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a)
Ta có :
\(x+y=3\)
\(x^2+y^2=5\Leftrightarrow\left(x+y\right)^2-2xy=5\Leftrightarrow9-2xy=5\Leftrightarrow2xy=4\Rightarrow xy=2\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3.\left(5-2\right)=9\)
b)
Ta có :
\(x-y=5\)
\(x^2+y^2=15\Leftrightarrow\left(x-y\right)^2+2xy=15\Leftrightarrow25+2xy=15\Rightarrow xy=-5\)
=> \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=\left(5\right)\left(15+-5\right)=50\)
x+ y = 2 => ( x + y)^2 = 2^2 = 4
=> x^2 + 2xy + y^2 = 4
=> 10 + 2xy = 4 => 2xy = - 6 => xy = - 3
thay vào ta có
x^3 + y^3 = ( x + y)(x^2 - xy + y^2)
= 2 ( 10 - ( - 3) ) = = 2 . 13 = 26
Ta có : x + y = 2
=> \(\left(x+y\right)^2=4\)
<=> x2 + 2xy + y2 = 4
=> 2xy + 10 = 4
=> 2xy = -6
=> xy = -6
P = x3 + y3 = (x + y)(x2 - xy + y2)
= 2(10 + 6)
= 2.16
=32
\(A=x^3+y^3+3xy=\left(x+y\right)^3-3xy\left(x+y\right)+3xy=1+0=1\)
\(B=\left(x-y\right)^3+3xy\left(x-y\right)-3xy=1\)
\(c,M=a^2-ab+b^2+3ab\left(a^2+b^2\right)+6a^2b^2=3ab\left(a^2+2ab+b^2\right)+a^2-ab+b^2\)
\(=3ab+a^2-ab+b^2=\left(a+b\right)^2=1\)
\(x+y=2;x^2+y^2=10\text{ do đó:}xy=-3\text{ nên }\left(x-y\right)^2=16\text{ do đó: }x-y=4\text{ hoặc }x-y=-4\)
\(\text{giải ra được:}x=3;y=-1\text{ hoặc ngược lại nên }x^3+y^3=-26\text{ hoặc }26\)
A = x3 + y3 + 3xy
= x3 + 3x2y + 3xy2 + y3 - 3x2y - 3xy2 + 3xy
= ( x3 + 3x2 + 3xy2 + y3 ) - ( 3x2y + 3xy - 3xy )
= ( x + y )3 - 3xy( x + y - 1 )
= 13 - 3xy( 1 - 1 )
= 13 - 3xy.0
= 1 - 0 = 1
Vậy A = 1
b) B = x3 - y3 - 3xy
= x3 - 3x2y + 3xy2 - y3 + 3x2y - 3xy2 - 3xy
= ( x3 - 3x2y + 3xy2 - y3 ) + ( 3x2y - 3xy2 - 3xy )
= ( x - y )3 + 3xy( x - y - 1 )
= 13 + 3xy( 1 - 1 )
= 1 + 3xy.0
= 1 + 0 = 1
Vậy B = 1
M = a3 + b3 + 3ab( a2 + b2 ) + 6a2b2( a + b )
= ( a + b )( a2 - ab + b2 ) + 3ab[ ( a + b )2 - 2ab ] + 6a2b2( a + b )
= ( a + b )[ ( a + b )2 - 3ab ] + 3ab[ ( a + b )2 - 2ab ] + 6a2b2( a + b )
= 1.( 1 - 3ab ) + 3ab( 1 - 2ab ) + 6a2b2.1
= 1 - 3ab + 3ab - 6a2b2 + 6a2b2
= 1
Vậy M = 1
d) x + y = 2
⇔ ( x + y )2 = 4
⇔ x2 + 2xy + y2 = 4
⇔ 10 + 2xy = 4 ( gt x2 + y2 = 10 )
⇔ 2xy = -6
⇔ xy = -3
x3 + y3 = x3 + 3x2y + 3xy2 + y3 - 3x2y - 3xy2
= ( x3 + 3x2y + 3xy2 + y3 ) - ( 3x2y + 3xy2 )
= ( x + y )3 - 3xy( x + y )
= 23 - 3.(-3).(2)
= 8 + 18 = 26
a) Ta có: A = x3 + y3 + 3xy = (x + y)(x2 - xy + y2) + 3xy = 1. (x2 - xy + y2) + 3xy = x2 - xy + y2 + 3xy = x2 + 2xy + y2 = (x + y)2 = 12 = 1
b)Ta có: B = x3 - y3 - 3xy = (x - y)(x2 + xy + y2) - 3xy = 1. (x2 + xy + y2) - 3xy = x2 + xy + y2 - 3xy = x2 - 2xy + y2 = (x - y)2 = 12 = 1
d) Ta có : D = x3 + y3 + 3xy(x2 + y2) + 6x2y2(x + y)
=> D = (x + y)(x2 - xy + y2) + 3xy(x2 + 2xy + y2) - 6x2y2 + 6x2y2
=> D = x2 - xy + y2 + 3xy(x + y)2
=> D = x2 - xy + y2 + 3xy.12
=> D = x2 + 2xy + y2
=> D = (x + y)2 = 12 = 1
\(A=x^2-2xy+y^2=\left(x-y\right)^2=\left(-3\right)^2=9\)
\(B=x^2+y^2=\left(x-y\right)^2+2xy=9+10.2=29\)
\(C=x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3=\left(-3\right)^3=-27\)
\(D=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=\left(-3\right)\left[x^2-2xy+y^2+3xy\right]=\left(-3\right)\left(\left(-3\right)^2.3.10\right)=-3.270=-810\)
Bài 1:
a) (x+y)2=92=81
=> x2+2xy+y2=81
=> x2+2.14+y2=81
=> x2+y2=53
=> x2-2xy+y2=81-2.14=25
=> (x-y)2=25
=> x-y=5 hoặc x-y=-5
b) Câu a đã tính được x2+y2=53
c) Ta có: x3+y3=(x+y)(x2-xy+y2)=9(53-14)=9.39=351
Bài 2:
Ta có: \(x^2+2xy+y^2-4x-4y+1=\left(x+y\right)^2-4\left(x+y\right)+1\)
Mà x+y=1
\(\Rightarrow1^2-4.1+1=-2\)
Bài 3:
Ta có: (x+y)3=x3+3x2y+3xy2+y3
= x3+y3+3xy(x+y)
Mà x+y=1 => (x+y)3=x3+y3+3xy=13=1
Bài 4:
Ta có: \(\left(x+y\right)^2=4^2=16\)
\(\Rightarrow x^2+2xy+y^2=16\Rightarrow10+2xy=16\)
\(\Rightarrow2xy=6\Rightarrow xy=3\)
Lại có: \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=4.\left(10-3\right)\)
\(=4.7=28\)
Bài 5:
Ta có: \(x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=1\left(x^2+xy+y^2\right)-3xy=x^2+xy+y^2-3xy\)
\(=x^2-2xy+y^2=\left(x-y\right)^2=1\)
Mấy bài này đầu hè làm hết rồi:))
Bài 1:
a) \(xy=14\Rightarrow x=\frac{14}{y}\)
Thay vào: \(\frac{14}{y}+y=9\)
\(\Leftrightarrow y^2+14-9y=0\)
\(\Leftrightarrow\left(y-2\right)\left(y-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=2\\y=7\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=2\end{cases}}\)
+ Nếu: \(\hept{\begin{cases}x=7\\y=2\end{cases}}\Rightarrow x-y=5\)
+ Nếu: \(\hept{\begin{cases}x=2\\y=7\end{cases}}\Rightarrow x-y=-5\)
b) Ta có: \(x+y=9\)
\(\Leftrightarrow\left(x+y\right)^2=81\)
\(\Leftrightarrow x^2+2xy+y^2=81\)
\(\Rightarrow x^2+y^2=81-2xy=81-2.14=53\)
c) Ta có: \(x+y=9\)
\(\Leftrightarrow\left(x+y\right)^3=9^3\)
\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=729\)
\(\Leftrightarrow x^3+y^3=729-3xy\left(x+y\right)=729-3.14.9=351\)
\(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\Leftrightarrow2^2=10+2xy\)
\(\Leftrightarrow xy=-3\)
\(\left(x+y\right)^3=x^3+y^3+3x^2y+3xy^2\)
\(\Leftrightarrow2^3=x^3+y^3+3xy\left(x+y\right)\)
\(\Leftrightarrow8=x^3+y^3+3.\left(-3\right).10\)
\(\Leftrightarrow x^3+y^3=98\)
Vì x + y = 2 Nên \(x^2\)+ \(y^2\)= 4
\(\Rightarrow\)\(x^2\)\(+\)\(y^2\)\(+\)\(2xy\)\(=\)\(4\)
\(\Rightarrow\)\(10\)\(+\)\(2xy\)\(=\)\(4\)
\(\Rightarrow\)\(2xy\)\(=\)\(-6\)
\(\Rightarrow\)\(xy\)\(=\) \(-3\)
Do đó \(x^3\)\(+\)\(y^3\)\(=\)\(\left(x+y\right).\left(x^2+y^2-xy\right)\) \(=\)\(2.\left[10-\left(-3\right)\right]=2.13=26\)
P = x3 + y3 - x2 - y2 + 3xy( x + y ) - 2xy + 3( x + y ) + 10
= ( x3 + y3 ) - ( x2 + 2xy + y2 ) + 3xy( x + y ) + 3.5 + 10
= ( x3 + 3x2y + 3xy2 + y3 - 3x2y - 3xy2 ) - ( x + y )2 + 3xy( x + y ) + 15 + 10
= [ ( x3 + 3x2y + 3xy2 + y3 ) - ( 3x2y + 3xy2 ) ] - 52 + 3xy( x + y ) + 25
= ( x + y )3 - 3xy( x + y ) - 25 + 3xy( x + y ) + 25
= 53 = 125
\(x+y=2\) nên \(\left(x+y\right)^2=4\)
\(\Rightarrow x^2+y^2+2xy=4\)
\(\Rightarrow10+2xy=4\)
\(\Rightarrow2xy=-6\)
\(\Rightarrow xy=-3\)
Do đó \(x^3+y^3=\left(x+y\right).\left(x^2+y^2-xy\right)=2.\left[10-\left(-3\right)\right]=2.13=26\)
ta có:
(x+y)2=x2+y2+2xy
=>2xy=(x+y)2-(x2+y2)
=4-10
=-6
=>xy=-3
ta lại có:
x3+y3=x3+y3+3x2y+3xy2-3x2y-3xy2
=(x+y)3-3xy.(x+y)
=8-3.(-3).2
=8+18
=26