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\(A=x^3+y^3+3xy=\left(x+y\right)^3-3xy\left(x+y\right)+3xy=1+0=1\)
\(B=\left(x-y\right)^3+3xy\left(x-y\right)-3xy=1\)
\(c,M=a^2-ab+b^2+3ab\left(a^2+b^2\right)+6a^2b^2=3ab\left(a^2+2ab+b^2\right)+a^2-ab+b^2\)
\(=3ab+a^2-ab+b^2=\left(a+b\right)^2=1\)
\(x+y=2;x^2+y^2=10\text{ do đó:}xy=-3\text{ nên }\left(x-y\right)^2=16\text{ do đó: }x-y=4\text{ hoặc }x-y=-4\)
\(\text{giải ra được:}x=3;y=-1\text{ hoặc ngược lại nên }x^3+y^3=-26\text{ hoặc }26\)
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
Bài 1:
Theo bài ra ta có:
\(\left(x-y\right)^2=x^2-2xy+y^2\)
\(=\left(5-y\right)^2-2\times2+\left(5-x\right)^2\)
\(=5^2-2\times5y+y^2-4+5^2-2\times5x+x^2\)
\(=25-10y+y^2+25-10x+x^2-4\)
\(=\left(25+25\right)-\left(10x+10y\right)+x^2+y^2-4\)
\(=50-10\left(x+y\right)+x^2+2xy+y^2-2xy-4\)
\(=50-10\times5+\left(x+y\right)^2-2\times2-4\)
\(=50-50+5^2-4-4\)
\(=25-8=17\)
Vậy giá trị của \(\left(x-y\right)^2\)là 17
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
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: \(x+y=9\)
\(\Leftrightarrow\left(x+y\right)^3=9^3\)
\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=729\)
\(\Leftrightarrow x^3+y^3=729-3xy\left(x+y\right)=729-3.14.9=351\)
\(x^2+y^2+2xy\)
\(=10+2.2\)
\(=14\)
\(\Rightarrow\left(x+y\right)^2=14\)
\(\Rightarrow x+y=\sqrt{14}\)
(x-y)2 = x2 - 2xy +y2 = 10 -4 = 6
x-y =\(\sqrt{6}\)
x2 -y2 =(x+y)(x-y) = \(\sqrt{14.6}\)= \(\sqrt{84}\)
a , Ta có: \(a+b=10\Rightarrow a=10-b\)
\(a+b=10\Rightarrow\left(a+b\right)^2=100\)
\(\Leftrightarrow a^2+2ab+b^2=100\)
\(\Leftrightarrow2ab=100-\left(a^2+b^2\right)=100-52=48\Rightarrow ab=24\)\(\Leftrightarrow\left(10-b\right)b=24\Leftrightarrow10b+b^2-24=0\)
\(\Leftrightarrow b^2+10b+25-49=0\)
\(\Leftrightarrow\left(b+5\right)^2=49\Rightarrow\left[{}\begin{matrix}b+5=7\\b+5=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}b=2\\b=-12\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=8\\a=-2\end{matrix}\right.\)b, ta có:
\(\left(x+y\right)=1\Rightarrow\left(x+y\right)^3=1\)
\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=1\)
\(\Leftrightarrow x^3+3xy\left(x+y\right)+y^3=1\)
\(\Leftrightarrow x^3+3xy+y^3=1\)
c, \(a+b=13\Rightarrow\left(a+b\right)^2=169\)
\(\Leftrightarrow a^2+2ab+b^2=169\Rightarrow a^2+b^2=169-2ab=169-2.9=151\)\(\Rightarrow a^3+b^3=\left(a+b\right)\left(a^2+b^2+ab\right)=13.\left(151+9\right)=2080\)
\(d,x+y=7\Rightarrow\left(x+y\right)^2=49\)
\(\Leftrightarrow x^2+2xy+y^2=49\Rightarrow2xy=49-\left(x^2+y^2\right)=16\Rightarrow xy=8\)
\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=7.\left(33-8\right)=175\)