Tìm x:
a) |x+4/15|-|-3,15|=-|-2,15|
b)2x +|x+4|=5
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`@` `\text {Ans}`
`\downarrow`
`a)`
`3x ( 12x - 4 ) - 9x( 4x - 3 ) = 30`
`=> 3x (12x-4) - 3*3x (4x - 3) = 30`
`=> 3x [12x - 4 - 3(4x-3)] = 30`
`=> 3x (12x - 4 - 12x + 9) = 30`
`=> 3x (-4+9)=30`
`=> 3x*5=30`
`=> 3x=6`
`=> x=2`
Vậy, `x=2`
`b)`
`x( 5 - 2x) + 2x( x - 1)`
`=> x(5-2x) + 2x^2 - 2x=15`
`=> 5x - 2x^2 + 2x^2 - 2x =15`
`=> 3x = 15`
`=> x=5`
Vậy, `x=5.`
a: =>36x^2-12x-36x^2+27x=30
=>15x=30
=>x=2
b: =>5x-2x^2+2x^2-2x=15
=>3x=15
=>x=5
Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
a) \(x+\dfrac{4}{15}=\dfrac{4}{12}\)
\(x=\dfrac{4}{12}-\dfrac{4}{15}\)
\(x=\dfrac{20}{60}-\dfrac{16}{60}\)
\(x=\dfrac{1}{15}\)
b) \(x-\dfrac{5}{8}=1\dfrac{2}{3}\)
\(x-\dfrac{5}{8}=\dfrac{5}{3}\)
\(x=\dfrac{5}{3}+\dfrac{5}{8}\)
\(x=\dfrac{40}{24}+\dfrac{15}{24}\)
\(x=\dfrac{55}{24}\)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
a) /x+\(\frac{4}{15}\)/ - / -3,75/ = -2,15
=> \(\orbr{\begin{cases}x+\frac{4}{15}+3,75=-2,15\\x+\frac{4}{15}+3,75=2,15\end{cases}}\)
=> ....v.....v giải ra ( từng th )
bài khác tương tự
a) 32 : (3.x - 2) = 8
3x - 2 = 32 : 8
3x - 2 = 4
3x = 4 + 2
3x = 6
x = 6 : 3
x = 2
b) 75 : (x - 18) = 25
x - 18 = 75 : 25
x - 18 = 3
x = 3 + 18
x = 21
c) (15 - 6.x) . 243 = 729
15 - 6x = 729 : 243
15 - 6x = 3
6x = 15 - 3
6x = 12
x = 12 : 6
x = 2
d) 4.(x - 12) + 9 = 17
4(x - 12) = 17 - 9
4(x - 12) = 8
x - 12 = 8 : 4
x - 12 = 2
x = 2 + 12
x = 14
e) 20 - 2.(x + 4) = 4
2(x + 4) = 20 - 4
2(x + 4) = 16
x + 4 = 16 : 2
x + 4 = 8
x = 8 : 2
x = 4
`32: ( 3xx x -2)=8`
`3xx x-2=32:8`
`3xx x-2=4`
`3 xx x=4+2`
`3xx x=6`
`x=6:3`
`x=2`
__
`75 : (x-18) =25`
`x-18=75:25`
`x-18= 3`
`x=3+18`
`x=21`
__
`(15-6 xx x ) xx 243 =729`
`15-6 xx x = 729 : 243`
`15-6 xx x = 3`
`6 xx x=15-3`
`6 xx x=12`
`x=12:6`
`x=2`
__
`4 xx (x-12)+9=17`
`4 xx (x-12)=17-9`
`4 xx (x-12)= 8`
`x-12=8:4`
`x-12=2`
`x=2+12`
`x=14`
__
`20-2xx(x+4)=4`
`2xx(x+4)=20-4`
`2xx(x+4)=16`
`x+4=16:2`
`x+4=8`
`x=8-4`
`x=4`
\(\frac{1}{2}x+\frac{1}{5}x+\frac{3}{5}=0\)
=> \(\left(\frac{1}{2}+\frac{1}{5}\right)x+\frac{3}{5}=0\)
=> \(\frac{7}{10}x+\frac{3}{5}=0\)
=> \(\frac{7}{10}x=-\frac{3}{5}\)
=> \(x=\left(-\frac{3}{5}\right):\frac{7}{10}=\left(-\frac{3}{5}\right)\cdot\frac{10}{7}=\left(-\frac{3}{1}\right)\cdot\frac{2}{7}=-\frac{6}{7}\)
b) \(\left|2\frac{1}{2}+x\right|-\left(-\frac{2}{3}\right)=3\)
=> \(\left|\frac{5}{2}+x\right|+\frac{2}{3}=3\)
=> \(\left|\frac{5}{2}+x\right|=\frac{7}{3}\)
=> \(\orbr{\begin{cases}\frac{5}{2}+x=\frac{7}{3}\\\frac{5}{2}+x=-\frac{7}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{29}{6}\end{cases}}\)
c) \(\left|x+\frac{4}{15}\right|-\left|-3.75\right|=-\left|-2,15\right|\)
=> \(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
=> \(\left|x+\frac{4}{15}\right|=\frac{8}{5}\)
=> \(\orbr{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=-\frac{8}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{28}{15}\end{cases}}\)
a, \(\frac{1}{2}x+\frac{1}{5}x+\frac{3}{5}=0\Leftrightarrow\frac{7}{10}x+\frac{3}{5}=0\Leftrightarrow x=-\frac{6}{7}\)
b, đề sai
c, \(\left|\frac{x+4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\frac{x+4}{15}-3,75=-2,15\Leftrightarrow\frac{x+4}{15}=\frac{8}{5}\Leftrightarrow x+4=24\Leftrightarrow x=28\)
\(a,\Rightarrow x^2+4x+25-x^2=3\\ \Rightarrow4x=-22\Rightarrow x=-\dfrac{11}{2}\\ b,\Rightarrow\left(2x-3-4x-3\right)\left(2x-3+4x+3\right)=0\\ \Rightarrow6x\left(-2x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\end{matrix}\right.\)