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(x+y)^2 =a^2
x^2 +2xy +y^2 =a^2
x^2+y^2 =a^2-2xy =a^2 -2b
x^3 +y^3 = (x+y)(x^2 -xy +y^2)
=a(a^2-2b-b)
=a(a^2-3b)
=a^3- 3ab
(x^2 +y^2)^2=(a^2-2b)^2 ( cái này tính cho x^4 + y^4)
tương tự như câu đầu tiên
x^5+ y^5 (cái đó mình không biết)
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a.
\(x^2+xy+x=x\left(x+y+1\right)\)
Tại \(x=77;y=22\Rightarrow x\left(x+y+1\right)=77\left(77+22+1\right)=77.100=7700\)
b.
\(x\left(x-y\right)+y\left(y-x\right)=x\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(x-y\right)=\left(x-y\right)^2\)
\(=\left(53-3\right)^2=50^2=2500\)
c.
\(x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x+y\right)\left(x-1\right)\)
\(=\left(2001+1999\right)\left(2001-1\right)=4000.2000=8000000\)
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Bài 2:
a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
hay \(x=\dfrac{2}{7}\)
b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow x^3=-8\)
hay x=-2
Bài 1:
a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)
\(=xy^2-x^2y^2+x^2y^2-xy^2+xy\)
\(=xy\)
=1
b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)
\(=x^2y^2+x^3y^2+x^2-x^3y^2-x^2y^2-y^2\)
\(=x^2-y^2\)
\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)
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\(x=\dfrac{1}{y}\Rightarrow\dfrac{1}{y}-y=4\\ \Rightarrow y^2+4y-1=0\\ \Leftrightarrow\left[{}\begin{matrix}y=-2-\sqrt{5}\Rightarrow x=2-\sqrt{5}\\y=-2+\sqrt{5}\Rightarrow x=2+\sqrt{5}\end{matrix}\right.\)
Với \(x=2-\sqrt{5};y=-2-\sqrt{5}\)
\(A=x^2+y^2=18\\ B=x^3-y^3=76\\ C=x^4+y^2=322\)
Với \(x=2+\sqrt{5};y=-2+\sqrt{5}\)
\(A=x^2+y^2=18\\ B=x^3-y^3=76\\ C=x^4+y^4=322\)
A=x^2+y^2
=(x-y)^2+2xy
=4^2+2=18
B=(x-y)^3+3xy(x-y)
=4^3+3*1*4
=64+12=76
C=(x^2+y^2)^2-2x^2y^2
=18^2-2
=322
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\(=\left[\left(\dfrac{-\left(x-y\right)}{x-2y}-\dfrac{x^2+y^2+y-2}{\left(x-2y\right)\left(x+y\right)}\right):\dfrac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}\right]:\dfrac{x+1}{2x^2+y+2}\)
\(=\dfrac{-x^2+y^2-x^2-y^2-y+2}{\left(x-2y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y-2\right)\left(2x^2+y+2\right)}\cdot\dfrac{2x^2+y+2}{x+1}\)
\(=\dfrac{-2x^2-y+2}{\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)}{\left(2x^2+y-2\right)\left(2x^2+y+2\right)}\cdot\dfrac{2x^2+y+2}{x+1}\)
\(=\dfrac{-1}{x-2y}\)
Thay $x=-1,76$ và $y=\dfrac{3}{25}$ vào $P=\dfrac{-1}{x-2y}$, ta được:
$P=\dfrac{-1}{-1,76-2.(\dfrac{3}{25})}=\dfrac{1}{2}$.
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Với mọi số thực ta luôn có:
`(x-y)^2>=0`
`<=>x^2-2xy+y^2>=0`
`<=>x^2+y^2>=2xy`
`<=>(x+y)^2>=4xy`
`<=>(x+y)^2>=16`
`<=>x+y>=4(đpcm)`
\(\dfrac{1}{x+3}+\dfrac{1}{y+3}=\dfrac{x+3+y+3}{\left(x+3\right)\left(y+3\right)}\)
\(=\dfrac{x+y+6}{3x+3y+13}\)(vì \(xy=4\))
=> \(\dfrac{x+y+6}{3x+3y+13}\)≤\(\dfrac{2}{5}\)
<=> \(5\left(x+y+6\right)\)≤\(2\left(3x+3y+13\right)\)
<=>\(6x+6y+26-5x-5y-30\)≥\(0\)
<=> \(x+y-4\)≥\(0\)
Áp dụng BĐT AM-GM \(\dfrac{a+b}{2}\)≥\(\sqrt{ab}\)
Ta có \(\dfrac{x+y}{2}\)≥\(\sqrt{xy}\)
<=>\(x+y\) ≥ 2\(\sqrt{xy}\)
=>2\(\sqrt{xy}-4\)≥\(0\)
<=> \(4-4\)≥0
<=>0≥0 ( Luôn đúng )
Vậy \(\dfrac{1}{x+3}+\dfrac{1}{y+3}\)≤\(\dfrac{2}{5}\)
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đối với các câu này bạn hãy khai triển phần nào dài bằng hàng dẳng thức rồi thu gọn lại nếu đúng thì vế trái bằng vế phải
B=6
C=8
1) ( x + y )3 = x3 + 3x2y + 3xy2 + y3
( x + y )3 = ( x3 + y3 ) + 3xy.( x + y )
x3 + y3 = ( x + y )3 - 3xy.( x + y )
x3 + y3 = 23 - ( 3.-1).2
x3 + y3 = 14
B = 14
2) x2 + y2 = 2xy
x2 + 2xy + y2 = 4xy
( x + y )2 = 4xy
xy = \(\frac{\left(x+y\right)^2}{4}\)
xy = 1
( x + y)2 = x2 + 2xy + y2
x2 + y2 = ( x + y )2 - 2xy
x2 + y2 = 22 - ( 2.-1)
x2 + y2 = 6
( x2 + y2 )2 = x4 + 2x2y2 + y4
x4 + y4 = ( x2 + y2 )2 - 2x2y2
x4 + y4 = ( 6)2 - 2.( 1 . 1)
x4 + y4 = 34
C = 34