a)
X x \(\dfrac{3}{4}\)+X=\(\dfrac{7}{8}\)
b)
1-(2\(\dfrac{3}{4}\)+X-1\(\dfrac{2}{3}\)):3 4/5=0
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\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)
b: Ta có: \(3^x+3^{x+2}=20\)
\(\Leftrightarrow3^x\cdot10=20\)
\(\Leftrightarrow3^x=2\left(loại\right)\)
\(a,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \Rightarrow\dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{26}{21}\\ \Rightarrow\dfrac{5}{2}-x=\dfrac{46}{105}\\ \Rightarrow x=\dfrac{433}{210}\\ b,-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow2x-\dfrac{4}{7}-x=\dfrac{3}{5}\\ \Rightarrow2x-x=\dfrac{41}{35}\\ \Rightarrow x=\dfrac{41}{35}\\ c,\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{40}+\dfrac{5}{8}-x=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{5}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{3}{5}.\)
\(a,\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\times\dfrac{8}{7}:2\\ =\left(\dfrac{31}{35}-\dfrac{4\times5}{35}\right)\times\dfrac{8}{7}:2\\ =\dfrac{11}{35}\times\dfrac{8}{7}:2\\ =\dfrac{88}{245}:2\\ =\dfrac{44}{245}\\ b,\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\left(\dfrac{2-1}{2}\right)\times\left(\dfrac{3-1}{3}\right)\times\left(\dfrac{4-1}{4}\right)\times\left(\dfrac{5-1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)
a, ( \(\dfrac{31}{35}\) - \(\dfrac{4}{7}\)) \(\times\) \(\dfrac{8}{7}\): 2
= \(\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) \(\times\) \(\dfrac{8}{7}\) : 2
= \(\dfrac{11}{35}\) \(\times\) \(\dfrac{8}{7}\) \(\times\) \(\dfrac{1}{2}\)
= \(\dfrac{44}{35}\) \(\times\) \(\dfrac{4}{7}\)
= \(\dfrac{44}{245}\)
b, ( 1 - \(\dfrac{1}{2}\)) \(\times\) ( 1 - \(\dfrac{1}{3}\)) \(\times\) ( 1 - \(\dfrac{1}{4}\)) \(\times\) ( 1 - \(\dfrac{1}{5}\))
= \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\)
= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{2\times3\times4}{2\times3\times4}\)
= \(\dfrac{1}{5}\)
\(\dfrac{3}{2}x-0,2=\dfrac{3}{5}\)
\(\dfrac{3}{2}x-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\dfrac{3}{2}x=\dfrac{3}{5}+\dfrac{1}{5}\)
\(\dfrac{3}{2}x=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}:\dfrac{3}{2}\)
\(x=\dfrac{4}{5}\cdot\dfrac{2}{3}\)
\(x=\dfrac{8}{15}\)
\(\dfrac{1}{3}+x=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{1}{3}\)
\(x=\dfrac{9}{12}-\dfrac{4}{12}\)
\(x=\dfrac{5}{12}\)
\(1\dfrac{1}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{3}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{3}{2}x=\dfrac{1}{4}+\dfrac{2}{5}\)
\(\dfrac{3}{2}x=\dfrac{13}{20}\)
\(x=\dfrac{13}{20}:\dfrac{3}{2}\)
\(x=\dfrac{13}{20}\cdot\dfrac{2}{3}\)
\(x=\dfrac{13}{30}\)
\(\dfrac{11}{8}-\dfrac{3}{8}\cdot x=\dfrac{1}{8}\)
\(\dfrac{3}{8}\cdot x=\dfrac{11}{8}-\dfrac{1}{8}\)
\(\dfrac{3}{8}\cdot x=\dfrac{5}{4}\)
\(x=\dfrac{5}{4}:\dfrac{3}{8}\)
\(x=\dfrac{5}{4}\cdot\dfrac{8}{3}\)
\(x=\dfrac{10}{3}\)
\(\dfrac{5}{4}+\left(2x-\dfrac{1}{2}\right)=\dfrac{5}{6}\\ =>2x-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{5}{4}\\ =>2x-\dfrac{1}{2}=\dfrac{10}{12}-\dfrac{15}{12}\\ =>2x-\dfrac{1}{2}=-\dfrac{5}{12}\\ =>2x=-\dfrac{5}{12}+\dfrac{1}{2}\\ =>2x=-\dfrac{5}{12}+\dfrac{6}{12}\\ =>2x=\dfrac{1}{12}\\ =>x=\dfrac{1}{12}:2\\ =>x=\dfrac{1}{12}\cdot\dfrac{1}{2}\\ =>x=\dfrac{1}{24}\)
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\(\dfrac{3}{2}-\left(x+\dfrac{1}{4}\right)=\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{3}{2}-\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{12}{8}-\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{7}{8}\\ =>x=\dfrac{7}{8}-\dfrac{1}{4}\\ =>x=\dfrac{7}{8}-\dfrac{2}{8}\\ =>x=\dfrac{5}{8}\)
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\(\dfrac{x}{3}=\dfrac{12}{x}\\ =>x^2=3\cdot12\\ =>x^2=36\\ =>x^2=6^2\\ =>x=\pm6\)
Tìm x:
a) \(\dfrac{5}{4}+\left(2x-\dfrac{1}{2}\right)=\dfrac{5}{6}\)
\(=>2x-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{5}{4}\)
\(=>2x-\dfrac{1}{2}=\dfrac{-5}{12}\)
\(=>2x=\dfrac{-5}{12}+\dfrac{1}{2}\)
\(=>2x=\dfrac{1}{12}\)
\(=>x=\dfrac{1}{12}:2\)
\(=>x=\dfrac{1}{24}\)
b) \(\dfrac{3}{2}-\left(x+\dfrac{1}{4}\right)=\dfrac{5}{8}\)
\(=>x+\dfrac{1}{4}=\dfrac{3}{2}-\dfrac{5}{8}\)
\(=>x+\dfrac{1}{4}=\dfrac{7}{8}\)
\(=>x=\dfrac{7}{8}-\dfrac{1}{4}\)
\(=>x=\dfrac{5}{8}\)
c) \(\dfrac{x}{3}=\dfrac{12}{x}\)
Ta có: \(x.x=3.12\)
\(\Rightarrow x^2=36\)
Vậy x = 6 hoặc x = -6
Chúc bạn học tốt
a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)
hay \(x=-\dfrac{17}{21}\)
Vậy: \(x=-\dfrac{17}{21}\)
b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)
Vậy: \(x=\dfrac{4}{5}\)
c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)
\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)
\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)
hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)
Vậy: \(x=-\dfrac{5}{7}\)
f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)
\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)
\(\Leftrightarrow-x-\dfrac{9}{60}=0\)
\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)
hay \(x=-\dfrac{3}{20}\)
Vậy: \(x=-\dfrac{3}{20}\)
g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)
a) \(\dfrac{2}{3}+\dfrac{3}{4}< x< 1\dfrac{1}{3}+\dfrac{4}{5}\)
\(\dfrac{2\times4}{3\times4}+\dfrac{3\times3}{4\times3}< x< \dfrac{\left(1\times3+1\right)\times5}{3\times5}+\dfrac{4\times3}{5\times3}\)
\(\dfrac{8}{12}+\dfrac{9}{12}< x< \dfrac{20}{15}+\dfrac{12}{15}\\ \dfrac{17}{12}< x< \dfrac{32}{15}\)
Ước tính: \(\dfrac{17}{12}=1,4\) và \(\dfrac{32}{15}=2,1\). Vậy số tự nhiên x = 2 sẽ thõa mãn 1,4 < x < 2,1
b)
\(\dfrac{5}{6}-\dfrac{1}{4}< x< 2\dfrac{1}{3}-\dfrac{2}{5}\\ \dfrac{5\times4}{6\times4}-\dfrac{1\times6}{4\times6}< x< \dfrac{\left(2\times3+1\right)\times5}{3\times5}-\dfrac{2\times3}{5\times3}\\ \dfrac{20}{24}-\dfrac{6}{24}< x< \dfrac{35}{15}-\dfrac{6}{15}\\ \dfrac{14}{24}< x< \dfrac{29}{15}\)
Ước tính \(\dfrac{14}{24}=0,5\) và \(\dfrac{29}{15}=1,9\)
Vậy với x là số tự nhiên x = 1 sẽ thõa mãn 0,5 < x < 1,9
a, \(x\cdot\dfrac{-5}{8}=\dfrac{15}{32}\)
\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)
\(x=\dfrac{-3}{4}\)
b, \(\dfrac{3}{10}:x=\dfrac{-9}{20}\)
\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)
\(x=-\dfrac{2}{3}\)
c, \(\dfrac{-1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)
\(\dfrac{-1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\dfrac{-1}{4}x=-\dfrac{1}{20}\)
\(x=-\dfrac{1}{20}:\dfrac{-1}{4}\)
\(x=\dfrac{1}{5}\)
d, \(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}\cdot\dfrac{-5}{12}\)
\(\dfrac{-7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)
\(\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{-7}{8}\)
\(\dfrac{2}{3}:x=\dfrac{5}{8}\)
\(x=\dfrac{2}{3}:\dfrac{5}{8}\)
\(x=\dfrac{16}{15}\)
#YVA6
\(a,x.\dfrac{-5}{8}=\dfrac{15}{32}\)
\(\Leftrightarrow x=\dfrac{15}{32}:\dfrac{-5}{8}\)
\(\Leftrightarrow x=\dfrac{15}{32}.\dfrac{-8}{5}\)
\(\Leftrightarrow x=-\dfrac{3}{4}\)
\(b,\dfrac{3}{10}:x=-\dfrac{9}{20}\)
\(\Leftrightarrow x=\dfrac{3}{10}:\dfrac{-9}{20}\)
\(\Leftrightarrow x=\dfrac{3}{10}.\dfrac{-20}{9}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c,-\dfrac{1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)
\(\Leftrightarrow-\dfrac{1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Leftrightarrow-\dfrac{1}{4}x=-\dfrac{1}{20}\)
\(\Leftrightarrow x=-\dfrac{1}{20}\times\left(-4\right)\)
\(\Leftrightarrow x=\dfrac{1}{5}\)
\(d,-\dfrac{7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)
\(\Leftrightarrow-\dfrac{7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{7}{8}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{16}{15}\)
a: \(x\cdot\dfrac{3}{4}+x=\dfrac{7}{8}\)
\(\Leftrightarrow x\cdot\dfrac{7}{4}=\dfrac{7}{8}\)
\(\Leftrightarrow x=\dfrac{1}{2}\)