b) Ta có: \(\left(5-x\right)^3+27=0\)

\(\Leftrightarrow\left(5-x\right)^3=-27\)

\(\Leftrightarrow5-x=-3\)

hay x=8

d) Ta có: \(\left(x^2-1\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=-7\end{matrix}\right.\)

f) Ta có: \(\left(x^2+81\right)\left(x-7\right)\left(x^2-2\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=27\)

\(\Rightarrow x^3+3^3-x\left(x^2-4\right)=27\)

\(\Rightarrow x^3+27-x^3+4x=27\)

\(\Rightarrow27+4x=27\)

\(\Rightarrow4x=0\)

\(\Rightarrow x=0\)

b) \(2x^2+7x+3=0\)

\(\Rightarrow2x^2+x+6x+3=0\)

\(\Rightarrow x\left(2x+1\right)+3\left(2x+1\right)=0\)

\(\Rightarrow\left(2x+1\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=-1\\x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

c) Trùng đề bài a

d) \(2x^2+11x+9=0\)

\(\Rightarrow2x^2+2x+9x+9=0\)

\(\Rightarrow2x\left(x+1\right)+9\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(2x+9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x+9=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\2x=-9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{9}{2}\end{matrix}\right.\)

a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)

b: =>x-1/2=1/3

=>x=5/6

c: =>2/3x-1=0 hoặc 3/4x+1/2=0

=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3

d =>4/9:x=10/3:9/4=10/3*4/9=40/27

=>x=4/9:40/27=4/9*27/40=108/360=3/10

a: \(\Leftrightarrow\left(\dfrac{1}{3}x-1\right)^3=\left(\dfrac{1}{5}x-1\right)^3\)

=>1/3x-1=1/5x-1

=>2/15x=0

hay x=0

b: Đặt 2x+1=a; 3x-1=b

Theo đề, ta có \(\left(a+b\right)^3-a^3-b^3=0\)

=>3ab(a+b)=0

=>5x(2x+1)(3x-1)=0

hay \(x\in\left\{0;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)

c: Đặt x-3=a; x+1=b

Theo đề, ta có: \(\left(a+b\right)^3=a^3+b^3\)

=>3ab(a+b)=0

=>(x-3)(x+1)(2x-2)=0

hay \(x\in\left\{3;-1;1\right\}\)

2. \(a+b+c=0\)

\(\Leftrightarrow\)\(\left(a+b+c\right)^3=0\)

\(\Leftrightarrow a^3+b^3+c^3+3a^2b+3ab^2+3a^{2c}+3ac^2+3b^2c+3bc^2+6abc\)

\(\Leftrightarrow a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3a^2c+3ac^2+3abc\right)+\left(3b^2c+3bc^2+3abc\right)-3abc\)

\(\Leftrightarrow a^3+b^3+c^3+3ab\left(a+b+c\right)+3ac\left(a+c+b\right)+3bc\left(b+c+a\right)-3abc\)

Ta có: \(a+b+c=0\)

\(a^3+b^3+c^3+3ab.0+3ac.0+3bc.0=3abc\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\)

Bài 2

\(a+b+c=0\Rightarrow a=-b-c\)

\(VT=a^3+b^3+c^3=\left(-b-c\right)^3+b^3+c^3\)

\(=\left(-b\right)^3-3\left(-b\right)^2c+3\left(-b\right)c^2-c^3+b^3+c^3\)

\(=\left(-b\right)^3-3b^2c-3bc^2-c^3+b^3+c^3\)

\(=-3b^2c-3bc^2=3bc\left(-b-c\right)=3abc=VP\)

bài 2

ta có a+b+c=0

=>a+b=-c

=>c=-(a+b)

thay -(a+b)=c vào 2 vế ta đc

a³+b³-(a+b)³=3ab[-a-b)]

=>a³+b³-(a³+3a²b+3ab²+b³)=-3a²b-3ab²

=>a3+b3-a3-3a2b-3ab2-b2=-3ab(a-b)

=>(a3-a3)+(b3-b3)+(-3a2b-3ab2)=-3ab(a-b)

=>0+0-3ab(a-b)=-3ab(a-b)(đpcm)