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a) \(A=3x\left(x^2-2x+3\right)-x^2.\left(3x-2\right)+5\left(x^2-x\right)\)
\(=3x^3-6x^2+9x-3x^3+2x^2+5x^2-5x\)
\(=x^2+4x\)
Thay \(x=5\)vào biểu thức ta có: \(A=5^2+4.5=25+20=45\)
b) \(B=x\left(x^2+xy+y^2\right)-y\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
Thay \(x=10\); \(y=-1\)vào biểu thức ta có:
\(B=10^3-\left(-1\right)^3=1000+1=1001\)
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) \(\left(x-1\right)\left(x^2+x+1\right)=x\left(x^2+x+1\right)-\left(x^2+x+1\right)\)
\(=x^3+x^2+x-x^2-x-1=x^3-1\) đpcm
b) \(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)=\left(x-y\right)\left[x\left(x^2+y^2\right)+y\left(x^2+y^2\right)\right]\)
\(=\left(x-y\right)\left(x^3+xy^2+x^2y+y^3\right)\) đpcm
a)\(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)^2\left(x^2-x+1\right)\)
b)\(x^4-x^3-x^2+1=\left(x^4-x^3\right)-\left(x^2-1\right)=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
c)\(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)
a) \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)
\(A=y\left(x^4-y^4\right)-y\left(y^4-y^4\right)=0\)
=> đpcm
b) \(B=\left(\frac{1}{3}+2x\right)\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)-\left(8x^3-\frac{1}{27}\right)\) (đã sửa đề)
\(B=\left(\frac{1}{27}+8x^3\right)-\left(8x^3-\frac{1}{27}\right)\)
\(B=\frac{2}{27}\)
=> đpcm
c) \(C=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x\) (đã sửa đề)
\(C=x^3-3x^2+3x-1-x^3+1+3x^2-3x\)
\(C=0\)
=> đpcm
1)a)x+y=60
<=>(x+y)^2=3600
<=>x^2+2xy+y^2=3600(1)
mà xy=35 nên 2xy=2.35=70
(1)<=>x^2+70+y^2=3600
<=>x^2+y^2=3530
<=>(x^2+y^2)^2=12460900
<=>x^4+2x^2.y^2+y^4=12460900(2)
mà xy=35 nên 2x.x.y.y=2450
(2)<=>x^4+y^4=123458450
b)x+y=1
<=>(x+y)^3=1
<=>x^3+3x^2y+3xy^2+y^3=1
<=>x^3+y^3+3xy(x+y)=1
<=>x^3+y^3+3xy=1
=>M=1
x+y=1
<=>x^2+2xy+y^2=1(1)
B=x^3+y^3+3xy(x^2+y^2)+3xy(2xy)
=x^3+y^3+3xy(x^2+2xy+y^2)
=M.1=1(từ(1)
c)
x-y=1
<=>(x-y)^3=1
<=>x^3-3x^2y+3xy^2-y^3=1
<=>x^3-y^3-3xy(x-y)=1
<=>x^3-y^3-3xy=1
=>N=1
a, (x+y)(x2-xy+y2)(x3-y3)=(x3+y3)(x3-y3)=x6-y6
b, (x-2)(x+2)(x2+4)-(x2+1)(x2-1)=(x2-4)(x2+4)-(x4-1)=x4-16-x4+1=-15