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a) \(2x+\frac{3}{15}=\frac{7}{5}\)
=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)
=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)
b) \(x-\frac{2}{9}=\frac{8}{3}\)
=> \(x=\frac{8}{3}+\frac{2}{9}\)
=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)
c) \(\frac{-8}{x}=\frac{-x}{18}\)
=> x(-x) = (-8).18
=> -x2 = -144
=> x2 = 144(bỏ dấu âm)
=> x = \(\pm\)12
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)
=> 5(2x + 3) = 6(x - 2)
=> 10x + 15 = 6x - 12
=> 10x + 15 - 6x + 12 = 0
=> 4x + 27 = 0
=> 4x = -27
=> x = -27/4
e) \(\frac{x+1}{22}=\frac{6}{x}\)
=> x(x + 1) = 132
=> x(x + 1) = 11.12
=> x = 11
f) \(\frac{2x-1}{2}=\frac{5}{x}\)
=> x(2x - 1) = 10
=> 2x2 - x = 10
=> 2x2 - x - 10 = 0
tới đây tự làm đi nhé
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
=> (2x - 1)(2x + 1) = 63
=> 4x2 - 1 = 63
=> 4x2 = 64
=> x2 = 16
=> x = \(\pm\)4
h) Tương tự
a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)
b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)
c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)
e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)
`=> (x-3)5 = (2x+1)3`
`=> 5x-15 = 6x+3`
`=> 5x-6x = 15+3`
`=> -x=18`
`=> x=-18`
\(\dfrac{x+1}{22}=\dfrac{6}{x}\)
`=> (x+1)x = 22*6`
`=> (x+1)x = 132`
`=> x^2 + x = 132`
`=> x^2+x-132=0`
`=> (x-11)(x+12)=0`
`=>`\(\left[{}\begin{matrix}x-11=0\\x+12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=11\\x=-12\end{matrix}\right.\)
\(\dfrac{2x-1}{2}=\dfrac{5}{x}\)
`=> (2x-1)x = 2*5`
`=> 2x^2 - x =10`
`=> 2x^2 - x - 10 =0`
`=> 2x^2 + 4x - 5x - 10 =0`
`=> (2x^2 + 4x) - (5x+10)=0`
`=> 2x(x+2) - 5(x+2)=0`
`=> (2x-5)(x+2)=0`
`=>`\(\left[{}\begin{matrix}2x-5=0\\x+2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)
`=> (2x-1)(2x+1)=21*3`
`=> 4x^2 + 2x - 2x - 1 = 63`
`=> 4x^2 - 1=63`
`=> 4x^2 - 1 - 63=0`
`=> 4x^2 - 64 = 0`
`=> 4(x^2 - 16)=0`
`=> 4(x^2 + 4x - 4x - 16)=0`
`=> 4[(x^2+4x)-(4x+16)]=0`
`=> 4[x(x+4)-4(x+4)]=0`
`=> 4(x-4)(x+4)=0`
`=>`\(\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)
`=> (2x+1)(x+1) = 9*5`
`=> (2x+1)(x+1)=45`
`=> 2x^2 + 2x + x + 1 = 45`
`=> 2x^2 + 3x + 1 =45`
`=> 2x^2 + 3x + 1 - 45 =0`
`=> 2x^2+3x-44=0`
`=> 2x^2 + 11x - 8x - 44=0`
`=> (2x^2 +11x) - (8x+44)=0`
`=> x(2x+11) - 4(2x+11)=0`
`=> (x-4)(2x+11)=0`
`=>`\(\left[{}\begin{matrix}x-4=0\\2x+11=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \left(x-3\right)\cdot5=\left(2x+1\right)\cdot3\\ x5-15=6x+3\\ x5-6x=3+15\\ -x=18\\ \Rightarrow x=-18\)
\(\dfrac{x+1}{22}=\dfrac{6}{x}\\ \left(x+1\right)\cdot x=6\cdot22\\ \left(x+1\right)\cdot x=2\cdot3\cdot2\cdot11\\ \left(x+1\right)\cdot x=12\cdot11\\ \Rightarrow x=11\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\\ \left(2x-1\right)\cdot\left(2x+1\right)=21\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot3\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot9\\ \Rightarrow2x+1=9\\ 2x=8\\ x=4\)
a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)
b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)
\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)
em muốn hỏi là tại sao 3,5 bên trên xuống dưới lại là 3 và -x +2/5 của em xuống dưới lại chuyển thành x-2/5 ạ mong anh giải đáp
a: =>|5/4x-7/2|=|5/8x+3/5|
=>5/4x-7/2=5/8x+3/5 hoặc 5/4x-7/2=-5/8x-3/5
=>5/8x=41/10 hoặc 15/8x=29/10
=>x=164/25 hoặc x=116/75
b: =>3:|x/4-2/3|=6-21/5=9/5
=>|1/4x-2/3|=5/3
=>1/4x-2/3=5/3 hoặc 1/4x-2/3=-5/3
=>1/4x=7/3 hoặc 1/4x=-1
=>x=28/3 hoặc x=-4
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(2x-x-9\right)\left(2x+x+9\right)=0\end{matrix}\right.\Leftrightarrow x=9\)
e: =>|2x-7|=2x-7
=>2x-7>=0
=>x>=7/2
a) \(\left|x+9\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=2x\\x+9=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
b) \(\left|5x\right|-3x=2\Leftrightarrow\left|5x\right|=3x+2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=3x+2\\-5x=3x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1}{4}\end{matrix}\right.\)
c) \(\left|x+6\right|-9=2x\Leftrightarrow\left|x+6\right|=2x+9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=2x+9\\-x-6=2x+9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
d) \(\left|2x-3\right|+x=21\Leftrightarrow\left|2x-3\right|=21-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=21-x\\2x-3=x-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-18\end{matrix}\right.\)
e) \(\left|2x+4\right|=-4x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=4x\\2x+4=-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{-2}{3}\end{matrix}\right.\)
i) \(\left|3x-1\right|+2=x\Leftrightarrow\left|3x-1\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=x-2\\3x-1=2-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{3}{4}\end{matrix}\right.\)
g) \(\left|x+15\right|+1=3x\Leftrightarrow\left|x+15\right|=3x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-1\\x+15=1-3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3,5\end{matrix}\right.\)
h) \(\left|2x-5\right|+x=2\Leftrightarrow\left|2x-5\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=2-x\\2x-5=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=3\end{matrix}\right.\)
a) |9+x|=2x
TH1: 9+x=2x
<=> 9=2x-x
<=> x=9
TH2: -9-x=2x
<=> -9=3x
<=> x=-3
b) |5x|-3x=2
TH1: 5x-3x=2
<=> 2x=2
<=> x=1
TH2: -5x-3x=2
<=> -8x=2
<=>x=-4
c) |x+6|-9=2x
TH1: x+6-9=2x
<=> -3=x
TH2: -x-6-9=2x
<=> -15=3x
<=>x=-5
d) |2x-3|+x=21
TH1: 2x-3+x=21
<=> 3x=24
<=> x=8
TH2: -2x+3+x=21
<=> -x=18
<=> x=-18
e,i,g,h tương tự
a) \(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\left(x\ne-1\right)\)
\(\Rightarrow\left(2x+1\right)\left(x+1\right)=9\cdot5=45\)
\(\Rightarrow2x^2+2x+x+1=45\)
\(\Rightarrow2x^2+3x-44=0\)
\(\Rightarrow2x^2+11x-8x-44=0\)
\(\Rightarrow x\left(2x+11\right)-4\left(2x+11\right)=0\)
\(\Rightarrow\left(x-4\right)\left(2x+11\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)
b) \(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\left(x\ne-\dfrac{1}{2}\right)\)
\(\Rightarrow\left(2x-1\right)\left(2x+1\right)=21\cdot3=63\)
\(\Rightarrow4x^2-1=63\)
\(\Rightarrow4x^2=64\)
\(\Rightarrow\left(2x\right)^2=8^2\)
\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
c) \(\dfrac{2x-1}{2}=\dfrac{5}{x}\left(x\ne0\right)\)
\(\Rightarrow x\left(2x-1\right)=5\cdot2=10\)
\(\Rightarrow2x^2-x=10\)
\(\Rightarrow2x^2-x-10=0\)
\(\Rightarrow2x^2+4x-5x-10=0\)
\(\Rightarrow2x\left(x+2\right)-5\left(x+2\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)
d) \(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)
\(\Rightarrow15\cdot\dfrac{x-3}{3}=15\cdot\dfrac{2x+1}{5}\)
\(\Rightarrow5\left(x-3\right)=3\left(2x+1\right)\)
\(\Rightarrow5x-15=6x+3\)
\(\Rightarrow6x-5x=-18\)
\(\Rightarrow x=-18\)