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\(5x^3-30x^2+45x\)
\(=5x\left(x^2-6x+9\right)\)
\(=5x\left(x-3\right)^2\)
5x\(^3\)-30x\(^2\)+45x
=5x ( \(x^2\) - 6x + 9 )
= 5x ( \(x^2\) - 6x + \(3^2\))
= 5x \(\left(x-3\right)^2\)
Chúc bạn học tốt
a/ 5x3-45x=5x(x2-9)=5x(x+3)(x-3)
b/7x3-7=7(x3-1)=7(x-1)(x2+x+1)
c/5x2y-30xy2+45y3=5y(x2-6xy+9y2)=5y(x-3y)2
\(\left(5x^4-45x^2y^2\right)+\left(-3x^3y+27xy^3\right)=5x^2\left(x^2-9y^2\right)-3xy\left(x^2-9y^2\right)=\left(x^2-9y^2\right)\left(5x^2-3xy\right)\)
\(5x^3-30x^2+45x=5x\left(x^2-6x+9\right)=5x\left(x-3\right)^2\)
a: =>5x-21-3x-12+(x-5)(x+4)=80+x2
\(\Leftrightarrow x^2-x-20+2x-33=x^2+80\)
=>x-53=80
hay x=133
b: \(\Leftrightarrow\left(\dfrac{1}{5}x-\dfrac{2}{3}\right)\cdot\left(\dfrac{4}{3}x^2+1\right)\cdot\dfrac{1}{6}=\dfrac{22}{45}:\dfrac{4}{5}=\dfrac{11}{18}\)
\(\Leftrightarrow\left(\dfrac{1}{5}x-\dfrac{2}{3}\right)\left(\dfrac{4}{3}x^2+1\right)=\dfrac{11}{3}\)
\(\Leftrightarrow\dfrac{4}{15}x^3+\dfrac{1}{5}x-\dfrac{8}{9}x^2-\dfrac{2}{3}-\dfrac{11}{3}=0\)
\(\Leftrightarrow\dfrac{4}{15}x^3-\dfrac{8}{9}x^2+\dfrac{1}{5}x-\dfrac{13}{3}=0\)
\(\Leftrightarrow12x^3-40x^2+9x-195=0\)
hay \(x\in\left\{\dfrac{10+\sqrt{685}}{6};\dfrac{10-\sqrt{685}}{6}\right\}\)
a) Ta có: \(9\left(2a-x\right)^2-4\left(3a-x\right)^2\)
\(=\left(6a-3x\right)^2-\left(6a-2x\right)^2\)
\(=\left(6a-3x-6a+2x\right)\left(6a-3x+6a-2x\right)\)
\(=-x\cdot\left(12a-5x\right)\)
b) Ta có: \(5x^3-45x\)
\(=5x\left(x^2-9\right)\)
\(=5x\left(x-3\right)\left(x+3\right)\)
\(5x^3-45x=0\)
\(\Leftrightarrow5x\left(x^2-9\right)=0\)
\(\Leftrightarrow5x\left(x-3\right)\left(x+3\right)=0\)
\(TH1:5x=0\)
\(\Leftrightarrow x=0\)
\(TH2:x-3=0\)
\(\Leftrightarrow x=3\)
\(TH3:x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy phương trình có nghiệm: \(S=\left\{0;\pm3\right\}\)
5x3 - 45x = 0
=> 5x(x2- 9)=0
=>\(\orbr{\begin{cases}5x^2=0\\x^2-9=0\end{cases}}\)=>\(\orbr{\begin{cases}x^2=0\\x^2=9\end{cases}}\)=>\(\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=3\\x=-3\end{cases}}\end{cases}}\)\(\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=3\\x=-3\end{cases}}\end{cases}}\)
5x^3- 45x=0
5x(x^2-9)=0
=> 5x =0 => x=0
hoặc x^2-9=0
=> x^2= 9 => x=-3
hoặc x=3
vậy x= 0 hoặc 3 hoặc -3