\(\left(x^2-9\right)^2-9\left(x-3\right)^2=0\)

\(< =>\left(x^2-9\right)^2-\left[3\left(x-3\right)\right]^2=0\)

\(< =>\left(x^2-9\right)^2-\left(3x-9\right)^2=0\)

\(< =>\left(x^2-9+3x-9\right)\left(x^2-9-3x+9\right)=0\)

\(< =>\left(x^2+3x-18\right)\left(x^2-3x\right)=0\)

\(=>\left[{}\begin{matrix}x^2+3x-18=0\\x^2-3x=0\end{matrix}\right.< =>\left[{}\begin{matrix}\left(x+6\right)\left(x-3\right)=0\\x\left(x-3\right)=0\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=-6\\x=3\\x=0\end{matrix}\right.\)

Nhanh đấy tốc độ đấy =))

a: \(3\frac{4}{27}+\frac{11}{21}+0,5-\frac{4}{27}+\frac{10}{21}\)

\(=\left(3+\frac{4}{27}-\frac{4}{27}\right)+0,5+\left(\frac{11}{21}+\frac{10}{11}\right)\)

=3+0,5+1

=4,5

b: \(6-3\cdot\left(-\frac13\right)^3-\sqrt9\)

\(=6-3\cdot\frac{-1}{27}-3\)

\(=3+\frac19=\frac{27}{9}+\frac19=\frac{28}{9}\)

c: \(\left(5,3-2,8\right)-\left(4+5,3\right)+0,2\)

=5,3-2,8-4-5,3+0,2

=-2,8-4+0,2

=-6,8+0,2

=-6,6

d: \(0,5\cdot\sqrt{100}-\sqrt{\frac14}=0,5\cdot10-\frac12=5-0,5=4,5\)

e: \(13\frac12\cdot\frac57-\frac57\cdot27\frac12\)

\(=\frac57\left(13+\frac12-27-\frac12\right)\)

\(=\frac57\cdot\left(-14\right)=-10\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27-28-\frac27\right):\left(-\frac35\right)\)

\(=-12\cdot\left(-\frac53\right)=20\)

g: \(\sqrt{0,01}-\sqrt{0,25}\)

=0,1-0,5

=-0,4

  5,3<x<5,4

=>x=5,31 ;5,32 ; 5;33

\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^x-\left(x-3\right)^{10}\right]=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-3=0\\\left(x-3\right)^x-\left(x-3\right)^{10}=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\\left(x-3\right)^x=\left(x-3\right)^{10}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=10\end{array}\right.\)

Vậy \(x\in\left\{3;10\right\}\)

\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^x-\left(x-3\right)^9\right]=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-3=0\\\left(x-3\right)^x-\left(x-3\right)^9=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\\left(x-3\right)^x=\left(x-3\right)^9\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=9\end{array}\right.\)

Vậy \(x\in\left\{3;9\right\}\)

PT bậc nhất có dạng `ax+b`

`=>` PT bậc nhất là a và b và c

a , b và c nhé bạn

{6;-6}