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d) x3-4x2-9x+36
=x2(x-4)-9(x-4)
=(x-4)(x2-9)
=(x-4)(x+3)(x-3)
e)(x+1)3+(2x-1)3
=x3+3x2+3x+1+8x3-12x2+6x-1
=9x3-9x2+9x
=9x(x2-x+1)
g)x3+3x2-4x-12
=x2(x+3)-4(x+3)
=(x+3)(x2-4)
=(x+3)(x+2)(x-2)
h) x3-4x2+4x-1
=x3-1-4x2+4x
=(x-1)(x2+x+1)-4x(x-1)
=(x-1)(x2+x+1-4x)
=(x-1)(x2-3x+1)
a: \(B=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{10}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)
b: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)
\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)
\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)
c: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=\dfrac{-5}{2}\)
d: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)
\(=\dfrac{1-4x^2}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x+4}\cdot\dfrac{3}{2\left(1-2x\right)}=\dfrac{3\left(2x+1\right)}{x+4}\)
\(a,a^3b-ab^3=ab\left(a^2-b^2\right)=ab\left(a-b\right)\left(a+b\right)\)
\(b,36x-x^3=x\left(36-x^2\right)=x\left(6^2-x^2\right)=x\left(6-x\right)\left(6+x\right)\)
\(c,x^2-12x+36=x^2-2\cdot x\cdot6+6^2=\left(x-6\right)^2\)
\(d,x^2-12x-x^2-36=-12x-36=-12\left(x+3\right)\)
\(e,4x^2-4x+1-y^2=\left(\left(2x\right)^2-2\cdot2x\cdot1+1^2\right)-y^2\)
\(=\left(2x-1\right)^2-y^2=\left(2x-1-y\right)\left(2x-1+y\right)\)
\(f,9-y^2+6x+x^2=\left(x^2+6x+9\right)-y^2=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)
CHÚC BẠN HỌC TỐT
a) \(a^3b-ab^3=ab\left(a^2-b^2\right)=ab\left(a-b\right)\left(a+b\right)\)
b) \(36x-x^3=x\left(36-x^2\right)=x\left(6-x\right)\left(6+x\right)\)
c) \(x^2-12x+36=x^2-6x-6x+36\)
\(=x\left(x-6\right)-6\left(x-6\right)\)
\(=\left(x-6\right)\left(x-6\right)=\left(x-6\right)^2\)
\(4x^2-28=0\)
\(\Leftrightarrow4\left(x^2-7\right)=0\)
\(\Leftrightarrow x^2-7=0\)
\(\Leftrightarrow x^2=7\)
\(\Leftrightarrow x=\pm\sqrt{7}\)
a)4(2x+1)\(^2\)+(4x+2)(2-6x)+(3x-1)\(^2\)=0
=>\(4\left(2x+1+3x-1\right)^2=0\)
=>\(4.\left(5x\right)^2=0\)
=>\(4.25x^2=0\)
=>\(100x^2=0\)
=>\(x^2=0\)
=>x=0
Vậy x =0
a) 2x2 - 98 = 0
2x2 = 0 + 98
2x2 = 98
x2 = 98 : 2
x2 = 49
x = \(\sqrt{49}\)
=> x = 7
Ta có : 2x2 - 98 = 0
=> 2(x2 - 49) = 0
Mà : 2 > 0
Nên x2 - 49 = 0
=> x2 = 49
=> x2 = -7;7