Tìm x a,x+1/2•1/3=3/4
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Những câu hỏi liên quan
18 tháng 8 2021
Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
KV
4 tháng 11 2023
Bài 1
a) 3 2/5 - 1/2
= 17/5 - 1/2
= 34/10 - 5/10
= 29/10
b) 4/5 + 1/5 × 3/4
= 4/5 + 3/20
= 16/20 + 3/20
= 19/20
c) 3 1/2 × 1 1/7
= 7/2 × 8/7
= 4
d) 4 1/6 : 2 1/3
= 25/6 : 7/3
= 25/14
KV
4 tháng 11 2023
Bài 2
a) 3 × 1/2 + 1/4 × 1/3
= 3/2 + 1/12
= 18/12 + 1/12
= 19/12
b) 1 4/5 - 2/3 : 2 1/3
= 9/5 - 2/3 : 7/3
= 9/5 - 2/7
= 63/35 - 10/35
= 53/35
T
22 tháng 8 2023
\(a,x+\dfrac{1}{4}\times\dfrac{1}{5}=\dfrac{4}{5}\)
\(x+\dfrac{1}{20}=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}-\dfrac{1}{20}\)
\(x=\dfrac{16}{20}-\dfrac{1}{20}\)
\(x=\dfrac{15}{20}=\dfrac{3}{4}\)
\(b,3\dfrac{2}{3}:x=2\dfrac{2}{3}-1\dfrac{1}{2}\)
\(\dfrac{11}{3}:x=\dfrac{8}{3}-\dfrac{3}{2}\)
\(\dfrac{11}{3}:x=\dfrac{7}{6}\)
\(x=\dfrac{11}{3}:\dfrac{7}{6}=\dfrac{22}{7}\)
22 tháng 8 2023
a: =>x+1/20=4/5
=>x=4/5-1/20=16/20-1/20=15/20=3/4
b: =>11/3:x=8/3-3/2=16/6-9/6=7/6
=>x=11/3:7/6=11/3*6/7=66/21=22/7
NH
17 tháng 8 2023
a)
`x+1/4xx1/5=4/5`
`x+1/20=4/5`
`x=4/5-1/20`
`x=16/20-1/20`
`x=15/20=3/4`
b)
`3 2/3:x=2 2/3-1 1/2`
`11/3:x=8/3-3/2`
`11/3:x=16/6-9/6`
`11/3:x=7/6`
`x=11/3:7/6`
`x=11/3xx6/7`
`x=22/7`
T
17 tháng 8 2023
\(a,x+\dfrac{1}{4}.\dfrac{1}{5}=\dfrac{4}{5}\)
\(\Rightarrow\)\(x+\dfrac{1}{20}=\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{4}{5}-\dfrac{1}{20}=\dfrac{16}{20}-\dfrac{1}{20}=\dfrac{3}{4}\)
\(b,3\dfrac{2}{3}:x=2\dfrac{2}{3}-1\dfrac{1}{2}\)
\(\Rightarrow\dfrac{11}{3}:x=\dfrac{7}{6}\)
\(\Rightarrow x=\dfrac{11}{3}:\dfrac{7}{6}=\dfrac{22}{7}\)
LT
3
14 tháng 7 2023
a) \(x-\dfrac{3}{4}=6.\dfrac{3}{8}=\dfrac{9}{4}\)
\(\Rightarrow x=\dfrac{9}{4}+\dfrac{3}{4}=3\)
c) \(x+\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{3}{4}\)
\(\Rightarrow x+\dfrac{1}{6}=\dfrac{3}{4}\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{9}{12}-\dfrac{2}{12}=\dfrac{7}{12}\)
Bạn xem lại đề c
NV
5 tháng 8 2023
a) \(2^x=8\)
⇔ \(2^x=2^3\)
⇒ \(x=3\)
b) \(3^x=27\)
⇔ \(3^x=3^3\)
⇒ \(x=3\)
c) \(\left(-\dfrac{1}{2}\right)x=\left(-\dfrac{1}{2}\right)^4\)
⇔ \(x=\left(-\dfrac{1}{2}\right)^4\div\left(-\dfrac{1}{2}\right)\)
⇔ \(x=\left(-\dfrac{1}{2}\right)^3\)
d) \(x\div\left(-\dfrac{3}{4}\right)=\left(-\dfrac{3}{4}\right)^2\)
⇔ \(x=\left(-\dfrac{3}{4}\right)^2\cdot\left(-\dfrac{3}{4}\right)\)
⇔ \(x=\left(-\dfrac{3}{4}\right)^3=-\dfrac{27}{64}\)
d) \(\left(x+1\right)^3=-125\)
⇔ \(\left(x+1\right)^3=\left(-5\right)^3\)
⇔ \(x+1=-5\)
⇔ \(x=-5-1=-6\)
5 tháng 8 2023
2:
a: (x-1,2)^2=4
=>x-1,2=2 hoặc x-1,2=-2
=>x=3,2(loại) hoặc x=-0,8(loại)
b: (x-1,5)^2=9
=>x-1,5=3 hoặc x-1,5=-3
=>x=-1,5(loại) hoặc x=4,5(loại)
c: (x-2)^3=64
=>(x-2)^3=4^3
=>x-2=4
=>x=6(nhận)
15 tháng 3 2021
Bài 1:
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)
\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)
Suy ra: \(-12x-3=8x-2-6x-8\)
\(\Leftrightarrow-12x-3-2x+10=0\)
\(\Leftrightarrow-14x+7=0\)
\(\Leftrightarrow-14x=-7\)
\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
24 tháng 8 2023
1: A=2
=>\(\sqrt{x}+1=2\left(\sqrt{x}-2\right)\)
=>\(2\sqrt{x}-4=\sqrt{x}+1\)
=>\(\sqrt{x}=5\)
=>x=25
2: A<1
=>A-1<0
=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
=>\(\dfrac{3}{\sqrt{x}-2}< 0\)
=>\(\sqrt{x}-2< 0\)
=>0<=x<4
3: A<1/3
=>A-1/3<0
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{1}{3}< 0\)
=>\(\dfrac{3\sqrt{x}+3-\sqrt{x}+2}{3\left(\sqrt{x}-2\right)}< 0\)
=>\(\dfrac{2\sqrt{x}+5}{3\left(\sqrt{x}-2\right)}< 0\)
=>\(\sqrt{x}-2< 0\)
=>0<=x<4
4:
A=căn x
=>\(\sqrt{x}+1=x-2\sqrt{x}\)
=>\(x-3\sqrt{x}-1=0\)
=>\(\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{13}}{2}\left(nhận\right)\\\sqrt{x}=\dfrac{3-\sqrt{13}}{2}\left(loại\right)\end{matrix}\right.\)
=>\(x=\dfrac{11+3\sqrt{13}}{2}\)
\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
=>\(x+\dfrac{1}{6}=\dfrac{3}{4}\)
=>\(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{9}{12}-\dfrac{2}{12}=\dfrac{7}{12}\)