Tìm |x| bít
a) x = -1/7
b) x = 1/7
c) x = -3/1/5
d) x = 0
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a) 3/35 - (3/5 + x) = 2/7
=> 3/5 + x= 3/35- 2/7
=> 3/5 +x = -1/5
=> x = -1/5 -3/5
=> x = -4/5
b) 3/7 +1/7 : x = 3/14
=> 1/7 : x= 3/14 -3/7
=> 1/7 : x = -3/14
=> x = 1/7 : -3/14
=> x = -2/3
c) (5x-1).(2x-1/3)=0
=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)
Học tốt :D
a)x=-4/5
b)x=-2/3
c)\(\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy.........
mik lười mong bn thông cảm
a) x2 - 5x - 6 = 0
=> x2 - 2x - 3x - 6 = 0
=> (x2 - 2x) + (-3x - 6) = 0
=> x(x - 2) - 3 (x - 2) = 0
=> (x - 2) (x - 3) = 0
=> x - 2 = 0 => x = 2
x - 3 = 0 => x = 3
còn lại tương tự nhé!! 46566578768698945635655675656788787868789789879789098089364556546
a)|x|=1/7
b)|x|=1/7
c)|x|=\(\left|-3\frac{1}{5}\right|=3\frac{1}{5}\)
d)|x|=|0|=0
c) \(\left(x-7\right).\left(y+2\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-7=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0+7\\y=0-2\end{cases}}\Rightarrow\hept{\begin{cases}x=7\left(TM\right)\\y=-2\left(TM\right)\end{cases}}\)
Vậy \(\left(x;y\right)\in\left\{7;-2\right\}.\)
Chúc bạn học tốt!
a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42
a; \(x\) - \(\dfrac{3}{5}\) = 1 - \(\dfrac{4}{5}\) + \(\dfrac{1}{6}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{30}{30}\) - \(\dfrac{24}{30}\) + \(\dfrac{5}{30}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{6}{30}\) + \(\dfrac{5}{30}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{11}{30}\) + \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{11}{30}\) + \(\dfrac{18}{30}\)
\(x\) = \(\dfrac{29}{30}\)
Vậy \(x\) = \(\dfrac{29}{30}\)
b; (- \(\dfrac{10}{4}\)) + \(\dfrac{1}{4}\) = \(\dfrac{3}{4}\) thế \(x\) của em đâu nhỉ???
c; - \(\dfrac{3}{2}\) + (\(x\) - \(\dfrac{1}{2}\)) = \(\dfrac{1}{2}\)
\(x\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\)
\(x\) - \(\dfrac{1}{2}\) = 2
\(x\) = 2 + \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{4}{2}\) + \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{5}{2}\)
Vậy \(x=\dfrac{5}{2}\)
a: 5-3x=6x+7
=>-3x-6x=7-5
=>-9x=2
=>\(x=-\dfrac{2}{9}\)
b: \(\dfrac{3x-2}{6}-5=3-\dfrac{2\left(x+7\right)}{4}\)
=>\(\dfrac{3x-2}{6}+\dfrac{x+7}{2}=8\)
=>\(\dfrac{3x-2+3\left(x+7\right)}{6}=8\)
=>3x-2+3x+14=48
=>6x+12=48
=>6x=36
=>\(x=\dfrac{36}{6}=6\)
c: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
=>\(\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
=>(x-1)(5x+3-3x+8)=0
=>(x-1)(2x+11)=0
=>\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
=>\(\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\dfrac{3x^6-4x^3}{x^3}-\dfrac{\left(3x+1\right)^2}{3x+1}-\dfrac{3x^7}{x^5}=0\)
\(\Leftrightarrow3x^3-4-3x-1-3x^2=0\)
\(\Leftrightarrow3x^3-3x^2-3x-5=0\)
\(\Leftrightarrow x\simeq1,9506\)
1.
a, => 21-x+3 < 0
=> 24-x < 0
=> x < 24
b, => 7+x > 0
=> x > -7
c, => x-1 < 0 ; x+2 > 0 ( vì x-1 < x+2 )
=> x < 1 ; x > -2
=> -2 < x < 1
Tk mk nha
a) / x / = / -1/7 / = 1/7
b) / x / = / 1/7 / = 1/7
c) / x / = / -3 / 1 / 5 / = 3 / 1 / 5
d ) /x/ = /0/ = 0