a)x ∈ ∅

b) x=3

\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)

h) \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow x=2\)

haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=

a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)

\(x=\dfrac{1}{2}+\dfrac{1}{3}\)

\(x=\dfrac{1}{5}\)

Bài 1:

a, \(\left(x-2\right)^2=9\)

\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)

b, \(\left(3x-1\right)^3=-8\)

\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)

\(\Rightarrow x=-\dfrac{1}{3}\)

c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)

\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)

d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)

Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)

e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)

Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)

 dề bài đâu mà tìm?

6:ĐKXĐ: x>=0; x<>1/25

BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)

=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)

=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)

7:

ĐKXĐ: x>=0

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)

=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)

=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)

=>\(-\sqrt{x}-2>=0\)(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)

=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)

=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9: 

ĐKXĐ: x>=0

\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)

=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)

=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)

=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)

10: 

ĐKXĐ: x>=0; x<>1/49

\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)

=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)

=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)

=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

a: \(\left(\sqrt{3}\right)^x=243\)

=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)

=>\(\dfrac{1}{2}\cdot x=5\)

=>x=10

b: \(0,1^x=1000\)

=>\(\left(\dfrac{1}{10}\right)^x=1000\)

=>\(10^{-x}=10^3\)

=>-x=3

=>x=-3

c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)

=>\(\left(0,2\right)^{x+3}< 0,2\)

=>x+3>1

=>x>-2

d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)

=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)

=>2x+1<-2

=>2x<-3

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

e: \(5^{x-1}+5^{x+2}=3\)

=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)

=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)

=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)

1.

\(\left(\dfrac{-2}{3}\right).0,75+1\dfrac{2}{3}:\left(\dfrac{-4}{9}\right)+\left(\dfrac{-1}{2}\right)^2\)

\(=\left(\dfrac{-2}{3}\right).\dfrac{3}{4}+\dfrac{5}{3}.\left(\dfrac{9}{-4}\right)+\dfrac{1}{4}\)

\(=-\dfrac{1}{2}+\dfrac{45}{-12}+\dfrac{1}{4}\)

\(=-\dfrac{6}{12}+\dfrac{-45}{12}+\dfrac{3}{4}\)

\(=\dfrac{-48}{12}\)

\(=-4\)

2.

a) \(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)

\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{10}{20}\)

\(\Leftrightarrow x=\dfrac{-11}{20}\)

b) \(\left|x-\dfrac{2}{5}\right|+\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{11}{4}-\dfrac{3}{4}\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=-2\Rightarrow x=-2+\dfrac{2}{5}=\dfrac{-8}{5}\\x-\dfrac{2}{5}=2\Rightarrow x=2+\dfrac{2}{5}=\dfrac{12}{5}\end{matrix}\right.\)

3.

a) \(\dfrac{16}{2^n}=2\)

\(\Leftrightarrow2^n=16:2\)

\(\Leftrightarrow2^n=8\)

\(\Leftrightarrow2^n=2^3\)

\(\Leftrightarrow n=3\)

b) \(\dfrac{\left(-3\right)^n}{81}=-27\)

\(\Leftrightarrow\left(-3\right)^n=\left(-27\right).81\)

\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^3.\left(-3\right)^4\)

\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^7\)

\(\Leftrightarrow n=7\)

4. Ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\) (1)

\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\) (2)

Từ (1) và (2) suy ra \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)

Vì \(x-y+x=-49\) ta có:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)

Vậy \(\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)