\(\dfrac{5}{6}-\left(\dfrac{2}{9}-x\right)=\dfrac{-15}{20}\)
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24 tháng 1 2022
=\(\left[\dfrac{\left(0,4.2\right)^5}{\left(0,4\right)^6}+\dfrac{2^9.2^6.3^8}{\left(3.2\right)^6.2^9}\right]=\left[\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}+\dfrac{2^6.3^8}{3^6.2^6}\right]\)
=\(\left[\dfrac{2^5}{0,4}+3^2\right]\)
=\(\left[80+9\right]=89\)
R
24 tháng 1 2022
\(\left[\dfrac{\left(2.0,4\right)^5}{0,4,0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^6.2^9}\right]\div\dfrac{3^{20}.5^{30}}{3^{15}.5^{30}}\)
\(=\left[\dfrac{2^5.0.4^5}{0,4.0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^{15}}\right]\div3^5\)
\(=\left[\dfrac{2^5}{0,4}+3^2\right]\div243\)
\(=80+\left(3^5\div3^2\right)\)
\(=80+3^3\)
\(=80+27\)
\(=107\)
27 tháng 6 2023
a: =11+3/4-6-5/6+4+1/2+1+2/3
=10+9/12-10/12+6/12+8/12
=10+13/12=133/12
b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)
=3-11/15
=34/15
c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)
d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 3 2023
\(F=\dfrac{5}{6}+6\dfrac{5}{6}\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(F=\dfrac{5}{6}+\dfrac{41}{6}\left(\dfrac{225}{20}-\dfrac{37}{4}\right):\dfrac{25}{3}\)
\(F=\dfrac{5}{6}+\dfrac{41}{6}.2.\dfrac{3}{25}\)
\(F=\dfrac{5}{6}+\dfrac{41}{25}.\dfrac{3}{25}\)
\(F=\dfrac{5}{6}+\dfrac{41}{25}\)
\(F=\dfrac{371}{150}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 3 2023
\(D=\left(\dfrac{136}{15}-\dfrac{28}{5}+\dfrac{62}{10}\right)\times\dfrac{21}{24}\)
\(D=\left(\dfrac{272}{30}-\dfrac{168}{30}+\dfrac{186}{30}\right)\times\dfrac{21}{24}\)
\(D=\dfrac{290}{30}\times\dfrac{21}{24}\)
\(D=\dfrac{29}{3}\times\dfrac{7}{8}\)
\(D=\dfrac{203}{24}\)
20 tháng 10 2023
d: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\left(3-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\dfrac{5}{6}\cdot\dfrac{6}{5}-17=1-17=-16\)
h: \(\dfrac{\left(-1\right)^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left|-\dfrac{5}{6}\right|\)
\(=-\dfrac{1}{15}+\dfrac{-8}{27}:\dfrac{8}{3}-\dfrac{5}{6}\)
\(=-\dfrac{1}{15}-\dfrac{1}{9}-\dfrac{5}{6}\)
\(=\dfrac{-6-10-75}{90}=\dfrac{-91}{90}\)
k: \(\dfrac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\)
\(=\dfrac{2^{10}\cdot3^9-2^5\cdot2^4\cdot3^8}{2^2\cdot2^8\cdot3^8}\)
\(=\dfrac{2^{10}\cdot3^9-2^9\cdot3^8}{2^{10}\cdot3^8}=\dfrac{2^9\cdot3^8\left(2\cdot3-1\right)}{2^{10}\cdot3^8}\)
\(=\dfrac{5}{2}\)
n: \(3-\left(-\dfrac{7}{8}\right)^0+\left(\dfrac{1}{2}\right)^3\cdot16\)
\(=3-1+\dfrac{1}{8}\cdot16\)
=2+2
=4
19 tháng 8 2021
a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)
\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)
b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)
\(\Leftrightarrow x+1=\dfrac{50}{9}\)
hay \(x=\dfrac{41}{9}\)
c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
hay \(x\in\left\{8;-8\right\}\)
NT
16 tháng 10 2022
c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x-1\right).\left(x+1\right)\)
\(63=x^2-1\)
\(x^2=63+1\)
\(x^2=64\)
\(x^2=8^2\)
\(x=8\)
16 tháng 2 2021
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
18 tháng 3 2023
a: \(\Leftrightarrow\dfrac{32}{x}=\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{99}\)
=>32/x=1/3-1/5+1/5-1/7+...+1/9-1/11
=>32/x=1/3-1/11=8/33
=>x=32:8/33=132
b: \(\Leftrightarrow1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{56}=\dfrac{x}{16}\)
\(\Leftrightarrow6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{x}{16}\)
=>x/16=6-1/2+1/8=11/2+1/8=45/8=90/16
=>x=90
c: \(\Leftrightarrow\dfrac{22}{x}=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\)
=>22/x=1/2*2/3*...*9/10*3/2*4/3*...*11/10
=>22/x=1/10*11/2=11/20=22/40
=>x=40
\(\dfrac{5}{6}\) - (\(\dfrac{2}{9}\) - \(x\)) = \(\dfrac{-15}{20}\)
\(\dfrac{2}{9}\) - \(x\) = \(\dfrac{5}{6}\) - (- \(\dfrac{15}{20}\))
\(\dfrac{2}{9}\) - \(x\) = \(\dfrac{19}{12}\)
\(x\) = \(\dfrac{2}{9}\) - \(\dfrac{19}{12}\)
\(x\) = - \(\dfrac{49}{36}\)