1/2+58/4= ??
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HM
0
EC
0
22 tháng 9 2015
a,\(\int\limits^{\frac{\Pi}{6}}_0\frac{sin\left(2x+x\right)}{cos^2x}dx=\int\limits^{\frac{\Pi}{6}}_0\frac{sin2x.cosx+cos2x.sinx}{cos^2x}dx=\int\limits^{\frac{\Pi}{6}}_0\frac{2cos^2x.sinx+\left(2cos^2x-1\right)sinx}{cos^2x}dx=\int\limits^{\frac{\Pi}{6}}_0\frac{4cos^2x.sinx}{cos^2x}dx+\int\limits^{\frac{\Pi}{6}}_0\frac{d\left(cosx\right)}{cos^2x}=\int\limits^{\frac{\Pi}{6}}_0sinxdx-\frac{1}{cosx}\)
thay cận vào nhé
DH
3
3 tháng 7 2016
1) 58× 75+ 58× 50 -58 × 25
= 58( 75 + 50 - 25 )
= 58 . 100
= 5800
3 tháng 7 2016
1/ 58 x 75 + 58 x 50 - 58 x 25
= 58 x ( 75 + 50 - 25 )
= 58 x 100
= 5800
2/= \(2\frac{3}{5}-\frac{3}{5}\cdot\frac{11}{84}\)
\(=\frac{13}{5}-\frac{3}{5}\cdot\frac{11}{84}\)
\(=\frac{13}{5}-\frac{11}{140}\)
\(=\frac{364-11}{140}\)
\(=\frac{353}{140}\)
11 tháng 2 2022
\(=4+\dfrac{5}{58}-3-\dfrac{1}{2}+8+\dfrac{15}{29}-3-\dfrac{5}{58}+7+\dfrac{14}{29}=13+\dfrac{1}{2}=\dfrac{27}{2}\)
9 tháng 1 2021
a) Ta có: \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow-2x+3+x+4=0\)
\(\Leftrightarrow-x+7=0\)
\(\Leftrightarrow-x=-7\)
hay x=7
Vậy: S={7}
b) Ta có: \(\dfrac{2+x}{5}-0.5x=\dfrac{1-2x}{4}+0.25\)
\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{0.5x\cdot20}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{20\cdot0.25}{20}\)
\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)
\(\Leftrightarrow8+4x-10x=5-10x+5\)
\(\Leftrightarrow-6x+8=-10x+10\)
\(\Leftrightarrow-6x+8+10x-10=0\)
\(\Leftrightarrow4x-2=0\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
d) Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-59}{1}+\dfrac{x-58}{2}+\dfrac{x-57}{3}\)
\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}-1=\dfrac{x-59}{1}-1+\dfrac{x-58}{2}-1+\dfrac{x-57}{3}-1\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{1}+\dfrac{x-60}{2}+\dfrac{x-60}{3}\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}\right)-\left(x-60\right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
mà \(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)
nên x-60=0
hay x=60
Vậy: S={60}
DN
0
\(\frac{1}{2}+\frac{58}{4}=\frac{1\times2}{2\times2}+\frac{58}{4}=\frac{2}{4}+\frac{58}{4}\)\(=\frac{2+58}{4}=\frac{60}{4}=60:4=15\)
Vậy \(\frac{1}{2}+\frac{58}{4}=15\)
_HT_
58/4=58:2/4:2=29/2
1/2+29/2=30/2
30/2=15