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Lời giải:
$2x^2-5x-6=2(x^2-\frac{5}{2}x+\frac{5^2}{4^2})-\frac{73}{8}$
$=2(x-\frac{5}{4})^2-\frac{73}{8}$
$=2(x-\frac{5}{4}-\frac{\sqrt{73}}{4})(x-\frac{5}{4}+\frac{\sqrt{73}}{4})$
\(x^4-5x^2+4=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
a: x^3-7x-6
=x^3-x-6x-6
=x(x-1)(x+1)-6(x+1)
=(x+1)(x^2-x-6)
=(x-3)(x+2)(x+1)
b: =2x^3+x^2-2x^2-x+6x+3
=x^2(2x+1)-x(2x+1)+3(2x+1)
=(2x+1)(x^2-x+3)
c: =2x^3-3x^2-2x^2+3x+2x-3
=x^2(2x-3)-x(2x-3)+(2x-3)
=(2x-3)(x^2-x+1)
d: =2x^3+x^2+2x^2+x+2x+1
=(2x+1)(x^2+x+1)
e: =3x^3+x^2-3x^2-x+6x+2
=(3x+1)(x^2-x+2)
f: =27x^3-9x^2-18x^2+6x+12x-4
=(3x-1)(9x^2-6x+4)
a) \(x^3-7x-6\)
\(=x^3-x-6x-6\)
\(=\left(x^3-x\right)-\left(6x+6\right)\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)\)
b) \(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(x^2-x+3\right)\left(2x+1\right)\)
c) \(2x^3-5x^2+5x+1\)
\(=2x^3-3x^2-2x^2+3x+2x-3\)
\(=\left(2x^3-3x^2\right)-\left(2x^2-3x\right)+\left(2x-3\right)\)
\(=x^2\left(2x-3\right)-x\left(2x-3\right)+\left(2x-3\right)\)
\(=\left(x^2-x+1\right)\left(2x-3\right)\)
d) \(2x^3+3x^2+3x+1\)
\(=2x^3+x^2+2x^2+x+2x+1\)
\(=\left(2x^3+x^2\right)+\left(2x^2+x\right)+\left(2x+1\right)\)
\(=x^2\left(2x+1\right)+x\left(2x+1\right)+\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(3x^3-2x^2+5x+2\)
\(=3x^3+x^2-3x^2-x+6x+2\)
\(=\left(3x^3+x^2\right)-\left(3x^2+x\right)+\left(6x+2\right)\)
\(=x^2\left(3x+1\right)-x\left(3x+1\right)+2\left(3x+1\right)\)
\(=\left(3x-1\right)\left(x^2-x+2\right)\)
f) \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
a) \(=\left(x^2-6\right)\left(x^2-1\right)=\left(x^2-6\right)\left(x-1\right)\left(x+1\right)\)
b) \(=\left(x^2-1\right)\left(x^2+3\right)=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
c) \(=x^2\left(x-1\right)-x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x^2-x+4\right)\)
\(=2x^2-6x+x-3=\left(x-3\right)\left(2x+1\right)\)
\(=2x^2-6x+x-3=2x\left(x-3\right)+\left(x-3\right)=\left(2x+1\right)\left(x-3\right)\)
= 2x^2 - x + 6x - 3
= x(2x - 1) + 3(2x - 1)
= (x + 3)(2x - 1)
\(2x^2+5x-3\)
\(=2x^2+6x-x-3\)
\(=2x\left(x+3\right)-\left(x+3\right)\)
\(=\left(2x-1\right)\left(x+3\right)\)
\(\left(5x^2-2x\right)^2+2x-5x^2-6=25x^4-2.5x^2.2x+4x^2+2x-5x^2-6.\)
\(=25x^4-20x^3-x^2+2x-6\)
\(=25x^4-25x^3+5x^3-5x^2+4x^2-4x+6x-6.\)
\(=\left(25x^4-25x^3\right)+\left(5x^3-5x^2\right)+\left(4x^2-4x\right)+\left(6x-6\right).\)
\(=25x^3\left(x-1\right)+5x^2\left(x-1\right)+4x\left(x-1\right)+6\left(x-1\right).\)
\(=\left(x-1\right)\left(25x^3+5x^2+4x+6\right)\)
\(=\left(x-1\right)\left[25x^3+15x^2-10x^2-6x+10x+6\right]\)
\(=\left(x-1\right)\left[5x^2\left(5x+3\right)-2x\left(5x+3\right)+2\left(5x+3\right)\right]\)
\(=\left(x-1\right)\left(5x+3\right)\left(5x^2-2x+2\right)\)
e ko bt phân tích đa thức thành nhân tử nên a tham khảo linh này nha
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwim29i-oIzyAhVSNKYKHZBdCJ4QFjAAegQIBRAD&url=https%3A%2F%2Fh7.net%2Fhoi-dap%2Ftoan-8%2Fphan-h-da-thuc-5x-2-2x-2-2x-5x-2-6-thanh-nhan-tu-faq341450.html&usg=AOvVaw2Kkix8idzI43uM1i2Mitp4
TL:
(5x^2 - 2x)^2+2x-5x^2-6
=(\(5x^2\)-2x+3)(\(5x^2\)-2x+2)