10-{[(x:3+17):10+3.24]:10}=5
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a: Ta có: \(10-\left\{\left[\left(x:3+17\right):10+3\cdot2^4\right]:10\right\}=5\)
\(\Leftrightarrow\left[\left(x:3+17\right):10+48\right]:10=5\)
\(\Leftrightarrow\left(x:3+17\right):10+48=50\)
\(\Leftrightarrow\left(x:3+17\right):10=2\)
\(\Leftrightarrow x:3+17=20\)
\(\Leftrightarrow x:3=3\)
hay x=9
b: Ta có: \(2448:\left[119-\left(x-6\right)\right]=24\)
\(\Leftrightarrow119-\left(x-6\right)=102\)
\(\Leftrightarrow x-6=17\)
hay x=23
c: Ta có: \(165-\left(35:x+3\right)\cdot19=13\)
\(\Leftrightarrow\left(35:x+3\right)\cdot19=152\)
\(\Leftrightarrow\left(35:x+3\right)=8\)
\(\Leftrightarrow35:x=5\)
hay x=7
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\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
<=> x = 15
a: Ta có: 10−{[(x:3+17):10+3·24]:10}=5
⇔[(x:3+17):10+48]:10=5
⇔(x:3+17):10+48=50
⇔(x:3+17):10=2
⇔x:3+17=20
⇔x:3=3
\(\Leftrightarrow\) X=3.3
\(\Leftrightarrow\) X= 9
Vậy x= 9
Hoktot~