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![](https://rs.olm.vn/images/avt/0.png?1311)
1) x=-7/5 mk k hiểu đề bài câu 5 của bạn
2)x=22/75
3)x=-4
4)x=29/60
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a: \(\dfrac{3}{5}-x=\dfrac{-1}{4}+\dfrac{7}{10}=\dfrac{-5}{20}+\dfrac{14}{20}=\dfrac{9}{20}\)
=>x=3/5-9/20=12/20-9/20=3/20
b: \(\dfrac{-5}{-8}-x=\dfrac{-5}{-6}+\dfrac{1}{8}\)
=>5/8-x=5/6+1/8
=>5/8-x=23/24
=>x=-1/3
c: \(8.25-x=3+\dfrac{1}{6}=\dfrac{19}{6}\)
=>x=33/4-19/6=99/12-38/12=61/12
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 7:
Để \(\dfrac{4}{2n-3}\) có giá trị là số nguyên
=> 4\(⋮\) 2n-3
=> 2n-3\(\in\) Ư(4)=\(\left\{\pm4;\pm1;\pm2\right\}\)
Ta có bảng sau:
2n-3 | 4 | -4 | 1 | -1 | 2 | -2 |
n | 3,5 | -0,5 | 2 | 1 | 2,5 | 0,5 |
mà n là số nguyên
=> n\(\in\left\{2;1\right\}\)
Vậy để \(\dfrac{4}{2n-3}\) có giá trị là số nguyên thì n\(\in\left\{2;1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: =>3/4x=-3/5-1/2=-11/10
\(\Leftrightarrow x=\dfrac{-11}{10}:\dfrac{3}{4}=\dfrac{-11}{10}\cdot\dfrac{4}{3}=-\dfrac{44}{30}=-\dfrac{22}{15}\)
b: \(\Leftrightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}=\dfrac{19}{12}\)
=>7/4x=19/12
=>x=19/21
c: \(\Leftrightarrow-\dfrac{2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
=>-4/3x=-1/3-1/6=-1/2
=>x=1/2:4/3=1/2x3/4=3/8
Bài 1
\(\dfrac{1}{7}:\dfrac{5}{17}-\dfrac{3}{2}.\left(\dfrac{1}{6}-\dfrac{7}{12}\right)\)
\(\dfrac{1}{7}.\dfrac{17}{5}-\dfrac{3}{2}.\left(-\dfrac{5}{12}\right)\)
\(\dfrac{17}{35}-\left(-\dfrac{5}{8}\right)\)
\(\dfrac{17}{35}+\dfrac{5}{8}\)
\(\dfrac{311}{280}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :Bỏ dấu ngoặc
2007-(7-3+4)
= 2007 -7+3-4
= 1999
6+[(-5) + 4 - 1 ]
= 6-5+4-1
=4
5-[(-6+8-2]
= 5+6-8+2
=5
-10+(7-3+1)
= -10 +7-3+1
= -5
Bài 3 Tìm x
\(\dfrac{1}{3} = \dfrac{x}{6}\)
\(<=> x= \dfrac{1.6}{3}\)
\(<=> x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a)
\(A=\dfrac{-5}{6}\cdot\dfrac{3}{10}\\ =\dfrac{\left(-5\right)\cdot3}{6\cdot10}\\ =\dfrac{-1}{4}\)
b)
\(B=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{12}\\ =\dfrac{4}{12}-\dfrac{3}{12}+\dfrac{1}{12}\\ =\dfrac{4-3+1}{12}\\ =\dfrac{1}{6}\)
Bài 2:
\(A=\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+\left|\dfrac{4}{5}-\dfrac{14}{5}\right|:\dfrac{8}{3}\\ =\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+\left|\dfrac{-10}{5}\right|\cdot\dfrac{3}{8}\\ =\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+2\cdot\dfrac{3}{8}\\ =\dfrac{-3}{4}+\dfrac{3}{4}\\ =0\)
\(B=\left(\dfrac{-1}{2}\right)^2:1\dfrac{3}{8}+25\%\cdot\dfrac{3}{11}\\ =\left(\dfrac{-1}{2}\right)^2:\dfrac{11}{8}+\dfrac{3}{4}\cdot\dfrac{3}{11}\\ =\dfrac{1}{4}\cdot\dfrac{8}{11}+\dfrac{3}{4}\cdot\dfrac{3}{11}\\ =\dfrac{8}{44}+\dfrac{9}{44}\\ =\dfrac{17}{44}\)
\(C=\dfrac{-8}{5}+0,6+\left|\dfrac{-1}{2}\right|+\dfrac{1}{2}\\ =\dfrac{-8}{5}+\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{1}{2}\\ =\left(\dfrac{-8}{5}+\dfrac{3}{8}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}\right)\\ =\left(-1\right)+1\\ =0\)
\(D=\dfrac{-5}{9}\cdot\dfrac{2}{13}+\dfrac{-5}{9}:\dfrac{13}{11}+1\dfrac{5}{9}\\ =\dfrac{-5}{9}\cdot\dfrac{2}{13}+\dfrac{-5}{9}\cdot\dfrac{11}{13}+\dfrac{14}{9}\\ =\dfrac{-5}{9}\cdot\left(\dfrac{2}{13}+\dfrac{11}{13}\right)+\dfrac{14}{9}\\ =\dfrac{-5}{9}\cdot1+\dfrac{14}{9}\\ =\dfrac{-5}{9}+\dfrac{14}{9}\\ =1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) x - 2 = -6
x = -6 + 2
x = -4
2) -5 . x - ( -3 ) =13
-5 . x = 13 + ( -3 )
-5 . x = 10
x = 10 : ( -5 )
x = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 4.|x-6|=28
<=> \(\left|x-6\right|=7\Leftrightarrow\left[{}\begin{matrix}x-6=7\left(x\ge6\right)\\6-x=7\left(x< 6\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(t.m\right)\\x=-1\left(t.m\right)\end{matrix}\right.\)
b, 15.|2.x-17|=225
\(\left|2x-7\right|=15\Leftrightarrow\left[{}\begin{matrix}2x-7=15\left(x\ge\frac{7}{2}\right)\\7-2x=15\left(x< \frac{7}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\left(t.m\right)\\x=-4\left(t.m\right)\end{matrix}\right.\)
c, |x-9|=15-(-2)+(-17)
<=> | x-9| = 0 <=> x=9
d, 72-3.x=5.x+8
<=> 8x=64 <=> x=8
e, (x+1).(x-3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
f, 127-5.(x+4)=42
<=> 5(x+4)=127-42=85 <=> x+4 =17 <=> x=13
g, 2.(x-5)-3.(x+7)=14
<=> 2x-10-3x-21=14 <=> -x = 14+10+21=45<=> x=-45
h, (3.x-24).75=2.76
<=>3x-24=152/75<=> 3x=1952/75<=> x=1952/225
i, 35-5.|x|=5.(24-4)
<=> \(5\left|x\right|\) = 35-100=-65 <=> \(\left|x\right|=-13\) ( vô nghiệm vì \(\left|x\right|\ge0,\) với mọi x thuộc R )
mik trùng đáp án vs bn các phần a, c, e, f nhưng cách lm gần giống và f ko giống. Thanks bn rất nhìu vì đã giải giúp mik nha!!!
1: =>x:7/3=-3/2
hay \(x=-\dfrac{3}{2}\cdot\dfrac{7}{3}=-\dfrac{7}{2}\)
2: \(x=\dfrac{11}{15}:\dfrac{2}{5}=\dfrac{11}{15}\cdot\dfrac{5}{2}=\dfrac{55}{30}=\dfrac{11}{6}\)
3: \(x=-\dfrac{6}{5}:\dfrac{3}{10}=\dfrac{-6}{5}\cdot\dfrac{10}{3}=\dfrac{-60}{15}=-4\)
4: \(x=\dfrac{1}{3}+\dfrac{3}{20}=\dfrac{20+9}{60}=\dfrac{29}{60}\)
6: \(\Leftrightarrow x\cdot\dfrac{13}{20}=\dfrac{5}{4}\)
hay \(x=\dfrac{5}{4}:\dfrac{13}{20}=\dfrac{5}{4}\cdot\dfrac{20}{13}=\dfrac{100}{52}=\dfrac{25}{13}\)