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\(15\left(2a^2-1\right)+5\left(3-\frac{1}{5a}-6a^2\right)\)
\(=30a^2-15+15-\frac{1}{a}-30a^2\)
\(=-\frac{1}{a}\)
tại \(a=2017\)=> M= \(\frac{-1}{a}=\frac{-1}{2017}\)
\(\left(x-y\right)\left(x^2+xy+y^2\right)+y^3\)
\(=x^3-y^3+y^3\)
\(=x^3\)
ại \(x=2\)=> N= \(x^3=2^3=8\)
\(\left(x+1\right)\left(x^2-x-x^2+x-1\right)=-\left(x+1\right)\)
\(\left(2a^2+1\right)^2-4a^2-\left(2a^2+1\right)^2=-4a^2\)
\(\left(a^2+b^2+c^2+a^2-b^2-c^2\right)\left(a^2+b^2+c^2-a^2+b^2+c^2\right)=2a^2\left(2b^2+2c^2\right)=4a^2b^2+4a^2c^2\)
\(\left(a-5\right)^2\left(a+5\right)^2=\left(a^2-25\right)^2\)
\(\left(3a^3+1\right)^2-9a^2-\left(3a^3+1\right)^2=-9a^2\)
k ) \(125x^3-1\)
\(=\left(5x\right)^3-1\)
\(=\left(5x-1\right)\left[\left(5x\right)^2+5x.1+1^2\right]\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
m ) \(x^6-y^3=\left(x^2\right)^3-y^3=\left(x^2-y\right).\left[\left(x^2\right)^2+x^2.y+y^2\right]=\left(x^2-y\right).\left(x^4+x^2y+y^2\right)\)
n ) \(a^4-2a^2+1\)
\(=\left(a^2\right)^2-2.a^2.1+1^2=\left(a^2-1\right)^2\)
i ) \(a^3+6a^2+12a+8\)
\(=\left(a+2\right)^3\)
k) \(125x^3-1=\left(5x\right)^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
m) \(x^6-y^3=\left(x^2\right)^3-y^3=\left(x^2-y\right)\left(x^4+x^2y+y^2\right)\)
n) \(a^4-2a^2+1=\left(a^2-1\right)^2=\left(a^2-1\right)\left(a^2-1\right)=\left(a-1\right)\left(a+1\right)\left(a-1\right)\left(a+1\right)\)
i) \(a^3+6a^2+12a+8=\left(a+2\right)^2\)
b)\(x^3-6x^2+12x-8-\left(x^3-6x^2\right)\)
<-> \(x^3-6x^2+12x-8-x^3+6x^2\)
<->12x-8
d)\(x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)\)
\(x^3+6x^2+12x+8-x^3+6x^2-12x+8\)
\(12x^2+16\)
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:
\(B=10^3-\left(-1\right)^3=1000+1=1001\)
a) \(A=\left(x-y\right)^2+\left(x+y\right)^2\)
\(=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)
\(=x^2-2xy+y^2+x^2+2xy+y^2\)
\(=\left(x^2+x^2\right)-\left(2xy-2xy\right)+\left(y^2+y^2\right)\)
\(=2x^2+2y^2\)
\(=2.\left(x^2+y^2\right)\)
b) \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)
\(=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(=\left(4a^2-4a^2\right)+\left(4ab+4ab\right)+\left(b^2-b^2\right)\)
\(=8ab\)\
c) \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(=x^2+2xy+y^2-x^2+2xy-y^2\)
\(=\left(x^2-x^2\right)+\left(2xy+2xy\right)+\left(y^2-y^2\right)\)
\(=4xy\)
d) \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+1-8x^2+24x-18+4\)
\(=\left(4x^2-8x^2\right)-\left(4x-24x\right)+\left(1-18+4\right)\)
\(=-4x^2+20x-13\)
\(=-4x^2+20x-25+12\)
\(=-\left(4x^2-20x+25\right)-8\)
\(=-\left[\left(2x\right)^2-2.4x.5+5^2\right]-8\)
\(=-\left(2x-5\right)^2-8\)
a,hđt số 3 = \(\left(a^2+2a\right)^2-9\)
b,hđt số 3=\(\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)(đổi dấu làm ngoặc khi trước nó là dấu trừ)=\(x^2-\left(y-6\right)^2\)
a) \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)
\(=\left(a^2+2a\right)^2+3.\left(-3\right)\)
\(=\left(a^2+2a\right)^2-9\)
b) \(\left(x-y+6\right)\left(x+y-6\right)\)
\(=\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)
\(=x^2-\left(y-6\right)^2\)