Bài 4:

a) \(\dfrac{x}{2}=\dfrac{2}{4}\)

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

\(\Rightarrow x=2\)

Vậy: \(x=2\)

b) \(-\dfrac{1}{5}=\dfrac{2}{x}\)

\(\Rightarrow x=\dfrac{-5.2}{1}=-10\)

Vậy: \(x=-10\)

c) \(\dfrac{x}{5}=\dfrac{5}{x}\)

\(\Rightarrow x^2=25\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{5;-5\right\}\)

\(\dfrac{x}{2}=\dfrac{2}{4}\\ =>x=\dfrac{2.2}{4}=1\)

\(\dfrac{-1}{5}=\dfrac{2}{x}\\ =>x=\dfrac{2.5}{-1}=-10\)

\(\dfrac{x}{5}=\dfrac{5}{x}\\ =>x^2=25\\ x=5;x=-5\)

a) 2

b) 5,-5

Giải:

a) \(2^5=4^x\) 

\(\Rightarrow2^5=\left(2^2\right)^x\) 

\(\Rightarrow2^5=2^{2x}\) 

\(\Rightarrow2x=5\) 

\(\Rightarrow x=\dfrac{5}{2}\) 

b) \(2.4^2.8^3.16^4=8^x\) 

\(\Rightarrow2.\left(2^2\right)^2.\left(2^3\right)^3.\left(2^4\right)^4=\left(2^3\right)^x\) 

\(\Rightarrow2.2^4.2^9.2^{16}=2^{3x}\) 

\(\Rightarrow2^{30}=2^{3x}\) 

\(\Rightarrow3x=30\) 

\(\Rightarrow x=30:3\) 

\(\Rightarrow x=10\) 

c) \(3^3:3^5=9^x\) 

\(\Rightarrow3^{-2}=\left(3^2\right)^x\) 

\(\Rightarrow3^{-2}=3^{2x}\) 

\(\Rightarrow2x=-2\) 

\(\Rightarrow x=-2:2\)

\(\Rightarrow x=-1\) 

Chúc bạn học tốt!

a) Ta có: \(2^5=4^x\)

nên \(2^{2x}=2^5\)

\(\Leftrightarrow2x=5\)

hay \(x=\dfrac{5}{2}\)

b) Ta có: \(2\cdot4^2\cdot8^3\cdot16^4=8^x\)

\(\Leftrightarrow2^{3x}=2\cdot2^5\cdot2^9\cdot2^{16}=2^{31}\)

\(\Leftrightarrow3x=31\)

hay \(x=\dfrac{31}{3}\)

c) Ta có: \(3^3:3^5=9^x\)

\(\Leftrightarrow3^{-2}=3^{2x}\)

\(\Leftrightarrow2x=-2\)

hay x=-1

a: =>1/3x-2/5x=5

=>-1/15x=5

=>x=-75
b: =>4x=4

=>x=1

c: =>6*3^x-5*3^x=243

=>3^x=243

=>x=5

a) Ta có: \(3\left(2-x\right)+1=4-2x\)

\(\Leftrightarrow6-3x+1-4+2x=0\)

\(\Leftrightarrow-x+3=0\)

\(\Leftrightarrow-x=-3\)

hay x=3

Vậy: S={3}

b) Ta có: \(2\left(x+4\right)=3-x\)

\(\Leftrightarrow2x+8-3+x=0\)

\(\Leftrightarrow3x+5=0\)

\(\Leftrightarrow3x=-5\)

hay \(x=-\dfrac{5}{3}\)

Vậy: \(S=\left\{-\dfrac{5}{3}\right\}\)

c) Ta có: \(7-3x=x-5\)

\(\Leftrightarrow7-3x-x+5=0\)

\(\Leftrightarrow-4x+12=0\)

\(\Leftrightarrow-4x=-12\)

hay x=3

Vậy: S={3}

d) Ta có: \(5x-\left(x-1\right)=7\)

\(\Leftrightarrow5x-x+1=7\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)

a, \(\left(x+2\right)\left(x+4\right)-x^2=24\\ \Rightarrow x^2+6x+8-x^2=24\\ \Rightarrow6x+8=24\\ \Rightarrow6x=16\\ \Rightarrow x=\dfrac{8}{3}\)

b, \(\left(x+5\right)\left(x-5\right)=x^2+x\)

\(\Rightarrow x^2+x-\left(x+5\right)\left(x-5\right)=0\)

\(\Rightarrow x^2+x-x^2+25=0\\ \Rightarrow x+25=0\\ \Rightarrow x=-25\)

\(a,< =>x^2+4x+2x+8-x^2=24< =>6x+8=24< =>x=\dfrac{24-8}{6}=\dfrac{8}{3}\)

b,\(< =>x^2-25-x^2-x=0< =>-25-x=0< =>x=-25\)

c,\(< =>4x^2-9-4x^2+4x=0< =>4x-9=0< =>x=\dfrac{9}{4}\)

d,\(< =>x^3+2^3=9< =>x^3=1=>x=1\)

c,(2x+3)(2x-3)=4x(x-1)

⇔ 4x²-9=4x²-4x

⇔ 4x=9

\(\Leftrightarrow x=\dfrac{9}{4}\)

d, (x+2)(x²-2x+4)=9

⇔ x³+8=9

⇔ x³=1

⇔ x=1

\(a,\dfrac{6}{5}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{6}{5}\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\\ \Rightarrow x=-21:\dfrac{3}{4}\\ \Rightarrow x=-28\\ c,\dfrac{2}{-7}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{2}{-7}\\ \Rightarrow x=-63\\ d,\dfrac{-5}{2}=\dfrac{10}{-x}\\ \Rightarrow x=-10:\dfrac{-5}{2}\\ \Rightarrow x=4\)

\(a,\dfrac{6}{5}=\dfrac{18}{x}\Rightarrow6.x=5.18=90\\ \Rightarrow6.x=90\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\Rightarrow3.x=4.21=84\\ \Rightarrow x=28\)

\(a,\dfrac{6}{5}=\dfrac{18}{x}=>x=\dfrac{18.5}{6}=\dfrac{90}{6}=15\)

\(b,\dfrac{3}{4}=-\dfrac{21}{x}=>x=-\dfrac{21.4}{3}=-\dfrac{84}{3}=24\)

\(c,\dfrac{2}{-7}=\dfrac{18}{x}=>x=\dfrac{18.-7}{2}=-\dfrac{126}{2}=-63\)

\(d,-\dfrac{5}{2}=\dfrac{10}{-x}=>-x=\dfrac{10.2}{-5}=\dfrac{20}{-5}=-4=>x=4\)

c) \(\left(34-2x\right)\left(2x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6-0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)

d) \(\left(2019-x\right)\left(3x-12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)

e) \(57\left(9x-27\right)=0\)

\(\Rightarrow9x-27=0\)

\(\Rightarrow9\left(x-3\right)=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

f) \(25+\left(15-x\right)=30\)

\(\Rightarrow25+15-x=30\)

\(\Rightarrow40-x=30\)

\(\Rightarrow x=40-30\)

\(\Rightarrow x=10\)

g) \(43-\left(24-x\right)=20\)

\(\Rightarrow43-24+x=20\)

\(\Rightarrow19+x=20\)

\(\Rightarrow x=20-19\)

\(\Rightarrow x=1\)

h) \(2\left(x-5\right)-17=25\)

\(\Rightarrow2\left(x-5\right)=17+25\)

\(\Rightarrow x-5=21\)

\(\Rightarrow x=21+5\)

\(\Rightarrow x=26\)

i) \(3\left(x+7\right)-15=27\)

\(\Rightarrow3\left(x+7\right)=27+15\)

\(\Rightarrow x+7=14\)

\(\Rightarrow x=14-7\)

\(\Rightarrow x=7\)

j) \(15+4\left(x-2\right)=95\)

\(\Rightarrow4\left(x-2\right)=95-15\)

\(\Rightarrow4\left(x-2\right)=80\)

\(\Rightarrow x-2=20\)

\(\Rightarrow x=20+2\)

\(\Rightarrow x=22\)

k) \(20-\left(x+14\right)=5\)

\(\Rightarrow x+14=20-5\)

\(\Rightarrow x+14=15\)

\(\Rightarrow x=15-14\)

\(\Rightarrow x=1\)

l) \(14+3\left(5-x\right)=27\)

\(\Rightarrow3\left(5-x\right)=27-14\)

\(\Rightarrow3\left(5-x\right)=13\)

\(\Rightarrow5-x=\dfrac{13}{3}\)

\(\Rightarrow x=5-\dfrac{13}{3}\)

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

b: \(y=-\dfrac{8}{3}x\)