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Bài 1:
\(A=\left(x-y\right)\left(x^2+xy+y^2\right)+2y^3\)
\(A=x^3-y^3+2y^3\)
\(A=x^3+y^3\)
Thay \(x=\dfrac{2}{3},y=\dfrac{1}{3}\) vào A, ta có:
\(A=\left(\dfrac{2}{3}\right)^3+\left(\dfrac{1}{3}\right)^3=\dfrac{8}{27}+\dfrac{1}{27}=\dfrac{9}{27}=\dfrac{1}{3}\)
a) ĐKXĐ : \(x+y\ne0\)
\(x^2-2y^2=xy\)
\(x^2-y^2-y^2-xy=0\)
\(\left(x-y\right)\left(x+y\right)-y\left(y+x\right)=0\)
\(\left(x+y\right)\left(x-2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\left(Loai\right)\\x-2y=0\left(Chon\right)\end{matrix}\right.\)
Với x - 2y = 0 ta có x = 2y
Thay x = 2y vào A ta có :
\(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
Cho hai số dương x,y thỏa mãn: 2x2+xy-y2=0. Tính giá trị biểu thức:
A = \(\frac{x^2y+xy^2}{x^3+y^3}\)
2.
a. Ta có: x + y = 5 ⇒ x = 5 - y
Thay vào A ta được:
\(A=3\left(5-y\right)^2+3y^2-2y+6\left(5-y\right).y-100\)
\(A=75-30y+3y^2+3y^2-2y+30y-6y^2-100\)
\(A=75-100=-25\)
b. Ta có: x - y = 7 ⇒ x = 7 + y
Thay x = 7 + y vào A ta được:
\(A=\left(7+y\right)\left(7+y+2\right)+y\left(y-2\right)-2\left(7+y\right).y+37\)
\(A=y^2+16y+63+y^2-2y-14y-2y^2+37\)
\(A=100\)
c. Ta có: x + 2y = 5 ⇒ x = 5 - 2y
Thay vào A ta có:
\(A=\left(5-2y\right)^2+4y^2-2\left(5-2y\right)+10+4\left(5-2y\right).y-4y\)
\(A=25-20y+4y^2+4y^2-19+4y+10+20y-8y^2-4y\)
\(A=16\)
a) \(\dfrac{1}{8}x^3y^3-27=\left(\dfrac{1}{2}xy\right)^3-3^3=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{1}{4}x^2y^2+\dfrac{1}{6}xy+9\right)\)
b)\(\dfrac{8}{125}x^3+27y^3=\left(\dfrac{2}{5}x\right)^3+\left(3y\right)^3=\left(\dfrac{2}{5}x+3y\right)\left(\dfrac{4}{25}x^2-\dfrac{6}{5}xy+9y^2\right)\)
c) \(0.008x^6-27y^3=\left(0.2x^2\right)^3-\left(3y\right)^3=\left(0.2x^2-3y\right)\left(0.04x^4+\dfrac{3}{5}x^2y+9y^2\right)\)
d)\(\left(2x+y\right)^3-\left(x-y\right)^3=\left(2x+y-x+y\right)[\left(2x+y\right)^2+\left(2x+y\right)\left(x-y\right)+\left(x-y\right)^2]\\ =\left(x+2y\right)\left(4x^2+4xy+y^2+2x^2-2xy+xy-y^2+x^2-2xy+y^2\right)\\ =\left(x+2y\right)\left(6x^2+xy+y^2\right)\)
Bài 1:
a) \(\dfrac{1}{8}x^3y^3-27\)
\(=\left(\dfrac{1}{2}xy\right)^3-3^3\)
\(=\left(\dfrac{1}{2}xy-3\right)\left[\left(\dfrac{1}{2}xy\right)^2+\dfrac{1}{2}xy.3+3^2\right]\)
\(=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{1}{4}xy+\dfrac{3}{2}xy+9\right)\)
\(=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{7}{4}xy+9\right)\)
b) \(\dfrac{8}{125}x^3+\dfrac{1}{8}y^3\)
\(=\left(\dfrac{2}{5}x\right)^3+\left(\dfrac{1}{2}y\right)^3\)
\(=\left(\dfrac{2}{5}x+\dfrac{1}{2}y\right)\left[\left(\dfrac{2}{5}x\right)^2-\dfrac{2}{5}x.\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\right]\)
\(=\left(\dfrac{2}{5}x+\dfrac{1}{2}y\right)\left(\dfrac{4}{25}x-\dfrac{1}{5}xy+\dfrac{1}{4}y\right)\)
c) \(0.008x^6-27y^3\)
\(=\left(\dfrac{1}{5}x^2\right)^3-\left(3y\right)^3\)
\(=\left(\dfrac{1}{5}x^2-3y\right)\left[\left(\dfrac{1}{5}x^2\right)^2+\dfrac{1}{5}x^2.3y+\left(3y\right)^2\right]\)
\(=\left(\dfrac{1}{5}x^2-3y\right)\left(\dfrac{1}{25}x^4+\dfrac{3}{5}x^2y+9y^2\right)\)
d) \(\left(2x+y\right)^3-\left(x-y\right)^3\)
\(=\left[\left(2x+y\right)-\left(x-y\right)\right]\left[\left(2x+y\right)^2+\left(2x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(2x+y-x+y\right)\left(4x^2+4xy+y^2+2x^3-2xy+xy-y^2+x^2-2xy+y^2\right)\)
\(=\left(x-2y\right)\left(4x^2+2x^3+xy\right)\)
a) A = ( 6x + 7)( 2x - 3) - ( 4x + 1)( 3x - \(\dfrac{7}{4}\))
A = 12x2 - 18x + 14x - 21 - ( 12x2 - 7x + 3x - \(\dfrac{7}{4}\))
A = \(\dfrac{-77}{4}\)
Vậy biểu thức trên ko phụ thuộc vào biến
b) x2 - 2y2 = xy
⇔ x2 - xy - 2y2 = 0
⇔ x2 + xy - 2xy - 2y2 = 0
⇔ x( x + y) - 2y( x + y) = 0
⇔ ( x - 2y )( x + y ) = 0
Do : x + y # 0
⇒ x - 2y = 0
⇔ x = 2y
Ta có : P = \(\dfrac{x-y}{x+y}\) ( x + y # 0 ; y # 0)
P = \(\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
KL....
\(A=2x+xy^2-x^2y-2y\)
\(=2\left(x-y\right)-xy\left(x-y\right)\)
\(=\left(x-y\right)\left(2-xy\right)\)
\(=\left(-\dfrac{1}{2}-\dfrac{-1}{3}\right)\left(2-\dfrac{-1}{2}\cdot\dfrac{-1}{3}\right)\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\cdot\left(2-\dfrac{1}{6}\right)\)
\(=\dfrac{-1}{6}\cdot\dfrac{11}{6}=-\dfrac{11}{36}\)