Tính\(2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
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a) \(\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
c) \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
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\(x^3+3x^2+6x+4=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)\)
\(=\left(x+1\right)x^2+2x.\left(x+1\right)+4.\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+4\right)\)
a) \(x^3+3x^2+6x+4\)
\(=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+4\right)\)
b) \(3a^2c^2+bd+3abc+acd\)
\(=\left(3a^2c^2+acd\right)+\left(3abc+bd\right)\)
\(=ac\left(3ac+d\right)+b\left(3ac+d\right)\)
\(=\left(ac+b\right)\left(d+3ac\right)\)
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a) \(13,5\times5,8-8,3\times4,2-5,8\times8,3+4,2\times13,5\)
\(=13,5\times\left(5,8+4,2\right)-8,3\times\left(4,2+5,8\right)\)
\(=10\times\left(13,5-8,3\right)=10\times5,2=52\)
b) \(7,8\times55,1+92,2\times55,1-7,8\times5,1-92,2\times5,1\)
\(=55,1\times\left(7,8+99,2\right)-5,1\times\left(7,8+99,2\right)\)
\(=100\times\left(55,1-5,1\right)=100\times50=5000\)
c) \(N=a^3-a^2b-ab^2+b^3=\left(a-b\right)^2\left(a+b\right)\)
\(=\left(5,75-4,25\right)^2.\left(5,75+4,25\right)=1,5^2\times10=22,5\)
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\(16x^3-16x^4+4x-8x^2-1=0\)
<=> \(-16x^4-4x^2+16x^3+4x-4x^2-1=0\)
<=> \(-4x^2\left(4x+1\right)+4x\left(4x^2+1\right)-\left(4x^2+1\right)=0\)
<=> \(-\left(4x^2+1\right)\left(4x^2-4x+1\right)=0\)
<=> \(-\left(4x^2+1\right)\left(2x-1\right)^2=0\)
<=> \(2x-1=0\) (do 4x2 + 1 > 0 )
<=> \(x=\frac{1}{2}\)
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a) \(x^3+3x^2+3x+2=0\)
<=> \(x^3+x^2+x+2x^2+2x+2=0\)
<=> \(x\left(x^2+x+1\right)+2\left(x^2+x+1\right)=0\)
<=> \(\left(x+2\right)\left(x^2+x+1\right)=0\)
tự làm
b) \(x^4-2x^3+2x-1=0\)
<=> \(\left(x^4-3x^3+3x^2-x\right)+\left(x^3-3x^2+3x-1\right)=0\)
<=> \(x\left(x^3-3x^2+3x-1\right)+\left(x^3-3x^2+3x-1\right)=0\)
<=> \(\left(x^3-3x^2+3x-1\right)\left(x+1\right)=0\)
<=> \(\left(x-1\right)^3\left(x+1\right)=0\)
tự làm
c) \(x^4-3x^3-6x^2+8x=0\)
<=> \(x\left(x^3-3x^2-6x+8\right)=0\)
<=> \(x\left[\left(x^3+x^2-2x\right)-\left(4x^2+4x-8\right)\right]=0\)
<=>\(x\left[x\left(x^2+x-2\right)-4\left(x^2+x-2\right)\right]=0\)
<=> \(x\left(x-4\right)\left(x^2+x-2\right)=0\)
<=> \(x\left(x-4\right)\left(x-1\right)\left(x+2\right)=0\)
tự làm
Gợi ý:
Đặt: \(\frac{1}{105}=a;\) \(\frac{1}{651}=b\)
Như vậy \(\frac{1}{315}=3a;\)\(\frac{650}{651}=1-b\)
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