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23 tháng 7 2020
Bài làm:
1) \(\frac{3}{5}\div\frac{2x}{15}=\frac{1}{2}\div\frac{4}{5}\)
\(\Leftrightarrow\frac{9}{2x}=\frac{5}{8}\)
\(\Rightarrow10x=72\)
\(\Leftrightarrow x=\frac{36}{5}\)
2) \(-\frac{4}{2,5}\div\frac{3}{5}=\frac{1}{5}\div x\)
\(\Leftrightarrow\frac{1}{5}\div x=-\frac{8}{3}\)
\(\Rightarrow x=-\frac{3}{40}\)
3) \(0,12\div3=2x\div\frac{3}{5}\)
\(\Leftrightarrow\frac{1}{25}=\frac{10}{3}x\)
\(\Rightarrow x=\frac{3}{250}\)
29 tháng 10 2017
a) \(\dfrac{1,2}{x+3}=\dfrac{5}{4}\)
\(\Rightarrow\left(x+3\right).5=1,2.4\)
\(\Rightarrow\left(x+3\right).5=4,8\)
\(\Rightarrow x+3=4,8:5\)
\(\Rightarrow x+3=0,96\)
\(\Rightarrow x=-2,04\)
vậy \(x=-2,04\)
b)\(\dfrac{3}{5}:\dfrac{2x}{15}=\dfrac{1}{2}:\dfrac{4}{5}\)
\(\Rightarrow\dfrac{3}{5}:\dfrac{2x}{15}=\dfrac{5}{8}\)
\(\Rightarrow\dfrac{2x}{15}=\dfrac{3}{5}:\dfrac{5}{8}\)
\(\Rightarrow\dfrac{2x}{15}=\dfrac{24}{25}\)
\(\Rightarrow15.24=\left(2x\right).25\)
\(\Rightarrow360=\left(2x\right).25\)
\(\Rightarrow360:25=2x\)
\(\Rightarrow14,4=2x\)
\(\Rightarrow x=7,2\)
vậy \(x=7,2\)
29 tháng 10 2017
\(a,\dfrac{1,2}{x+3}=\dfrac{5}{4}\\ \left(x+3\right).5=1,2.4\\ 5x+8=4,8\\ 5x=4,8-8\\ 5x=-3,2\\ x=-3,2:5=-0,64\)
\(b,\dfrac{3}{5}:\dfrac{2x}{15}=\dfrac{1}{2}:\dfrac{4}{5}\\ \dfrac{2x}{15}=\dfrac{3}{5}\cdot\dfrac{4}{5}:\dfrac{1}{2}\\ \dfrac{2x}{15}=\dfrac{12}{25}.2\\ \dfrac{2x}{25}=\dfrac{24}{25}\\ 2x=\dfrac{24}{25}.5\\ 2x=\dfrac{24}{5}\\ x=\dfrac{24}{5}\cdot\dfrac{1}{2}=\dfrac{12}{5}\)
\(c,-\dfrac{4}{2,5}:3,5=1,5:x\\ x=3,5.1,5:\left(-\dfrac{4}{25}\right)\\ x=\dfrac{21}{4}\cdot\left(-\dfrac{25}{4}\right)=-\dfrac{525}{16}\)
\(d,0,12:3=2x:\dfrac{3}{5}\\ 2x=0,12\cdot\dfrac{3}{5}:3\\ 2x=\dfrac{9}{125}\cdot\dfrac{1}{3}\\ 2x=\dfrac{3}{125}\\ x=\dfrac{3}{125}\cdot\dfrac{1}{2}=\dfrac{3}{250}\)
ND
Nguyễn Đức Trí
VIP
13 tháng 7 2023
a) \(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{72}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{72}=1-1+\dfrac{1}{72}=\dfrac{1}{72}\)
b) \(=\dfrac{1}{5}-\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{5}{9}-\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{7}{13}-\dfrac{7}{13}-\dfrac{9}{16}\)
\(=\dfrac{9}{16}\)
ND
Nguyễn Đức Trí
VIP
25 tháng 8 2023
a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)
\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)
\(\Leftrightarrow x-4=25\)
\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)
b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)
\(\Leftrightarrow x\left(x+1\right)=18.4\)
\(\Leftrightarrow x\left(x+1\right)=72\)
vì \(72=8.9=\left(-8\right).\left(-9\right)\)
\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)
c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)
\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)
\(\Leftrightarrow2x+3-2x-8⋮x+4\)
\(\Leftrightarrow-5⋮x+4\)
\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)
\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)
2 tháng 4 2020
1: Ta có: \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x-3x^2+x=0\)
\(\Leftrightarrow-2x^2-2x=0\)
\(\Leftrightarrow-2x\left(x+1\right)=0\)
Vì -2≠0
nên \(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy: x∈{0;-1}
2) Ta có: \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x-12+x=0\)
\(\Leftrightarrow11x-2=0\)
\(\Leftrightarrow11x=2\)
hay \(x=\frac{2}{11}\)
Vậy: \(x=\frac{2}{11}\)
3) Ta có: \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}-5=0\)
\(\Leftrightarrow\frac{-13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow\frac{4}{3}x=\frac{-13}{3}\)
hay \(x=\frac{-13}{3}:\frac{4}{3}=\frac{-13}{4}\)
Vậy: \(x=\frac{-13}{4}\)
4) Ta có: \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{5}\\x-\frac{4}{5}=\frac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{5}\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{5};\frac{7}{5}\right\}\)
BL
2 tháng 4 2020
1. \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x=3x^2-x\)
\(\Leftrightarrow\left(2x-5x+x\right)+\left(x^2-3x^2\right)=0\)
\(\Leftrightarrow-2x-2x^2=0\)
\(\Leftrightarrow-2x\left(1+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\1+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
2. \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x=12-x\)
\(\Leftrightarrow\left(15-5-12\right)+\left(10x+x\right)=0\)
\(\Leftrightarrow-2+11x=0\)
\(\Leftrightarrow11x=2\Leftrightarrow x=\frac{2}{11}\)
3. \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\Leftrightarrow\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-5\right)-\left(\frac{1}{3}x+x\right)=0\)
\(\Leftrightarrow-\frac{13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow-\frac{4}{3}x=\frac{13}{3}\Leftrightarrow x=-\frac{13}{4}\)
4. \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Rightarrow x-\frac{4}{5}=-\frac{3}{5}\) hoặc \(x-\frac{4}{5}=\frac{3}{5}\)
\(TH1:x-\frac{4}{5}=-\frac{3}{5}\Rightarrow x=\frac{1}{5}\)
\(TH2:x-\frac{4}{5}=\frac{3}{5}\Rightarrow x=\frac{7}{5}\)
VT
30 tháng 9 2019
Ta cố bdt \(|a|+|b|\ge|a+b|\), dễ dàng chứng mình bằng bình phương 2 vế. Dấu = sảy ra <=>IaI.IbI=a.b <=> a.b>=0
áp dụng vào từng câu
a)A=Ix+1I+Ix+2I+Ix+3I+I-x-4I+I-x-5I ( vì Ix+4I=I-x=4I, Ix+5I=I-x-5I
A>=I(x+1)+(-x-5)I+I(x+2)+(-x-4)I +Ix+3I=4+2+Ix+3I=6+Ix+3I>=6
Dấu bằng khi (x+1)(-x-5)>=0;(x+2)(-x-4)>=0;Ix+3I=0 =>x=-3
b) LÀm tương tự MinB=18
Dấu = khi (2x+1)(-2x-11)>=0;(2x+3)(-2x-9)>=0;(2x+5)(-2x-7)>=0 <=>-7/2<=x<=-5/2
...
11 tháng 1 2022
a: =>2x>-6
hay x>-3
e: =>(5-x)/x<0
=>0<x<5
h: \(\Leftrightarrow\dfrac{x+5-x-3}{x+3}< 0\)
\(\Leftrightarrow x+3< 0\)
hay x<-3
g: \(\Leftrightarrow\dfrac{2x+7}{x+4}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{7}{2}\\x< -4\end{matrix}\right.\)
LM
Lê Minh Vũ
VIP
2 tháng 8 2023
\(\dfrac{1}{5}\times x-\dfrac{2}{3}=\dfrac{1}{10}\times x+\dfrac{5}{6}\)
\(\dfrac{1}{5}x-\dfrac{2}{3}-\dfrac{1}{10}x-\dfrac{5}{6}=0\)
\(\dfrac{1}{5}x-\dfrac{1}{10}x-\dfrac{2}{3}-\dfrac{5}{6}=0\)
\(\dfrac{1}{10}x-\dfrac{3}{2}=0\)
\(\dfrac{1}{10}x=\dfrac{3}{2}\)
\(x=15\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
2 tháng 8 2023
\(\dfrac{1}{5}\).x - \(\dfrac{2}{3}\) = \(\dfrac{1}{10}\).x + \(\dfrac{5}{6}\)
⇒ \(\dfrac{1}{5}\).x - \(\dfrac{1}{10}\).x = \(\dfrac{5}{6}\) + \(\dfrac{2}{3}\)
⇒ \(\dfrac{2}{10}\).x - \(\dfrac{1}{10}\).x = \(\dfrac{5}{6}\) + \(\dfrac{4}{6}\)
⇒ \(\dfrac{1}{10}\).x = \(\dfrac{9}{6}\)
⇒ x = \(\dfrac{9}{6}\) : \(\dfrac{1}{10}\)
⇒ x = \(\dfrac{9}{6}\) . 10
⇒ x = \(\dfrac{90}{6}\)
⇒ x = 15
Vậy x = 15
TT
0
3/5 : 2x/15=1/2:4/5
= 3/5 : 2x/15 = 5/8
= 45/10x = 5/8
= 9/2x = 5/8
= 10x = 72
= x = 72:10
= x = 36/5
b, -4/2,5:3/5=1/5:x
= 1/5 : x = -8/3
= x = 1/5 : (-8/3)
= x = -3/40
0,12:3=2x:3/5
= 1/25 = 10/3 x
= x = 10/3:1/25
= x = 100/3
= x = 10/3:1/25\(|\)
= x = 100/3 | Oi zuiiii, chết mk nhầm bạn sửa lại thành như thế này nhé.
=> 1/25:10/3
=> 3/250