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LG
1
28 tháng 7 2020
a) \(\frac{0,5}{0,2}=\frac{1,25}{0,1x}\Leftrightarrow0,1x.0,5=0,2.1,25\)
\(\Leftrightarrow0,1x.0,5=0,25\Leftrightarrow0,1x=0,5\Leftrightarrow x=5\)
b) \(x-\frac{3}{2}=2x-\frac{4}{3}\Leftrightarrow x-2x=\frac{-4}{3}+\frac{3}{2}\)
\(\Leftrightarrow x-2x=\frac{1}{6}\Leftrightarrow-x=\frac{1}{6}\Leftrightarrow x=\frac{-1}{6}\)
c) \(x+\frac{13}{14}=\frac{4}{7}\Rightarrow x=\frac{4}{7}-\frac{13}{14}\Rightarrow x=\frac{-5}{14}\)
d)\(-3\left(x-2\right)=2x+1\)
\(\Leftrightarrow-3x+6=2x+1\Leftrightarrow-3x-2x=1-6\)
\(\Leftrightarrow-5x=-5\Leftrightarrow x=1\)
e) \(\left(x-1\right)^2-4=0\Leftrightarrow\left(x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
cậu có thể tham khảo bài trên ạ, nếu thấy đúng thì cho mk 1 t.i.c.k ạ, thank nhiều
PN
28 tháng 7 2020
\(d,-3\left(x-2\right)=2x+1\)
\(< =>-3x+6=2x+1\)
\(< =>-3x-2x+6-1=0\)
\(< =>5-5x=0\)
\(< =>5\left(1-x\right)=0< =>x=1\)
\(e,\left(x-1\right)^2-4=0\)
\(< =>\left(x-1+2\right)\left(x-1-2\right)=\left(x+1\right)\left(x-3\right)=0\)
\(< =>\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}< =>\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
VN
Võ Ngọc Phương
CTVHS
VIP
11 tháng 9 2023
Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
TP
16 tháng 7 2017
a, 2 -|3/2x -1/4| =|-1,25 |
=>2 -|3/2x-1/4 | = 1,25
=> |3/2x -1/4| = 2-1,25
=> 3/2x -1/4 = 0,75 hoac 3/2x -1/4 = -0,75
=> 3/2 x = 3/4 -1/4 hoac 3/2 x = -3/4 -1/4
=> 3/2 x = 1/2 hoac 3/2 x = -1
=> x = 1/2 :3/2 hoac x = -1 : 3/2
=> x = 1/3 hoac x = -2/3
vay
PP
16 tháng 7 2017
cảm ơn nhé nhưng mk bt làm òi
mk chỉ đăng câu hỏi cho vv thui
21 tháng 8 2017
a. \(\dfrac{2x}{3}:\dfrac{1}{5}=1\dfrac{1}{3}:\dfrac{1}{4}\Rightarrow\dfrac{2x}{3}\cdot\dfrac{1}{4}=\dfrac{1}{5}\cdot1\dfrac{1}{3}\Rightarrow\dfrac{2x}{3}=\dfrac{1}{5}\cdot1\dfrac{1}{3}:\dfrac{1}{4}=\dfrac{16}{15}\Rightarrow2x=\dfrac{16}{15}\cdot3=\dfrac{16}{5}\Rightarrow x=\dfrac{16}{5}:2=\dfrac{8}{5}\)
Vậy \(x=\dfrac{8}{5}\)
b. \(1,35:0,2=1,25:0,1x\Rightarrow1,35\cdot0,1x=0,2\cdot1,25\Rightarrow0,1x=\dfrac{0,2\cdot1,25}{1,35}=\dfrac{5}{27}\Rightarrow x=\dfrac{5}{27}:0,1=\dfrac{50}{27}\)
Vậy \(x=\dfrac{50}{27}\)
c. \(2:1\dfrac{1}{4}=\dfrac{1}{2}:2x\Rightarrow2\cdot\dfrac{4}{5}=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot x\Rightarrow\dfrac{8}{5}=\dfrac{1}{4}x\Rightarrow x=\dfrac{8}{5}:\dfrac{1}{4}=\dfrac{32}{5}\)
Vậy \(x=\dfrac{32}{5}\)
1 tháng 2 2022
a: \(=\left(1.25\right)^{16}\cdot8^{16}\cdot8=8\cdot10^{16}\)
b: \(=\left(\dfrac{5}{2}\right)^{13}\cdot4^{13}\cdot4^2=10^{13}\cdot4^2\)
c: \(=\left(0.25\right)^4\cdot8^4\cdot8^2=2^4\cdot8^2=64\cdot16=1024\)
d: \(=\left(\dfrac{1}{2}\right)^{15}\cdot2^{18}=2^3=8\)
e: \(=\left(\dfrac{1}{3}\cdot6\right)^7\cdot\left(\dfrac{1}{2}\right)^7\cdot\dfrac{1}{2}=2^7\cdot\left(\dfrac{1}{2}\right)^7\cdot\dfrac{1}{2}=\dfrac{1}{2}\)
6 tháng 12 2021
\(=\left(\dfrac{1}{2}-\dfrac{6}{5}\right):\dfrac{21}{20}-\dfrac{25}{4}+\dfrac{1}{2}\)
\(=\dfrac{-7}{10}\cdot\dfrac{20}{21}-\dfrac{25}{4}+\dfrac{2}{4}\)
\(=\dfrac{-2}{3}-\dfrac{23}{4}=\dfrac{-8-69}{12}=\dfrac{-77}{12}\)
R
6 tháng 12 2021
\(\left(\sqrt{\dfrac{1}{4}}-1,2\right):1\dfrac{1}{20}-\left(-\dfrac{5}{2}\right)^2+\left|1,25-\dfrac{3}{4}\right|\)
\(=-\dfrac{7}{10}:\dfrac{21}{20}-\dfrac{25}{4}+\left|\dfrac{1}{2}\right|\)
\(=-\dfrac{7}{10}.\dfrac{20}{21}-\dfrac{25}{4}+\dfrac{1}{2}\)
\(=-\dfrac{2}{3}-\dfrac{25}{4}+\dfrac{1}{2}\)
\(=-\dfrac{77}{12}\)
|2x+1/3|-1,25=(-1/2)2
|2x+1/3| - 1\(\frac{1}{4}\) =1/4
|2x+1/3| = 3/2
Ta có 2TH:
TH1: 2x+1/3=3/2
2x= 7/6
x= 7/12
TH2: 2x+1/3= -3/2
2x= -11/6
x= -11/12
Vậy x=7/12 hoặc x=-11/12