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T
1 tháng 9 2023
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
T
1 tháng 9 2023
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
29 tháng 10 2020
TÌM X:
a) 2x - 3 = \(\frac{1}{2}\)
2x = \(\frac{1}{2}+3\)
2x = \(\frac{7}{2}\)
x = 2 : \(\frac{7}{2}\)
x = 2 . \(\frac{2}{7}\)
x = \(\frac{4}{7}\)
b) /x+1/ = 0.25
/x+1/ = \(\frac{1}{4}\)
\(\orbr{\begin{cases}x+1=\frac{1}{4}\\x+1=-\frac{1}{4}\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{4}-1\\x=-\frac{1}{4}-1\end{cases}}\)
\(\orbr{\begin{cases}x=-\frac{3}{4}\\x=-\frac{5}{4}\end{cases}}\)
c) 32 : 2x = 2
\(2x=32:2\)
\(2x=16\)
\(x=16:2\)
\(x=8\)
~GOOD STUDY~
P
3
14 tháng 12 2016
|x+3/5|-1/2=1/2
|x+3/5|=1/2+1/2
|x+3/5|=1
=>x+3/5=1
x=1-3/5
x=2/5
hay
x+3/5=-1
x=(-1)-3/5
x=-8/5
Vậy x=2/5 hay-8/5
14 tháng 12 2016
1)vì |x-4/5|=3/4 =>x-4/5=3/4 hay x-4/5=-3/4 x=3/4+4/5 x= -3/4+4/5 x=31/20 x=1/20 vậy x=31/20 hay 1/20
TG
4
25 tháng 10 2016
a) |2x - 5| + |3x + 1| = 6
- Với \(x< \frac{-1}{3}\) thì |2x - 5| = 5 - 2x; |3x + 1| = -(3x + 1) = -3x - 1
Ta có: (5 - 2x) + (-3x - 1) = 6
=> 4 - 5x = 6
=> 5x = 4 - 6 = -2
\(\Rightarrow x=\frac{-2}{5}\), thỏa mãn \(x< \frac{-1}{3}\)
- Với \(\frac{-1}{3}\le x< \frac{5}{2}\) thì |2x - 5| = 5 - 2x; |3x + 1| = 3x + 1
Ta có: (5 - 2x) + (3x + 1) = 6
=> 6 + x = 6
=> x = 6 - 6 = 0, thỏa mãn \(\frac{-1}{3}\le x< \frac{5}{2}\)
- Với \(x\ge\frac{5}{2}\) thì |2x - 5| = 2x - 5; |3x + 1| = 3x + 1
Ta có: (2x - 5) + (3x + 1) = 6
=> 5x - 4 = 6
=> 5x = 6 + 4 = 10
=> x = 10 : 5 = 2, không thỏa mãn \(x\ge\frac{5}{2}\)
Vậy \(\left[\begin{array}{nghiempt}x=\frac{-2}{5}\\x=0\end{array}\right.\) thỏa mãn đề bài
TM
0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
9 giờ trước (20:38)
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
9 giờ trước (20:48)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
F=|x+2|+|x+4|+|x+6| = ( |x+2|+|x+6) + |x+4| = ( |x+2|+|-x-6) + |x+4|
ta có \(\hept{\begin{cases}\left|x+2\right|+\left|-x-6\right|\ge\left|x+2-x-6\right|=4\\\left|x+4\right|\ge0\end{cases}}\)
=> F > 4+0=4
=> Fmin=4
<=> x+4=0 => x=-4
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