Tìm x, biết:
a ) 3 8 . x = 3 5 b ) x . 5 3 = 11 3 c ) x : 7 4 = 2 7 d ) 3 5 : x = 6 5
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a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)
b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)
Bài 7:
a, \(x\) = \(\dfrac{1}{5}\) + \(\dfrac{2}{11}\)
\(x\) = \(\dfrac{11}{55}\) + \(\dfrac{10}{55}\)
\(x=\dfrac{21}{55}\)
b, \(\dfrac{x}{15}\) = \(\dfrac{3}{5}\) - \(\dfrac{2}{3}\)
\(\dfrac{x}{15}\) = \(\dfrac{9}{15}\) - \(\dfrac{10}{15}\)
\(\dfrac{x}{15}\) = \(\dfrac{1}{15}\)
\(x\) = 1
c, \(\dfrac{11}{8}\) + \(\dfrac{13}{6}\)= \(\dfrac{85}{x}\)
\(\dfrac{33}{24}\) + \(\dfrac{52}{24}\) = \(\dfrac{85}{x}\)
\(\dfrac{85}{24}\) = \(\dfrac{85}{x}\)
24 = \(x\)
1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
\(a.\dfrac{1}{2}:3+x=\dfrac{14}{5}\)
\(\dfrac{1}{6}+x=\dfrac{14}{5}\)
\(=>x=\dfrac{79}{30}\)
\(b.\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\)
\(\left(\dfrac{8}{5}\cdot\dfrac{4}{7}\right):x=\dfrac{11}{6}\)
\(\dfrac{32}{35}:x=\dfrac{11}{6}\)
\(x=\dfrac{192}{385}\)
\(c.\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\)
\(x:\dfrac{3}{4}=\dfrac{11}{3}-\dfrac{24}{10}\)
\(x:\dfrac{3}{4}=\dfrac{38}{30}\)
\(=>x=\dfrac{19}{20}\)
\(a,\dfrac{1}{2}:3+x=\dfrac{14}{5}\\ \Leftrightarrow x+\dfrac{1}{6}=\dfrac{14}{5}\\ \Leftrightarrow x=\dfrac{79}{30}\\ b,\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\\ \Leftrightarrow x=\dfrac{192}{385}\\ c,\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\\ \Leftrightarrow\dfrac{4}{3}x=\dfrac{19}{15}\\ \Leftrightarrow x=\dfrac{19}{20}\)
***c,4(x-5)-3(x+7)=-19
Suy ra (4x-20)-(3x+21)=-19
Suy ra (4x-3x)-(20+21)=-19
Suy ra x-41=-19
suy ra x=-19+41
Suy ra x=22
Vậy x=22
a) \(\left(2x+3\right)^3=\left(2x+3\right)^8\)
TH1 \(2x+3=1\)
\(2x=1-3=-2\)
\(x=-1\)
TH2 \(2x+3=0\)
\(2x=-3\Rightarrow x=-\frac{3}{2}\)
b) ? sai đề
c) \(\left|5-3\right|=\left|11+2x\right|\Rightarrow\left|2\right|=\left|11+2x\right|\)
\(\hept{\begin{cases}11+2x=-2\\11+2x=2\end{cases}\Rightarrow}\hept{\begin{cases}2x=13\\2x=9\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{13}{2}\\x=\frac{9}{2}\end{cases}}\)
d) \(\left(x-5\right)^4=\left(x-5\right)^6\Rightarrow\hept{\begin{cases}x-5=0\\x-5=1\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=6\end{cases}}\)
a) \(\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
=> \(\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x+\frac{1}{3}=0\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}x\right)+\left(-\frac{2}{5}+\frac{1}{3}\right)=0\)
=> \(\frac{1}{6}x-\frac{1}{15}=0\Rightarrow\frac{1}{6}x=\frac{1}{15}\Rightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{2}{5}\)
Vậy x = 2/5
b) \(\frac{1}{3}x+\frac{2}{5}\left(x+1\right)=0\)
=> \(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
=> \(\frac{11}{15}x+\frac{2}{5}=0\Rightarrow\frac{11}{15}x=-\frac{2}{5}\)
=> \(x=\left(-\frac{2}{5}\right):\frac{11}{15}=\left(-\frac{2}{5}\right)\cdot\frac{15}{11}=-\frac{6}{11}\)
Vậy x = -6/11
c) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
=> \(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
=> \(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{1}{3}x-x\right)=5\)
=> \(\frac{2}{3}-\frac{4}{3}x=5\)
=> \(\frac{4}{3}x=-\frac{13}{3}\Rightarrow x=\left(-\frac{13}{3}\right):\frac{4}{3}=\left(-\frac{13}{3}\right)\cdot\frac{3}{4}=-\frac{13}{4}\)
Vậy x = -13/4
d) \(\frac{11}{5}-\left(\frac{7}{9}-x\right)\cdot\frac{3}{8}=\frac{61}{90}+\frac{x}{3}\)
=> \(\frac{11}{5}-\frac{3}{8}\left(\frac{7}{9}-x\right)=\frac{61}{90}+\frac{30x}{90}\)
=> \(\frac{11}{5}-\frac{7}{24}+\frac{3}{8}x=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{3}{8}x=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{3x}{8}=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{45x}{120}=\frac{61+30x}{90}\)
=> \(\frac{229+45x}{120}=\frac{61+30x}{90}\)
=> \(\frac{3\left(229+45x\right)}{360}=\frac{4\left(61+30x\right)}{360}\)
=> \(3\left(229+45x\right)=4\left(61+30x\right)\)
=> \(687+135x=244+120x\)
=> \(687+135x-244-120x=0\)
=> \(\left(687-244\right)+\left(135x-120x\right)=0\)
=> \(443+15x=0\)
=> \(15x=-443\Rightarrow x=-\frac{443}{15}\)
Vậy x = -443/15