=>(x+8)(x-1)=0

=>x=-8 hoặc x=1

a) không có số nào

B) chịu

sao lạ vại men, ko còn ý kiến nào hay hơn à

U LÀ TRỜI

\(3^{x+1}-2.3^x=243\\ \Rightarrow3^x.3-2.3^x=243\\ \Rightarrow3^x=3^5\\ \Rightarrow x=5\)

\(d.3^{x+1}-2.3^x=243\)

\(x+\left\{\left\{x+3\right\}-\left[\left(x+3\right)-\left(-x-2\right)\right]\right\}=x\)

\(x+\left\{\left\{x+3\right\}-\left[x+3-x+2\right]\right\}=x\)

\(x+\left\{\left\{x+3\right\}-5\right\}=x\)

\(x-2=0\)

\(x=2\)

X=2  nhé bạn

\(x\left(x+2\right)\left(x+4\right)\left(x+6\right)=9\)

\(\Leftrightarrow x\left(x+2\right)\left(x+4\right)\left(x+6\right)-9=0\)

Đặt \(A=x\left(x+2\right)\left(x+4\right)\left(x+6\right)-9\)

\(A=x\left(x+6\right)\left(x+2\right)\left(x+4\right)-9\)

\(=\left(x^2+6x\right)\left(x^2+6x+8\right)-9\)(1)

Đặt \(a=x^2+6x\)

\(\Rightarrow\left(1\right)=a\left(a+8\right)-9=a^2+8a-9\)

\(=\left(a+4\right)^2-25=\left(a+4-5\right)\left(a+4+5\right)\)

\(=\left(a-1\right)\left(a+9\right)=\left(x^2+6x-1\right)\left(x^2+6x+9\right)\)

\(=\left(x^2+6x-1\right)\left(x+3\right)^2\)

\(pt\Leftrightarrow\left(x^2+6x-1\right)\left(x+3\right)^2=0\)

\(TH1:x^2+6x-1=0\)

\(\Leftrightarrow\left(x+3\right)^2-10=0\)

\(\Leftrightarrow\left(x+3\right)^2=10\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=\sqrt{10}\\x+3=-\sqrt{10}\end{cases}}\Leftrightarrow x=\pm\sqrt{10}+3\)

\(TH2:\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

Vậy \(x\in\left\{\pm\sqrt{10}+3;-3\right\}\)

\(\left(\dfrac{1}{8}\right)^9\cdot8^9-\left(\dfrac{3}{4}-2x\right)^2=\dfrac{41}{9}-\dfrac{72^2}{36^2}\\ \Rightarrow\left(\dfrac{1}{8}\cdot8\right)^9-\left(\dfrac{3}{4}-2x\right)^2=\dfrac{41}{9}-\left(\dfrac{72}{36}\right)^2\\ \Rightarrow1^9-\left(\dfrac{3}{4}-2x\right)^2=\dfrac{41}{9}-2^2=\dfrac{5}{9}\\ \Rightarrow\left(\dfrac{3}{4}-2x\right)^2=1-\dfrac{5}{9}=\dfrac{4}{9}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}-2x=\dfrac{2}{3}\\2x-\dfrac{3}{4}=\dfrac{2}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{12}\\2x=\dfrac{17}{12}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{24}\\x=\dfrac{17}{24}\end{matrix}\right.\)

\(x^3=x^2\)

\(\Leftrightarrow x^3-x^2=0\)

\(\Leftrightarrow x^2\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Vậy x = 0 hoặc x = 1

x=o hoặc x=1

Hok

tốt 

nha 

bạn

Nếu 

x^3=x^2

Thì có 2 trường hợp

TH1: x=1

x^3=x^2=1

TH2: x=0

0^3=0^2=0

Vậy x=1,0