(75%-\(\frac{21}{10}\)) : (-5,4)-(\(\left(\frac{3}{2}\right)^2\)
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26 tháng 2 2020
a) Ta có: \(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)=\frac{2}{3}\)
⇒\(\frac{2}{3}:\left(x-1\right)=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)
⇒\(x-1=\frac{2}{3}:\frac{1}{6}=\frac{2}{3}\cdot6=4\)
hay x=5
Vậy: x=5
b) \(5,4-3\left[x-120\%\right]=\frac{3}{10}\)
⇔\(\frac{27}{5}-3\cdot\left(x-\frac{6}{5}\right)=\frac{3}{10}\)
⇔\(3\left(x-\frac{6}{5}\right)=\frac{27}{5}-\frac{3}{10}=\frac{51}{10}\)
hay \(x-\frac{6}{5}=\frac{51}{10}\cdot\frac{1}{3}=\frac{17}{10}\)
⇔\(x=\frac{17}{10}+\frac{6}{5}=\frac{29}{10}\)
Vậy: \(x=\frac{29}{10}\)
c) \(10\cdot3^{x+2}-3^x=89\)
\(\Leftrightarrow10\cdot3^2\cdot3^x-3^x=89\)
\(\Leftrightarrow3^x\left(90-1\right)=89\)
\(\Leftrightarrow3^x=1\)
hay x=0
Vậy: x=0
d) \(5\cdot\left(x-0,2\right)=3x+\left(\frac{-2}{3}\right)^3\)
⇒\(5\cdot\left(x-\frac{1}{5}\right)=3x+\frac{-8}{27}\)
\(\Leftrightarrow5x-1-3x-\frac{-8}{27}=0\)
\(\Leftrightarrow2x-\frac{19}{27}=0\)
\(\Leftrightarrow2x=\frac{19}{27}\)
hay \(x=\frac{\frac{19}{27}}{2}=\frac{19}{27}\cdot\frac{1}{2}=\frac{19}{54}\)
Vậy: \(x=\frac{19}{54}\)
e) \(\left(2x+\frac{3}{4}\right)^2-1,5=2\frac{1}{2}\)
\(\Leftrightarrow\left(2x+\frac{3}{4}\right)^2=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{3}{2}=-2\\2x+\frac{3}{2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2-\frac{3}{2}\\2x=2-\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{7}{2}\\2x=\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{2}\cdot\frac{1}{2}\\x=\frac{1}{2}\cdot\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{4}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{7}{4};\frac{1}{4}\right\}\)
26 tháng 6 2019
\(\frac{-2}{3}-\left(\frac{-2}{5}\right)-\frac{7}{10}\)
\(=\frac{-10}{15}-\frac{-6}{15}-\frac{7}{10}\)
\(=\frac{-4}{15}-\frac{7}{10}\)
\(=\frac{-4}{15}+\frac{\left(-7\right)}{10}\)
\(=\frac{-40}{150}+\frac{-105}{150}\)
\(=\frac{-29}{30}\)
26 tháng 6 2019
\(\left[\frac{11}{24}:\frac{55}{36}\right]\cdot\frac{10}{3}\)
\(=\left[\frac{11}{24}\cdot\frac{36}{55}\right]\cdot\frac{10}{3}\)
\(=\left[\frac{1}{2}\cdot\frac{3}{5}\right]\cdot\frac{10}{3}\)
\(=\frac{3}{10}\cdot\frac{10}{3}=1\)
HP
15 tháng 6 2016
a) Dễ thấy VT > 0;mà VT=VP
=>VP > 0 => 4x > 0=> x > 0
=>\(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)
=>BT đầu tương đương \(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{6}\right)=4x\)
\(=>3x+1=4x=>x=1\)
15 tháng 6 2016
a) Để đẳng thức xảy ra thì: x>0 (vì: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|>0\) )
Khi đó: \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)
=>\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x\)
<=>x=1
Vậy x=1
b)Điều kiện: \(x\ne-3;-10;-21;-34\)
\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
<=>\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
<=>\(\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
=>x+34-x-3=x
<=>x=31 (nhận)
Vậy x=31
T
8 tháng 3 2018
P/s: Bài này chỉ tính được giá trị "gần đúng" của biểu thức thôi nhé!
\(\frac{21}{54}+\frac{3}{75}:\frac{\left(\frac{39}{65}+0,415-\frac{33}{600}\right)\frac{21}{91}}{7^2-18,25+13\frac{15}{36}-16\frac{17}{102}}\)
\(\Leftrightarrow\frac{21}{54}+\frac{3}{75}:\frac{\left(0,6+0,415-0,055\right)0,23}{49-18,25+\frac{483}{36}-\frac{1649}{102}}\)
\(\Leftrightarrow\frac{21}{54}+\frac{3}{75}:\frac{\left(0,6+0,415-0,055\right)0,23}{49-18,25+13,41-16,1}\)
\(\Leftrightarrow\frac{21}{54}+\frac{3}{75}:\frac{0,96.0,23}{28,06}\Leftrightarrow\frac{21}{54}+\frac{3}{75}:\frac{0,2208}{28,06}\)
\(\Leftrightarrow\frac{21}{54}+\left(\frac{3}{75}:\frac{0,2208}{28,06}\right)\Leftrightarrow\frac{21}{54}+\left(\frac{3}{75}.\frac{28,06}{0,2208}\right)=\frac{21}{54}+\frac{61}{12}=\frac{197}{36}\)
P/s: Giải bài này mà mệt cả đầu =((
\(=\left(\frac{75}{100}-\frac{21}{10}\right):\left(-5,4\right)-\left(\frac{3}{2}\right)^2\)\(=\frac{-27}{20}.\frac{-5}{27}-\frac{9}{4}=\frac{1}{4}-\frac{9}{4}=\frac{-8}{4}=-2\)