C = 5x2(x - y2) + 3x(xy2 - y) - 5x3
C = 5x3 - 5x2y2 + 3x2y2 - 3xy - 5x3
C = -2x2y2 - 3xy (1)
Thay x = -2; y = -5 vào (1), ta có:
-2.(-2)2.(-5)2 - 3.(-2).(-5) = -230
D = 6x2(y2 - xy + 2x2y) - 3xy(2xy - x2 + 4x3)
D = 6x2y2 - 6x3y + 12x4y - 6x2y2 + 3x3y - 12x4y
D = -3x3y (1)
Thay x = 11; y = -1 vào (1), ta có:
-3.113.(-1) = 3993
1: \(B=10x^2-15xy-xy+5xy=10x^2-11xy\)
\(=10\cdot\dfrac{1}{25}-11\cdot\dfrac{-1}{5}\cdot\dfrac{1}{2}\)
\(=\dfrac{5}{2}+\dfrac{11}{10}=\dfrac{18}{5}\)
2: \(C=x^2y^2-xy^2-2x^3+2x^2y^2\)
\(=-xy^2+3x^2y^2-2x^3\)
\(=-\dfrac{1}{2}\cdot2^2+3\cdot\left(\dfrac{1}{2}\cdot2\right)^2-2\cdot\dfrac{1}{8}\)
\(=-2+3-\dfrac{1}{4}=1-\dfrac{1}{4}=\dfrac{3}{4}\)
1: \(B=10x^2-15xy-xy+5xy=10x^2-11xy\)
\(=10\cdot\dfrac{1}{25}-11\cdot\dfrac{-1}{5}\cdot\dfrac{1}{2}\)
\(=\dfrac{5}{2}+\dfrac{11}{10}=\dfrac{18}{5}\)
2: \(C=x^2y^2-xy^2-2x^3+2x^2y^2\)
\(=-xy^2+3x^2y^2-2x^3\)
\(=-\dfrac{1}{2}\cdot2^2+3\cdot\left(\dfrac{1}{2}\cdot2\right)^2-2\cdot\dfrac{1}{8}\)
\(=-2+3-\dfrac{1}{4}=1-\dfrac{1}{4}=\dfrac{3}{4}\)
Bài 1 :
a) \(M=\dfrac{1}{2}x^2y.\left(-4\right)y\)
\(\Rightarrow M=-2x^2y^2\)
Khi \(x=\sqrt[]{2};y=\sqrt[]{3}\)
\(\Rightarrow M=-2.\left(\sqrt[]{2}\right)^2.\left(\sqrt[]{3}\right)^2\)
\(\Rightarrow M=-2.2.3=-12\)
b) \(N=xy.\sqrt[]{5x^2}\)
\(\Rightarrow N=xy.\left|x\right|\sqrt[]{5}\)
\(\Rightarrow\left[{}\begin{matrix}N=xy.x\sqrt[]{5}\left(x\ge0\right)\\N=xy.\left(-x\right)\sqrt[]{5}\left(x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}N=x^2y\sqrt[]{5}\left(x\ge0\right)\\N=-x^2y\sqrt[]{5}\left(x< 0\right)\end{matrix}\right.\)
Khi \(x=-2< 0;y=\sqrt[]{5}\)
\(\Rightarrow N=-x^2y\sqrt[]{5}=-\left(-2\right)^2.\sqrt[]{5}.\sqrt[]{5}=-4.5=-20\)
2:
Tổng của 4 đơn thức là;
\(A=11x^2y^3+\dfrac{10}{7}x^2y^3-\dfrac{3}{7}x^2y^3-12x^2y^3=0\)
=>Khi x=-6 và y=15 thì A=0
Lời giải:
a, Ta có: A = 5x ( 4x2 - 2x + 1) - 2x ( 10x2 - 5x - 2 ) = 5x.4x2 - 5x.2x + 5x - 2x.10x2 + 2x.5x + 2x.2
= 20x3 - 10x2 + 5x - 20x3 + 10x2 + 4x
= (20x3 - 20x3) - (10x2 - 10x2 ) + (5x + 4x) = 9x
Thay x = 15 vào A => A= 9 . 15 = 135 .Vậy: A = 135
b, Ta có: B = 5x(x - 4y) - 4y(y - 5x) = 5x2 - 20xy - 4y2 + 20xy
= 5x2 - 4y2
Thay x = \(-\frac{1}{5}\) , y = \(-\frac{1}{2}\)vào B => B \(=5\left(-\frac{1}{5}\right)^2-4\left(-\frac{1}{2}\right)^2\) \(=-\frac{4}{5}\).Vậy: B = \(-\frac{4}{5}\)
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1: \(B=10x^2-15xy-xy+5xy=10x^2-11xy\)
\(=10\cdot\dfrac{1}{25}-11\cdot\dfrac{-1}{5}\cdot\dfrac{-1}{2}\)
\(=\dfrac{2}{5}-\dfrac{11}{10}=\dfrac{-7}{10}\)
2: \(C=x^2y^2-xy^3-2x^3+2x^2y^2\)
\(=-xy^3+3x^2y^2-2x^3\)
\(=-\dfrac{-1}{2}\cdot2^3+3\cdot\left(\dfrac{-1}{2}\cdot2\right)^2-2\cdot\left(-\dfrac{1}{2}\right)^3\)
\(=\dfrac{1}{4}\cdot8+3-2\cdot\dfrac{-1}{8}\)
\(=2+3+\dfrac{1}{4}=5+\dfrac{1}{4}=\dfrac{21}{4}\)