phân tích đa thuc thành nhân tử chung
\(4x^8+1\)
nhanh mình tick
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\(x^3-4x^2-12x+27=\left(x^3+27\right)-\left(4x^2+12x\right)\)\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\\ =\left(x+3\right)\left(x^2-3x+9-4x\right)=\left(x+3\right)\left(x^2-7x+9\right)\)
\(x^3-4x^2-12x+27\)
\(=x^3+3x^2-7x^2-21x+9x+27\)
\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
\(9x^6-12x^7+4x^8\)
\(=x^6\left(4x^2-12x+9\right)\)
\(=x^6.\left(2x-3\right)^2\)
hk
tốt
\(a,\left(a+b\right)^3+\left(a-b\right)^3\)
\(=\left(a+b+a-b\right)[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2]\)
\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)\)
\(=2a\left(a^2+3b^2\right)\)
\(b,9x^2+6xy+y^2\)
\(=\left(3x\right)^2+2.3x.y+y^2\)
\(=\left(3x+y\right)^2\)
\(c,4x^2-25\)
\(=\left(2x\right)^2-5^2\)
\(=\left(2x-5\right)\left(2x+5\right)\)
gợi ý thui, làm mêt lắm
4x2 - 1 = (2x)2 - 12
1- 2x = -(2x-1)
\(\left(4x^2-4x+1\right)-\left(x-1\right)^2\)
\(=\left(2x-1\right)^2-\left(x-1\right)^2\)
\(=\left(2x-1-x+1\right)\left(2x-1+x-1\right)\)
\(=x\left(3x-2\right)\)
\(4x^2-4x-35\) \(=\left(2x\right)^2-2.2x.1+1-36\)
\(=\left(2x-1\right)^2-6^2\)
\(=\left(2x-7\right)\left(2x+5\right)\)
\(18x^2-5x-2\) \(=\left(x-\frac{1}{2}\right)\left(x+\frac{2}{9}\right)\)
\(8x^3-26x^2+13x+5=\) \(8x^3-8x^2-18x^2+18x-5x+5\)
\(=8x^2\left(x-1\right)-18x\left(x-1\right)-5\left(x-1\right)\)
\(=\) \(\left(8x^2-18x-5\right)\left(x-1\right)\)
\(=\left(x-\frac{5}{2}\right)\left(x+\frac{1}{4}\right)\)\(\left(x-1\right)\)
4x2 - 4y2 + 4x - 12y - 8
= (2x)2 + 4x - (2y)2 - 12y - 8
= [ (2x)2 + 2.2x + 1 ] - [ (2y)2 + 2.2y.3 + 32 ]
= ( 2x + 1 )2 - ( 2y + 3)2
= ( 2x + 1 - 2y - 3 ) ( 2x + 1 + 2y + 3 )
= ( 2x - 2y - 2 ) ( 2x + 2y + 4 )
a, x2-7x-14y+2x
=x(x+2)-7(x-2y)
b, x3-4x2y+4xy2-25x
=x3-4x2y+4xy2-y3-25x+y3
=(x-y)3-25x+y3
a ) = x(x+2) - 7(x+2y)
b) = -4 xy ( x-y) + (x^3-25x) [ câu này mk , chaqcs là làm đúng đâu ]
\(4x^8+1\)
\(=\left(2x^4\right)^2+1\)
\(=\left(2x^4\right)^2+4x^4+1-4x^4\)
\(=\left(2x^4+1\right)^2-\left(2x^2\right)^2\)
\(=\left(2x^4+1-2x^2\right)\left(2x^4+1+2x^2\right)\)
\(4\times^8+1\)
\(=4\times^8+2\cdot2\times^4\cdot1+1-2\cdot2\times^4\cdot1\)
\(=\left(2\times^4+1\right)^2-\left(2\times^2\right)^2\)
\(=\left(2\times^4+1-2\times^2\right)\left(2\times^4+1+2\times^2\right)\)