tìm x
a) |x|=3,5 b) |x|=2,7 c) |1-x|+0,73=3 d) |x+3/4|-5=-2
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c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\)
l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)
Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)
a) \(\left|x\right|=3,5\)
\(\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
b) \(\left|1-x\right|+0,73=3\)
\(\Rightarrow\left|1-x\right|=3-0,73\)
\(\Rightarrow\left|1-x\right|=2,27\)
\(\Rightarrow\left[{}\begin{matrix}1-x=2,27\\1-x=-2,27\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1,27\\x=3,27\end{matrix}\right.\)
c) \(\dfrac{-5}{8}+x=\dfrac{4}{9}\)
\(\Rightarrow x=\dfrac{4}{9}-\left(\dfrac{-5}{8}\right)\)
\(\Rightarrow x=\dfrac{77}{72}\)
a/ \(\left|x\right|=3\dfrac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\dfrac{1}{5}\\x=-3\dfrac{1}{5}\end{matrix}\right.\)
Vậy ............
b/ \(\left|x\right|=-2,7\) \(\Leftrightarrow x\in\varnothing\)
Vậy ....
c/ \(\left|x-3,5\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)
Vậy ....
a. \(\left|x\right|=\dfrac{31}{5}\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{31}{5}\\x=\dfrac{-31}{5}\end{matrix}\right.\)
Vậy....
b. \(\left|x\right|=-2,7\)
không có giá trị nào thỏa mãn vì \(\left|x\right|\ge0\) với \(\forall x\in Q\)
c. \(\left|x-3,5\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)
Vậy......
A. x - 3,5 = 2,4 + 1,5
x = 3,9 + 3,5
x = 7,4
B. x + 6,14 = 27,8 - 8,6
x = 19,2 - 6,14
x = 12,8
C. 4,3 x x = 3,8 x 4,3
x = 3,8 x 4,3 : 4,3
x = 3,8
D. 7, 3 x x + 2,7 x x = 10
x x ( 7, 3 + 2,7 ) = 10
x = 10 : 10
x = 1
Bài 2:
a,\(|x-2|-3=-7\)
\(\Rightarrow|x-2|=-4\)
\(\Rightarrow x\in\varnothing\)
b,\(|x-5|+3=8\)
\(\Rightarrow|x-5|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-5=5\\x-5=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=0\end{matrix}\right.\)
c,\(|x+3|-2+2x=5\)
\(\Rightarrow|x+3|+2x=7\)
\(\Rightarrow\left[{}\begin{matrix}x+3=2x=7,x+3\ge0\\-\left(x+3\right)+2x=7,x+3< 0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=10,x< -3\end{matrix}\right.,x\ge-3\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x\in\varnothing\end{matrix}\right.\)
d,\(|x-3,1|+|y-1,2|=0\)
Thay y =0
\(\Rightarrow|x-3,1|+|0-1,2|=0\)
\(\Rightarrow|x-3,1|+|-1,2|=0\)
\(\Rightarrow|x-3,1|+1,2=0\)
\(\Rightarrow|x-3,1|=-1,2\)
\(\Rightarrow x\in\varnothing\)
Bài 3:
a,(-2,7)+9-3,5)-1,2(-2)
=>-2,7-3,5+2,4
=>-3,8
b, (-2,3)(-0,4)-(-3,1)(-2,1)-13,2
=>0,92+3,1(-2,1)-13,2
=>0,92-6,51-13,2
=-18,79
a: =>(2x+1/2)^2=1/4
=>2x+1/2=1/2 hoặc 2x+1/2=-1/2
=>x=-1/2 hoặc x=0
b: =>(x-1/5)^2=49
=>x-1/5=7 hoặc x-1/5=-7
=>x=-6,8 hoặc x=7,2
c: =>1,2x=12
=>x=10
d: =>3/4x+1/2x+1/2=-11/4
=>5/4x=-11/4-2/4=-13/4
=>x=-13/5
e: =>-0,25x+1,25x=0,2
=>x=0,2
a) \(x+\frac{1}{5}-\frac{2}{7}=4\frac{1}{2}\)
=> \(x+\frac{1}{5}-\frac{2}{7}=\frac{9}{2}\)
=> \(x=\frac{9}{2}+\frac{2}{7}-\frac{1}{5}=\frac{321}{70}\)
b) \(1\frac{3}{4}+x-\frac{7}{8}=5\)
=> \(\frac{7}{4}+x-\frac{7}{8}=5\)
=> \(\frac{7}{4}+x=\frac{47}{8}\)
=> \(x=\frac{33}{8}\)
c) \(\frac{1}{2}-\frac{1}{3}\cdot x=\frac{3}{4}\)
=> \(\frac{1}{3}\cdot x=\frac{1}{2}-\frac{3}{4}=-\frac{1}{4}\)
=> \(x=\left(-\frac{1}{4}\right):\frac{1}{3}=\left(-\frac{1}{4}\right)\cdot3=-\frac{3}{4}\)
Y÷ 3,5 =14,6 + 2,7