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Bài 2:
a: =>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
a \(\dfrac{2}{3}x+\dfrac{1}{3}=\dfrac{1}{5}\\ \dfrac{2}{3}x=\dfrac{1}{5}-\dfrac{1}{3}\\ \dfrac{2}{3}x=\dfrac{-2}{15}\\ x=-\dfrac{2}{15}:\dfrac{2}{3}\\ x=-\dfrac{1}{5}\) b) \(\dfrac{4}{5}-\dfrac{5}{3}x=-2\\ \dfrac{5}{3}x=\dfrac{4}{5}+2\\ \dfrac{5}{3}x=\dfrac{14}{5}\\ x=\dfrac{14}{5}:\dfrac{5}{3}\\ x=\dfrac{42}{25}\)c) \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\\ \dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}\\ \dfrac{5}{3}:x=\dfrac{3}{10}\\ x=\dfrac{5}{3}:\dfrac{3}{10}\\ x=\dfrac{50}{9}\)d) \(\dfrac{5}{7}:x-3=-\dfrac{2}{7}\\ \dfrac{5}{7}:x=3-\dfrac{2}{7}\\ \dfrac{5}{7}:x=\dfrac{19}{7}\\ x=\dfrac{5}{7}:\dfrac{19}{7}\\ x=\dfrac{5}{19}\)
`@` ` \text {Ans}`
`\downarrow`
`a,`
`1/4+3/4*x=3/2-x`
`=> 1/4 + 3/4x - 3/2 + x = 0`
`=> (1/4 - 3/2) + (3/4x + x) = 0`
`=> -5/4 + 7/4x = 0`
`=> 7/4x = 5/4`
`=> x = 5/4 \div 7/4`
`=> x = 5/7`
Vậy, `x=5/7`
`b,`
`3/5*x-1/4=1/10*x-1/2`
`=> 3/5x - 1/4 - 1/10x + 1/2 = 0`
`=> (3/5x - 1/10x) + (-1/4 + 1/2)=0`
`=> 1/2x + 1/4 = 0`
`=> 1/2x = -1/4`
`=> x = -1/4 \div 1/2`
`=> x = -1/2`
Vậy, `x=-1/2`
`c,`
`3x-3/5=x-1/4`
`=> 3x - 3/5 - x + 1/4 = 0`
`=> (3x - x) - (3/5 - 1/4) = 0`
`=> 2x - 7/20 = 0`
`=> 2x = 0,35`
`=> x = 0,35 \div 2`
`=> x = 7/40`
Vậy, `x=7/40`
`d,`
`3/2*x-2/5=1/3*x-1/4`
`=> 3/2x - 2/5 - 1/3x + 1/4 = 0`
`=> (3/2x - 1/3x) - (2/5 - 1/4) = 0`
`=> 7/6x - 3/20 = 0`
`=> 7/6x = 3/20`
`=> x = 3/20 \div 7/6`
`=> x = 9/70`
Vậy, `x=9/70`
`@` `\text {Kaizuu lv uuu}`
Bài 1
a) \(x=x^5\)
\(x^5-x=0\)
\(x\left(x^4-1\right)=0\)
\(x=0\) hoặc \(x^4-1=0\)
* \(x^4-1=0\)
\(x^4=1\)
\(x=1\)
Vậy x = 0; x = 1
b) \(x^4=x^2\)
\(x^4-x^2=0\)
\(x^2\left(x^2-1\right)=0\)
\(x^2=0\) hoặc \(x^2-1=0\)
*) \(x^2=0\)
\(x=0\)
*) \(x^2-1=0\)
\(x^2=1\)
\(x=1\)
Vậy \(x=0\); \(x=1\)
c) \(\left(x-1\right)^3=x-1\)
\(\left(x-1\right)^3-\left(x-1\right)=0\)
\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)
*) \(x-1=0\)
\(x=1\)
*) \(\left(x-1\right)^2-1=0\)
\(\left(x-1\right)^2=1\)
\(x-1=1\) hoặc \(x-1=-1\)
**) \(x-1=1\)
\(x=2\)
**) \(x-1=-1\)
\(x=0\)
Vậy \(x=0\); \(x=1\); \(x=2\)
\(a,\dfrac{x}{2}=\dfrac{8}{x}\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\\ b,\dfrac{x+1}{5}=\dfrac{x+1}{5}\left(luôn.đúng\right)\\ c,\dfrac{x+1}{5}=\dfrac{x+3}{10}\\ \Rightarrow\dfrac{2x+2}{10}=\dfrac{x+3}{10}\\ \Rightarrow2x+2=x+3\\ \Rightarrow2x-x=3-2\\ \Rightarrow x=1\\ d,\dfrac{x}{4}=\dfrac{18}{x+1}\\ \Rightarrow x\left(x+1\right)=4.18\\ \Rightarrow x^2+x=72\\ \Rightarrow x^2+x-72=0\\ \Rightarrow\left(x^2+9x\right)-\left(8x+72\right)=0\\ \Rightarrow x\left(x+9\right)-8\left(x+9\right)=0\\ \Rightarrow\left(x-8\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-9\end{matrix}\right.\)
\(a,x\in\left\{-5;-4;-3;-2;-1\right\}\\ b,x\in\left\{-3;-2;-1;...;5;6\right\}\\ c,x\in\left\{-4;-3;...;3;4\right\}\\ d,x\in\left\{-3;-2;-1;0;1;2\right\}\)
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
a) |x+2|=0 => x =-2
b) |x-5| = 7 => x-5 = 7 hoac -7 => x =12 hoac -2
c) 3|x-1|+2|x-1| = 3|x-1|+4
=> 5|x-1| - 3|x-1| = 4
=> 2|x-1| = 4
=> |x-1| =2
=> x-1 = 2 hoac -2 => x = 3 hoac -1
d) 1<|x-2|<4
=> |x-2| = 2 và 3
=> x-2 = 2 hoac -2 va x-2 = 3 hoac -3
=> x = 4 hoac 0 va x = 5 hoac -1
vay x = { -1 ; 0 ; 4 ; 5}