\(\frac{5^5}{5^x}=5^{18}\)

=>\(5^{5-x}=5^{18}\)

=>\(5-x=18\)

=>\(-x=13\)

=>\(x=-13\)

\(\frac{5^5}{5^x}=5^{18}\)

=> \(5^x=5^{18}:5^5=5^{13}=>x=13\)

\(-2^x=4^5.4^3=4^8=2^{16}\)

k có gt x nào

a) \(2x+5-\left(x-7\right)=18\)

\(2x+5-x+7=18\)

\(x+12=18\)

\(x=18-12\)

\(x=6\)

b) \(2\left(x+1\right)+4^2=2^4\)

\(2\left(x+1\right)+16=16\)

\(2\left(x+1\right)=16-16\)

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

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

\(x=0-1\)

\(x=-1\)

c) \(\frac{x-3}{x+5}=\frac{5}{7}\)

\(\Rightarrow\left(x-3\right)7=\left(x+5\right)5\)

\(7x-21=5x+25\)

\(7x-5x=25+21\)

\(2x=46\)

\(x=\frac{46}{2}\)

\(x=23\)

a. \(\frac{2x+3}{15}=\frac{7}{5}\)

\(\Leftrightarrow5\left(2x+3\right)=15.7\)

\(\Leftrightarrow10x+15=105\)

\(\Leftrightarrow10x=90\)

\(\Leftrightarrow x=9\)

b. \(\frac{x-2}{9}=\frac{8}{3}\)

\(\Leftrightarrow3\left(x-2\right)=9.8\)

\(\Leftrightarrow3x-6=72\)

\(\Leftrightarrow3x=78\)

\(\Leftrightarrow x=26\)

c. \(\frac{-8}{x}=\frac{-x}{18}\)

\(\Leftrightarrow-x^2=-144\)

\(\Leftrightarrow x^2=12^2\)

\(\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)

Mấy câu kia tương tự

d, \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=6x-12\Leftrightarrow4x=-27\Leftrightarrow x=-\frac{27}{4}\)

e, \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x=132\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)

f, \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x=10\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x+2\right)\left(2x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{5}{2}\end{cases}}\)

g, \(\left(2x-1\right)\left(2x+1\right)=63\Leftrightarrow4x^2+2x-2x-1=63\Leftrightarrow4x^2-64=0\)

\(\Leftrightarrow x^2=16\Leftrightarrow x=\pm4\)

h, \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow\left(10x+5\right)\left(x+1\right)=30\Leftrightarrow10x^2+10x+5x+5=30\)

\(\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(2x+5\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=1\end{cases}}\)

1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)

\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu

\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)

\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)

Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)

Bài 1b) có thể giải gọn hơn nhuư thế này

\(\frac{x}{x+4}=\frac{5}{6}=>6x=5\left(x+4\right)=5x+20\)

\(=>6x-5x=20=>x=20\)

áp dụng \(\frac{a}{b}=\frac{c}{d}< =>a.d=b.c\)
 

a) \(\frac{5^5}{5^x}=5^{18}\)

\(\Rightarrow5^x=5^{18}:5^5\)

\(\Rightarrow5^x=5^{13}\)

\(\Rightarrow x=13\)

Vậy \(x=13.\)

b) \(\frac{2^{4-x}}{16^5}=32^6\)

\(\Rightarrow2^{4-x}=32^6.16^5\)

\(\Rightarrow2^{4-x}=\left(2^5\right)^6.\left(2^4\right)^5\)

\(\Rightarrow2^{4-x}=2^{30}.2^{20}\)

\(\Rightarrow2^{4-x}=2^{50}\)

\(\Rightarrow4-x=50\)

\(\Rightarrow x=4-50\)

\(\Rightarrow x=-46\)

Vậy \(x=-46.\)

Chúc bạn học tốt!

Từ x:3=y:5 suy ra 4x:12=y:5 và 4x-y=14

Áp dụng tính chất của dãy tỉ số bằng nhau

x:3=y:5=4x-y:12-5=14:7=2

+)x:3=2 suy ra x=6

+)y:7=2 suy ra y=14

Vậy x=6;y=7

a, \(\left|x+\frac{1}{3}\right|=0\Leftrightarrow x=-\frac{1}{3}\)

b, \(\left|\frac{5}{18}-x\right|-\frac{7}{24}=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{18}-x=\frac{7}{24}\\\frac{5}{18}-x=-\frac{7}{24}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{72}\\x=\frac{41}{72}\end{cases}}\)

c, \(\frac{2}{5}-\left|\frac{1}{2}-x\right|=6\Leftrightarrow\left|\frac{1}{2}-x\right|=-\frac{28}{5}\)vô lí 

Vì \(\left|\frac{1}{2}-x\right|\ge0\forall x\)*luôn dương* Mà \(-\frac{28}{5}< 0\)

=> Ko có x thỏa mãn 

\(|x+\frac{1}{3}|=0\)

\(< =>x+\frac{1}{3}=0< =>x=-\frac{1}{3}\)

\(|x+\frac{3}{4}|=\frac{1}{2}\)

\(< =>\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)