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HQ
Hà Quang Minh
Giáo viên
24 tháng 8 2023

\(a,3^{x-1}=27\\ \Leftrightarrow3^{x-1}=3^3\\ \Leftrightarrow x-1=3\\ \Leftrightarrow x=4\\ b,100^{2x^2-3}=0,1^{2x^2-18}\\ \Leftrightarrow10^{4x^2-6}=10^{-2x^2+18}\\ \Leftrightarrow4x^2-6=-2x^2+18\\ \Leftrightarrow6x^2=24\\ \Leftrightarrow x^2=4\\ \Leftrightarrow x=\pm2\)

HQ
Hà Quang Minh
Giáo viên
24 tháng 8 2023

\(c,\sqrt{3}e^{3x}=1\\ \Leftrightarrow e^{3x}=\dfrac{1}{\sqrt{3}}\\ \Leftrightarrow3x=ln\left(\dfrac{1}{\sqrt{3}}\right)\\ \Leftrightarrow x=\dfrac{1}{3}ln\left(\dfrac{1}{\sqrt{3}}\right)\)

\(d,5^x=3^{2x-1}\\ \Leftrightarrow2x-1=log_35^x\\ \Leftrightarrow2x-1-xlog_35=0\\ \Leftrightarrow x\left(2-log_35\right)=1\\ \Leftrightarrow x=\dfrac{1}{2-log_35}\)

29 tháng 8 2019

a) x - 26 : 13 = 2017

             x - 2 = 2017

                  x = 2017 + 2

                  x = 2019

b) \(\frac{2}{3}+\frac{1}{3}.\left(x+28\right)=18\)

                \(\frac{1}{3}.\left(x+28\right)=18-\frac{2}{3}\)

                \(\frac{1}{3}.\left(x+28\right)=\frac{54}{3}-\frac{2}{3}\)

                \(\frac{1}{3}.\left(x+28\right)=\frac{52}{3}\)

                             \(x+28=\frac{52}{3}:\frac{1}{3}\)

                             \(x+28=52\)

                                         \(x=52-28\)

                                         \(x=24\) 

c) \(\frac{7}{8}.\left(x-27\right)=\frac{9}{8}-0,125\)

   \(\frac{7}{8}.\left(x-27\right)=1\)

                \(x-27=1:\frac{7}{8}\)

                \(x-27=\frac{8}{7}\)

                            \(x=\frac{8}{7}+27\)

                            \(x=\frac{8}{7}+\frac{189}{7}\)

                           \(x=\frac{197}{7}\)

\(a,9\left(2x+1\right)=4\left(x-5\right)^2\)

\(4x^2-40x+100=18x+9\)

\(4x^2-58x+91=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{29+3\sqrt{53}}{4}\\x=\frac{29-3\sqrt{53}}{4}\end{cases}}\)

\(b,x^3-4x^2-12x+27=0\)

\(\left(x+3\right)\left(x^2-7x+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}}\)

\(c,x^3+3x^2-6x-8=0\)

\(\left(x+4\right)\left(x-2\right)\left(x+1\right)=0\)

\(Th1:x+4=0\Leftrightarrow x=-4\)

\(Th2:x-2=0\Leftrightarrow x=2\)

\(Th3:x+1=0\Leftrightarrow x=-1\)

5 tháng 3 2020

\(a,9.\left(2x+1\right)=4.\left(x-5\right)^2\)

\(< =>4x^2-40x+100=18x+9\)

\(< =>4x^2+58x+91=0\)

\(< =>\orbr{\begin{cases}x=\frac{29-3\sqrt{53}}{4}\\x=\frac{29+3\sqrt{53}}{4}\end{cases}}\)

\(b,x^3-4x^2-12x+27=0\)

\(< =>\left(x+3\right)\left(x^2-7x+9\right)=0\)

\(< =>\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}\)

a: \(27^{2-x}< =9\)

=>\(\left(3^3\right)^{2-x}< =3^2\)

=>\(3^{6-3x}< =3^2\)

=>6-3x<=2

=>-3x<=-4

=>\(x>=\dfrac{4}{3}\)

b: \(7^{3-x}< 49\)

=>\(7^{3-x}< 7^2\)

=>3-x<2

=>-x<2-3=-1

=>x>1

c: \(27^{3-x}>9\)

=>\(\left(3^3\right)^{3-x}>3^2\)

=>\(3^{9-3x}>3^2\)

=>9-3x>2

=>-3x>-7

=>\(x< \dfrac{7}{3}\)

d: \(2^{3-x}< 2^3\)

=>3-x<3

=>-x<0

=>x>0

e: \(27^{3-x^2}< 27^{x+1}\)

=>\(3-x^2< x+1\)

=>\(-x^2-x+2< 0\)

=>\(x^2+x-2>0\)

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

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

27 tháng 5 2021

Đk: \(x\ge-3\)

Pt \(\Leftrightarrow4\left(x^2+18\right)^2=49\left(x^3+27\right)\)

\(\Leftrightarrow4x^4-49x^3+144x^2-27=0\)

\(\Leftrightarrow\left(x^2-7x-3\right)\left(4x^2-21x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+\sqrt{61}}{2}\\x=\dfrac{7-\sqrt{61}}{2}\\x=\dfrac{21+3\sqrt{33}}{8}\\x=\dfrac{21-3\sqrt{33}}{8}\end{matrix}\right.\)

Vậy...

27 tháng 5 2021

ĐKXĐ: \(x\ge-3\).

\(PT\Leftrightarrow2\left(x^2+18\right)=7\sqrt{\left(x+3\right)\left(x^2-3x+9\right)}\). (*)

Đặt \(\sqrt{x+3}=a;\sqrt{x^2-3x+9}=b\left(a,b\ge0\right)\).

\(\left(\cdot\right)\Leftrightarrow2\left(b^2+3a^2\right)=7ab\Leftrightarrow6a^2-7ab+2b^2=0\)

\(\Leftrightarrow\left(3a-2b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}3a=2b\\2a=b\end{matrix}\right.\).

+) \(3a=2b\Leftrightarrow3\sqrt{x+3}=2\sqrt{x^2-3x+9}\Leftrightarrow4\left(x^2-3x+9\right)=9\left(x+3\right)\Leftrightarrow4x^2-12x+36=9x+27\Leftrightarrow4x^2-21x+9=0\Leftrightarrow x=\dfrac{21\pm3\sqrt{33}}{8}\). (TMĐK)

+) \(2a=b\Leftrightarrow4\left(x+3\right)=x^2-3x+9\Leftrightarrow x^2-7x-3=0\Leftrightarrow x=\dfrac{7\pm\sqrt{61}}{2}\left(TMĐK\right)\)

Vậy...

24 tháng 3 2020

\(\frac{1}{x^2+3}+\frac{1}{x^2+9x+18}+\frac{1}{x^2+15x+54}=\frac{1}{2}\left(27-\frac{1}{x+9}\right)\)

\(\Leftrightarrow\frac{3}{x\left(x+3\right)}+\frac{3}{\left(x+3\right)\left(x+6\right)}+\frac{3}{\left(x+6\right)\left(x+9\right)}=27-\frac{1}{x+9}\)

Mà 

\(\frac{3}{x\left(x+3\right)}+\frac{3}{\left(x+3\right)\left(x+6\right)}+\frac{3}{\left(x+6\right)\left(x+9\right)}\)

\(=\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\)

\(=\frac{1}{x}-\frac{1}{x+9}\)

\(\Rightarrow\frac{1}{x}=27\Rightarrow x=\frac{1}{27}\)

\(\Leftrightarrow3\sqrt{x+2}-x-6+5\sqrt{x+18}-21=0\)

=>\(3\sqrt{x+2}-9+5\sqrt{x+18}-x-18=0\)

=>\(3\left(\sqrt{x+2}-3\right)+\sqrt{x+18}\left(5-\sqrt{x+18}\right)=0\)

=>\(3\cdot\dfrac{x+2-9}{\sqrt{x+2}+3}+\sqrt{x+18}\cdot\dfrac{25-x-18}{5+\sqrt{x+18}}=0\)

=>\(\left(x-7\right)\cdot\left(\dfrac{3}{\sqrt{x+2}+3}-\dfrac{\sqrt{x+18}}{5+\sqrt{x+18}}\right)=0\)

=>x-7=0

=>x=7