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Bài 1.
A = x2 + 2xy + y2 = ( x + y )2 = ( -1 )2 = 1
B = x2 + y2 = ( x2 + 2xy + y2 ) - 2xy = ( x + y )2 - 2xy = (-1)2 - 2.(-12) = 1 + 24 = 25
C = x3 + 3xy( x + y ) + y3 = ( x3 + y3 ) + 3xy( x + y ) = ( x + y )( x2 - xy + y2 ) + 3xy( x + y )
= -1( 25 + 12 ) + 3.(-12).(-1)
= -37 + 36
= -1
D = x3 + y3 = ( x3 + 3x2y + 3xy2 + y3 ) - 3x2y - 3xy2 = ( x + y )3 - 3xy( x + y ) = (-1)3 - 3.(-12).(-1) = -1 - 36 = -37
Bài 2.
M = 3( x2 + y2 ) - 2( x3 + y3 )
= 3( x2 + y2 ) - 2( x + y )( x2 - xy + y2 )
= 3( x2 + y2 ) - 2( x2 - xy + y2 )
= 3x2 + 3y2 - 2x2 + 2xy - 2y2
= x2 + 2xy + y2
= ( x + y )2 = 12 = 1
1) 2x2-8xy-5x+20y
=2x(x-4y)-5(x-4y)
=(2x-5)(x-4y)
2) x3-x2y-xy+y2
=x2(x-y)-y(x-y)
=(x2-y)(x-y)
3) x2-2xy-4z2+y2
=(x-y)2-(2z)2
=(x-y-2z)(x-y+2z)
4) a3+a2b-a2c-abc
=a2(a+b)-ac(a+b)
=(a2-ac)(a+b)
=a(a-c)(a+b)
5) x3+y3+3x2y+3xy2-x-y
=(x+y)(x2-xy+y2)+3xy(x+y)-(x+y)
=(x+y)(x2-xy+y2+3xy-1)
=(x+y)[(x+y)2-1)]
=(x+y)(x+y+1)(x+y-1)
6) x3+x2y-x2z-xyz
=x2(x+y)-xz(x+y)
=(x2-xz)(x+y)
=x(x-z)(x+y)
7) =[x(y+z)2-2xyz]+[y(z+x)2-2xyz]+z(x+y)2
=x(y2+z2)+y(z2+x2)+z(x+y)2
=xy(x+y)+z2(x+y)+z(x+y)2
=(x+y)(xy+z2+zx+zy)
=(x+y)(x+z)(y+z)
8) x3(z-y)+y3(x-z)+z3(y-x)
Tách x-z= -[z-y+y-x]
a. Có \(x+y=2\Rightarrow x^2+2xy+y^2=4\Rightarrow x^2+y^2=4-2.\left(-3\right)=10\)
\(x^4+y^4=\left(x^2\right)^2+\left(y^2\right)^2=\left(x^2+y^2\right)^2-2x^2y^2\)
\(=10^2-2.\left(-3\right)^2=82\)
b. Ta có \(x+y=1\Rightarrow x^2+y^2=1-2xy\)
\(x^3+y^3+3xy=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\)
\(=1.\left(1-2xy-xy\right)+3xy=1\)
Các câu còn lại tương tự
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: \(x+y=9\)
\(\Leftrightarrow\left(x+y\right)^3=9^3\)
\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=729\)
\(\Leftrightarrow x^3+y^3=729-3xy\left(x+y\right)=729-3.14.9=351\)
1, 2x2 - 8xy - 5x + 20y
= (2x2 - 5x) - (8xy - 20y)
= x(2x - 5) - 4y(2x - 5)
= (2x - 5) (x - 4y)
2, x3 - x2y - xy + y2
= (x3 - xy) - (x2y - y2)
= x(x2 - y) - y(x2 - y)
= (x2 - y) (x - y)
3, x2 - 2xy - 4z2 + y2
= (x2 - 2xy + y2) - 4z2
= (x - y)2 - (2z)2
= (x - y - 2z) (x - y + 2z)
4, a3 + a2b - a2c - abc
= (a3 - a2c) + (a2b - abc)
= a2(a - c) + ab(a - c)
= (a - c) (a2 + ab)
5, x3 + y3 + 3x2y + 3xy2 - x - y
= (x3 + 3x2y + 3xy2 + y3) - (x + y)
= (x + y) 3 - (x + y)
= (x + y) [(x + y)2 - 1]
= (x + y) (x + y - 1) (x + y + 1)
1/Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=81\)
\(\Rightarrow M=ab+bc+ca=\frac{\left(81-141\right)}{2}\)
a) \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left[\left(y-z\right)+\left(x-y\right)\right]+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left(y-z\right)-y^2\left(x-y\right)+z^2\left(x-y\right)\)
\(=\left(y-z\right)\left(x^2-y^2\right)-\left(x-y\right)\left(y^2-z^2\right)\)
\(=\left(y-z\right)\left(x-y\right)\left(x+y\right)-\left(x-y\right)\left(y-z\right)\left(y+z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x+y-y-z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
c) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)^3-x^3-y^3\right]+3z\left(x+y\right)\left(x+y+z\right)\)
\(=3xy\left(x+y\right)+3\left(x+y\right)\left(xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
d) \(\left(x^2+y^2-5\right)^2-4x^2y^2-16xy-16\)
\(=\left(x^2+y^2-5\right)^2-\left(4x^2y^2+16xy+16\right)\)
\(=\left(x^2+y^2-5\right)^2-\left[\left(2xy\right)^2+2.2xy.4+16\right]\)
\(=\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2\)
\(=\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\)
\(=\left(x^2-2xy+y^2-9\right)\left(x^2+2xy+y^2-1\right)\)
\(=\left[\left(x-y\right)^2-3^2\right]\left[\left(x+y\right)^2-1\right]\)
\(=\left(x-y-3\right)\left(x-y+3\right)\left(x+y-1\right)\left(x+y+1\right)\)
e) \(\left(x^2+4y^2-5\right)^2-16\left(x^2y^2+2xy+1\right)\)
\(=\left(x^2+4y^2-5\right)^2-4^2\left(xy+1\right)^2\)
\(=\left(x^2+4y^2-5\right)^2-\left[4\left(xy+1\right)\right]^2\)
\(=\left(x^2+4y^2-5\right)-\left(4xy+4\right)^2\)
\(=\left(x^2+4y^2-5-4xy-4\right)\left(x^2+4y^2-5+4xy+4\right)\)
\(=\left(x^2+4y^2-4xy-9\right)\left(x^2+4y^2+4xy-1\right)\)
\(=\left[\left(x-2y\right)^2-3^2\right]\left[\left(x+2y\right)^2-1\right]\)
\(=\left(x-2y-3\right)\left(x-2y+3\right)\left(x+2y-1\right)\left(x+2y+1\right)\)
f) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1\)
\(=\left(x-y+5-1\right)^2\)
\(=\left(x-y+4\right)^2\)
1/ Ta có : P\left(x\right)=-x^2+13x+2012=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}P(x)=−x2+13x+2012=−(x−213)2+48217≤48217
Dấu "=" xảy ra khi x = 13/2
Vậy Max P(x) = 8217/4 tại x = 13/2
1/ Ta có : P\left(x\right)=-x^2+13x+2012=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}P(x)=−x2+13x+2012=−(x−213)2+48217≤48217
Dấu "=" xảy ra khi x = 13/2
Vậy Max P(x) = 8217/4 tại x = 13/2
2/ Ta có : x^3+3xy+y^3=x^3+3xy.1+y^3=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1x3+3xy+y3=x3+3xy.1+y3=x3+y3+3xy(x+y)=(x+y)3=1
3/ a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0a+b+c=0⇔(a+b+c)2=0⇔a2+b2+c2+2(ab+bc+ac)=0
\Leftrightarrow ab+bc+ac=-\frac{1}{2}⇔ab+bc+ac=−21 \Leftrightarrow\left(ab+bc+ac\right)^2=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}⇔(ab+bc+ac)2=41⇔a2b2+b2c2+c2a2+2abc(a+b+c)=41
\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}⇔a2b2+b2c2+c2a2=41(vì a+b+c=0)
Ta có : a^2+b^2+c^2=1\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1a2+b2+c2=1⇔(a2+b2+c2)2=1⇔a4+b4+c4+2(a2b2+b2c2+c2a2)=1
\Leftrightarrow a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+c^2a^2\right)=1-\frac{2.1}{4}=\frac{1}{2}⇔a4+b4+c4=1−2(a2b2+b2c2+c2a2)=1−42.1=21
a) Ta có x + y = 25
=> (x + y)2 = 625
=> x2 + y2 + 2xy = 625
=> x2 + y2 + 10 = 625
=> x2 +y2 = 615
b) Ta có x + y = 3
=> (x + y)3 = 27
=> x3 + 3x2y + 3xy2 + y3 = 27
=> x3 + y3 + 3xy(x + y) = 27
=> x3 + y3 + 9xy = 27
Lại có x + y = 3
=> (x + y)2 = 9
=> x2 + y2 + 2xy = 9
=> 2xy = 4
=> xy = 2
Khi đó x3 + y3 + 9xy + 27
=> x3 + y3 + 18 = 27
=> x3 + y3 = 9
c) Ta có x - y = 5
=> (x - y)2 = 25
=> x2 + y2 - 2xy = 25
=> 2xy = -10
=> xy = -5
Khi đó : x3 - y3 = (x - y)(x2 + xy + y2) = 5(15 - 5) = 5.10 = 50
Bài 4.
a) x2 + y2 = x2 + 2xy + y2 - 2xy
= ( x2 + 2xy + y2 ) - 2xy
= ( x + y )2 - 2xy
= 252 - 2.136
= 625 - 272
= 353
b) x + y = 3
⇔ ( x + y )2 = 9
⇔ x2 + 2xy + y2 = 9
⇔ 5 + 2xy = 9 ( gt x2 + y2 = 5 )
⇔ 2xy = 4
⇔ xy = 2
x3 + y3 = x3 + 3x2y + 3xy2 + y3 - 3x2y - 3xy2
= ( x3 + 3x2y + 3xy2 + y3 ) - ( 3x2y + 3xy2 )
= ( x + y )3 - 3xy( x + y )
= 33 - 3.2.3
= 27 - 18
= 9