8x3 - 1 phần 8
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\(\frac{7}{8}+\frac{1}{8}\times\frac{3}{8}+\frac{1}{8}\times\frac{5}{8}=\frac{7}{8}+\frac{1}{8}\times\left(\frac{3}{8}+\frac{5}{8}\right)=\frac{7}{8}+\frac{1}{8}\times1\)
\(=\frac{7}{8}+\frac{1}{8}=1\)
`a) (243756000)/(-487512000)=243756000/(-2.243756000)=1/(-2)=(-1)/2`
`b) (3.16-3.14)/(-12)=(3.2)/(3.(-4))=2/(-4)=(-1)/2`
`c) (-8.4+8.3)/(8.6-8.4)=(-8)/(8.2)=(-1.8)/(2.8)=(-1)/2`
`d) (-15)/(-90)=(-15.1)/(-15.6)=1/6`
\(8x^3-\frac{1}{8}\)
\(=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)
\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
tích hộ
a) \(A=-x^3+6x^2-12x+8\)
\(A=-\left(x^3-6x^2+12x-8\right)\)
\(A=-\left(x-2\right)^3\)
Thay x=-28 vào A ta có:
\(A=-\left(-28-2\right)^3=27000\)
Vậy: ...
b) \(B=8x^3+12x^2+6x+1\)
\(B=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3\)
\(B=\left(2x+1\right)^3\)
Thay \(x=\dfrac{1}{2}\) vào B ta có:
\(B=\left(2\cdot\dfrac{1}{2}+1\right)^3=8\)
Vậy: ...
1, \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)
2, \(x^2-10x+25=\left(x-5\right)^2\)
3, \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
4, \(x^2+4xy+4y^2=\left(x+2y\right)^2\)
1) \(a^6+b^3=\left(a^2\right)^3+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)
2) \(x^2-10x+25=\left(x-5\right)^2\)
3) \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{3}\right)^3=\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2x}{3}+\dfrac{1}{4}\right)\)
4) \(x^2+4xy+4y^2=\left(x+2y\right)^2\)
\(=8\left(x^3-\dfrac{1}{64}\right)=8\left(x-\dfrac{1}{4}\right)\left(x^2+\dfrac{1}{4}x+\dfrac{1}{16}\right)\)