tìm x,biết
\(\left(b\right)\frac{-2}{x}=\frac{-x}{\frac{8}{25}}\)
giải chi tiết nhé mk tick cho
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a) Lm r nkoa!
b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)
\(=\left(0,25x\right):3=\frac{20}{3}\)
\(\Rightarrow0,25x=\frac{20}{3}\cdot3=20\)
\(\Rightarrow x=20:0,25=80\)
\(\Rightarrow x=80\)
c) \(0,01:2,5=\left(0,75x\right):0,75\)
\(=\frac{1}{250}=\left(0,75x\right):0,75\)
\(\Rightarrow0,75x=\frac{1}{250}\cdot0,75=\frac{3}{1000}\)
\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)
\(\Rightarrow x=\frac{1}{250}\)
d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)
\(=\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow0,1x=\frac{2}{3}:\frac{5}{3}=\frac{2}{5}\)
\(\Rightarrow x=\frac{2}{5}:0,1=4\)
\(\Rightarrow4\)
a) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)\(5\)
=> \(\frac{2}{3}-\left(\frac{1}{3}x-\frac{1}{2}\right)-\left(x+\frac{1}{2}\right)=5\)
=>\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
=>\(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5\)
=>\(\frac{2}{3}-\frac{4}{3}x=5\)
=>\(\frac{4}{3}x=\frac{2}{3}-5=-\frac{13}{3}\)
=>\(x=-\frac{13}{3}:\frac{4}{3}=-\frac{13}{4}\)
b)\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
=>\(4x-x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)
=> \(3x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)
=>\(x=-\left(-\frac{9}{2}\right)+\frac{1}{2}=5\)
\(2-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)\)\(:16\frac{2}{3}=0\)
\(\Rightarrow\)\(2-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)=0\)
\(\Rightarrow\)\(5\frac{3}{8}+x-7\frac{5}{24}=2\)
\(\Rightarrow5\frac{3}{8}+x=9\frac{5}{24}\)
\(\Rightarrow x=3\frac{5}{6}\)
( Cách làm thì đúng rồi nhưng đáp án tớ ko chắc chắn )
a) \(\left(x+\frac{1}{3}\right)^3=\frac{-8}{27}\)
\(\left(x+\frac{1}{3}\right)^3=\left(\frac{-2}{3}\right)^3\)
\(x+\frac{1}{3}=\frac{-2}{3}\)
\(x=-1\)
b) \(\left(\frac{1}{3}x+\frac{4}{3}\right)^2=\frac{25}{9}\)
\(\left(\frac{1}{3}x+\frac{4}{3}\right)^2=\left(\frac{5}{3}\right)^2\)
\(\frac{1}{3}x+\frac{4}{3}=\frac{5}{3}\)
\(\frac{1}{3}x=\frac{1}{3}\)
\(x=1\)
c) \(2^x+2^{x+1}=24\)
\(2^x+2^x.2=24\)
\(2^x.\left(1+2\right)=24\)
\(2^x.3=24\)
\(2^x=8\)
\(2^x=2^3\)
\(x=3\)
a, (x+1/3)^3 = -8/27
=>(x+1/3)^3 = (-2/3)^3
=>x+1/3 = -2/3
=>x = -1
b, (1/3x+4/3)^2 = 25/9
=>(1/3x+4/3)^2 = (5/3)^2
=>(1/3x+4/3) = 5/3
=>1/3x = 1/3
=> x = 1
c, 2^x + 2^x+1 = 24
=>2^x + 2^x . 2 = 24
=>2^x.(1+2) = 24
=>2^x . 3 = 24
=>2^x =8
=>2^x = 2^3
=> x = 3
a.
\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)
TH1:
\(x+\frac{1}{2}=0\)
\(x=-\frac{1}{2}\)
TH2:
\(x-\frac{3}{4}=0\)
\(x=\frac{3}{4}\)
Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{3}{4}\)
b.
\(\left(\frac{1}{2}x-3\right)\times\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)
TH1:
\(\frac{1}{2}x-3=0\)
\(\frac{1}{2}x=3\)
\(x=3\div\frac{1}{2}\)
\(x=3\times2\)
\(x=6\)
TH2:
\(\frac{2}{3}x+\frac{1}{2}=0\)
\(\frac{2}{3}x=-\frac{1}{2}\)
\(x=-\frac{1}{2}\div\frac{2}{3}\)
\(x=-\frac{1}{2}\times\frac{3}{2}\)
\(x=-\frac{3}{4}\)
Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)
c.
\(\frac{2}{3}-\frac{1}{3}\times\left(x-\frac{3}{2}\right)-\frac{1}{2}\times\left(2x+1\right)=5\)
\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5-\frac{2}{3}\)
\(-\frac{4}{3}x=\frac{13}{3}\)
\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)
\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)
\(x=-\frac{13}{4}\)
d.
\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)
\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)
\(x=5\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{\left(x+1\right)-x}{x\left(x+1\right)}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1004}{2010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2010}\)
\(\Leftrightarrow x+1=2010\)
\(\Leftrightarrow x=2009\)
\(\frac{-2}{x}=\frac{-x}{\frac{8}{25}}\)
=>\(\left(-x\right)x=\left(-2\right).\frac{8}{25}\)
=>\(-x^2=-\frac{16}{25}\)
=>\(x^2=\frac{16}{25}\)
=>\(x=-\frac{4}{5}\) hoặc \(x=\frac{4}{5}\)
\(-\frac{2}{x}=-\frac{x}{\frac{8}{25}}\Rightarrow-2.-\frac{8}{25}=x^2\Rightarrow\frac{16}{25}=x^2\Rightarrow x=\frac{\sqrt{16}}{\sqrt{25}}=\frac{4}{5}\)