Phan tích da thuc thanh nhan tu
0.125(a+2)^3-1
x^6-1
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\(x^6-2x^3+1=\left(x^3-1\right)^2\)
\(x^4+2x^2+1=\left(x^2+1\right)^2\)
a) x6 - 2x3 + 1
= (x3)2 - 2x3 + 1
= ( x3 - 1)2
b) x4 + 2x2 + 1
= ( x2)2 + 2x2 + 1
= ( x2 + 1)2
\(1-3x-x^3+3x^2\)\(=\left(1-x^3\right)+\left(3x^2-3x\right)\)
\(=\left(1-x\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-x^2-x-1\right)=\left(x-1\right)\left(2x-x^2-1\right)\)
\(a^6+a^4+a^2b^2+b^4-b^6\)
\(=(a^2)^3-(b^2)^3+(a^4+a^2b^2+b^4)\)
\(=(a^2-b^2)(a^4+a^2b^2+b^4)+(a^4+a^2b^2+b^4)\)
\(=(a^2-b^2+1)(a^4+a^2b^2+b^4)\)
\(=(a^4+2a^2b^2+b^4-a^2b^2)(a^2-b^2+1)\)
\(=(a^2+ab+b^2)(a^2-ab+b^2)(a^2-b^2+1)\)
\(a^6+a^2b^2+a^4+b^2-b^6\)
\(=a^4\left(a^2+b^2\right)+a^2\left(a^2+b^2\right)-b^6\)
\(=\left(a^2+b^2\right)+\left(a^4+a^2\right)-b^6\)
\(125\left(a+2\right)^3-1\)
\(=\left[5\left(a+2\right)\right]^3-1\)
\(=\left(5a+10\right)^3-1\)
\(=\left(5a+10-1\right)\left[\left(5a+10\right)^2-\left(5a+10\right)+1\right]\)
\(=\left(5a+10-1\right)\left[25a^2+100a+100-5a-10+1\right]\)
\(=\left(5a+9\right)\left[25a^2+95a+91\right]\)
b) \(x^6-1=\left(x^3\right)^2-1=\left(x^3-1\right)\left(x^3+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
0,125(a+2)3-1
= (0,5)3(a+2)3-1
= (0,5)3(a+2)3-13
k bt làm nx ==
x6-1 = (x3)2-12=(x3-1)(x3+1)