Tìm x biết: -\(\frac{2}{3}\)x (X-\(\frac{1}{4}\))=\(\frac{1}{3}\)x (2X-1)
Giúp mk nha. Mk tik cho.
K
Khách
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NX
31 tháng 7 2018
\(\frac{1}{3}\) + \(\frac{5}{6}\): \(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)
<=> \(\frac{5}{6}\):\(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)- \(\frac{1}{3}\)
<=> \(\frac{5}{6}\) : \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{12}\)
<=> \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{6}\) : \(\frac{5}{12}\)
,<=> \(\left(x-2\frac{1}{5}\right)\)= 2
<=. x = 2 + \(\frac{11}{5}\)
<=> x = \(\frac{21}{5}\)
KT
1
22 tháng 3 2017
\(\frac{-6}{3}\left[x-\frac{1}{4}\right]=2x-1\)
\(-2x-\left[\frac{1}{4}.-2\right]=2x-1\)\
\(-2x-\frac{-1}{2}=2x-1\)
\(2x--2x=1-\frac{-1}{2}\)
\(\)\(4x=\frac{3}{2}\)
\(x=\frac{3}{2}:4\)
\(x=\frac{3}{8}\)
T
28 tháng 7 2018
\(\frac{-11}{9}\le x+\frac{11}{18}\Leftrightarrow x\ge\frac{-11}{6}\)
28 tháng 7 2018
\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow\frac{-13}{9}\le x-\frac{-11}{18}\)
\(\Leftrightarrow\frac{-13}{9}\le x+\frac{11}{18}\)
\(\Rightarrow x\ge\frac{-37}{18}\)
Học tốt nhé !! ^-^
7 tháng 9 2018
\(\frac{x-1}{4}=\frac{2x+1}{5}\)
\(\Rightarrow5\left(x-1\right)=4\left(2x+1\right)\)
\(\Rightarrow5x-5=8x+4\)
\(\Rightarrow5x-8x=4+5\)
\(\Rightarrow-3x=9\)
\(\Rightarrow x=-3\)
vậy_
\(\frac{x+2}{x-1}=\frac{x-3}{x+1}\)
\(\Rightarrow\left(x+2\right)\left(x+1\right)=\left(x-1\right)\left(x-3\right)\)
\(\Rightarrow x^2+x+2x+2=x^2-3x-x+3\)
\(\Rightarrow x^2+x+2x-x^2+3x+x=3-2\)
\(\Rightarrow7x=1\)
\(\Rightarrow x=\frac{1}{7}\)
vậy_
VT
17 tháng 9 2016
A=5-3(2x+1)^2
Ta có : (2x+1)^2\(\ge\)0
\(\Rightarrow\)-3(2x-1)^2\(\le\)0
\(\Rightarrow\)5+(-3(2x-1)^2)\(\le\)5
Dấu = xảy ra khi : (2x-1)^2=0
=> 2x-1=0 =>x=\(\frac{1}{2}\)
Vậy : A=5 tại x=\(\frac{1}{2}\)
Ta có : (x-1)^2 \(\ge\)0
=> 2(x-1)^2\(\ge\)0
=>2(x-1)^2+3 \(\ge\)3
=>\(\frac{1}{2\left(x-1\right)^2+3}\)\(\le\)\(\frac{1}{3}\)
Dấu = xảy ra khi : (x-1)^2 =0
=> x = 1
Vậy : B = \(\frac{1}{3}\)khi x = 1
\(\frac{x^2+8}{x^2+2}\)= \(\frac{x^2+2+6}{x^2+2}=1+\frac{6}{x^2+2}\)
Làm như câu B GTNN = 4 khi x =0
k vs nha
Ta có\(-\frac{2}{3}\) \(X\) (\(X\) \(-\frac{1}{4}\) ) = \(\frac{1}{3}\)\(X\) (\(2X-1\) )
\(\Rightarrow\) \(\frac{-2}{3}\) \(X^2\)\(+\) \(\frac{1}{6}\) \(X\) = \(\frac{2}{3}\) \(X^2\) \(-\) \(\frac{1}{3}\) \(X\)
\(\Rightarrow\) \(\frac{-2}{3}\) \(X\) \(+\) \(\frac{1}{6}\) = \(\frac{2}{3}\) \(X\) \(-\) \(\frac{1}{3}\)
\(\Rightarrow\) \(\frac{-2}{3}\) \(X\) \(+\) \(\frac{1}{6}\) \(+\) \(\frac{1}{3}\) = \(\frac{2}{3}\) \(X\)
\(\Rightarrow\) \(\frac{-2}{3}\) \(X\) \(+\) \(\frac{1}{2}\) = \(\frac{2}{3}\) \(X\)
\(\Rightarrow\) \(\frac{1}{2}\) = \(\frac{2}{3}\) \(X\) \(+\) \(\frac{2}{3}\) \(X\)
\(\Rightarrow\) \(\frac{1}{2}\) = \(X\) (\(\frac{2}{3}\) \(+\) \(\frac{2}{3}\) )
\(\Rightarrow\) \(\frac{1}{2}\) = \(\frac{4}{3}\) \(X\)
\(\Rightarrow\) \(X\) = \(\frac{1}{2}\) \(\div\) \(\frac{4}{3}\)
\(\Rightarrow\) \(X\) = \(\frac{3}{8}\)
Có gì không hiểu cứ hỏi tớ nhá !
Thank you