Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
P = ( xy + 1 ) ( x2y2 - xyt + 1 )
= x3y3 + 1
= \(\left(5.\frac{3}{5}\right)^3+1\)
= \(27+1\)
= 28
a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
\(B=\dfrac{1}{x}+\dfrac{1}{y}\\ =\dfrac{x+y}{xy}=\dfrac{5}{6}\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\\ =5^3-3.6.5\\ =125-90\\ =35\)
A = x2 + y2
= (x2 + 2xy + y2) - 2xy
= (x + y)2 - 2xy
= 52 - 2.6
= 25 - 12
= 13
F = x3 + y3
= (x + y)3 - 3xy(x + y)
= 53 - 3.6.5
= 125 - 90
= 35
\(1,\left(\frac{a}{3}+4y\right)^2=\frac{a^2}{9}+\frac{8ay}{3}+16y^2\)
\(2,\)Bạn xem lại đề bài giùm mk nhé
\(\left(x^2+\frac{2}{5}y\right).\left(x^2-\frac{2}{5}y\right)=\left(x^2\right)^2-\left(\frac{2}{5}y\right)^2=x^4-\frac{4}{25}y^2\)
\(x^2+y^2-xy-2x-2y+9=x^2+y^2+2xy-2x-2y+9-3xy\)
\(=\left(x+y\right)^2-2\left(x+y\right)+9-3xy=\left(x+y-2\right)\left(x+y\right)+9-3xy.\)
\(đếnđâytịt\)
b
c, =3 dễ
\(\frac{3x^2-6x+9}{x^2-2x+3}=\frac{3\left(x^2-2x+3\right)}{x^2-2x+3}=3\)
a) \(A=\left(3x-2\right)^2+\left(x+1\right)^2-2\left(x+1\right)\left(3x-2\right)\)
\(\Leftrightarrow A=\left(x+1\right)^2-2\left(x+1\right)\left(3x-2\right)+\left(3x-2\right)^2\)
\(\Leftrightarrow A=\left[\left(x+1\right)-\left(3x-2\right)\right]^2\)
\(\Leftrightarrow A=\left(x+1-3x+2\right)^2\)
\(\Leftrightarrow A=\left(3-2x\right)^2\)
Thay \(x=\dfrac{3}{2}\) vào biểu thức A ta được:
\(\left(3-2.\dfrac{3}{2}\right)^2=\left(3-3\right)^2=0^2=0\)
Vậy giá trị của biểu thức A tại \(x=\dfrac{3}{2}\) là 0
b) \(B=\dfrac{x^2y\left(y-x\right)-xy^2\left(x-y\right)}{3y^2-3x^2}\)
\(\Leftrightarrow B=\dfrac{x^2y\left(y-x\right)+xy^2\left(y-x\right)}{3\left(y^2-x^2\right)}\)
\(\Leftrightarrow B=\dfrac{\left(y-x\right)\left(x^2y+xy^2\right)}{3\left(y-x\right)\left(y+x\right)}\)
\(\Leftrightarrow B=\dfrac{xy\left(y-x\right)\left(x+y\right)}{3\left(y-x\right)\left(y+x\right)}\)
\(\Leftrightarrow B=\dfrac{xy\left(y-x\right)\left(y+x\right)}{3\left(y-x\right)\left(y+x\right)}\)
\(\Leftrightarrow B=\dfrac{xy}{3}\)
Thay \(x=-3\) và \(y=\dfrac{1}{2}\) vào biểu thức B ta được:
\(\dfrac{\left(-3\right).\dfrac{1}{2}}{3}=\dfrac{\dfrac{-3}{2}}{3}=\dfrac{\dfrac{-3}{2}}{3}=\dfrac{-1}{2}\)
Vậy giá trị của biểu thức B tại \(x=-3\) và \(y=\dfrac{1}{2}\) là \(\dfrac{-1}{2}\)
c) \(C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{2x\left(1-x\right)}{9-x^2}\)
\(\Leftrightarrow C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}+\dfrac{2x\left(1-x\right)}{x^2-9}\)
\(\Leftrightarrow C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}+\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\) MTC: \(\left(x-3\right)\left(x+3\right)\)
\(\Leftrightarrow C=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{\left(x+1\right)\left(x+3\right)-\left(x-3\right)\left(1-x\right)+2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{\left(x^2+3x+x+3\right)-\left(x-x^2-3+3x\right)+\left(2x-2x^2\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{x^2+3x+x+3-x+x^2+3-3x+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{2}{x-3}\)
Thay \(x=5\) vào biểu thức C ta được:
\(\dfrac{2}{5-3}=\dfrac{2}{2}=1\)
Vậy giá trị của biểu thức C tại \(x=5\) là 1
\(A=\left(x-y\right)\left(x^2+xy+y^2\right)+2y^3\)
\(=x^3-y^3+2y^3=x^3+y^3\)
Khi x=2/3 và y=1/3 thì \(A=\left(\dfrac{2}{3}\right)^3+\left(\dfrac{1}{3}\right)^3=\dfrac{8}{27}+\dfrac{1}{27}=\dfrac{9}{27}=\dfrac{1}{3}\)
Ta có:
\(A=\left(x-y\right)\left(x^2+xy+y^2\right)+2y^3\)
\(A=x^3-y^3+2y^3\)
\(A=x^3+y^3\)
Thay x = \(\dfrac{2}{3}\) và \(y=\dfrac{1}{3}\) vào A ta có:
\(A=\left(\dfrac{2}{3}\right)^3+\left(\dfrac{1}{3}\right)^3=\dfrac{8}{27}+\dfrac{1}{27}=\dfrac{9}{27}=\dfrac{1}{3}\)