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Bài 1 :
S = \(\frac{6}{2.5}+\frac{6}{5.8}+...+\frac{6}{29.32}\)
= 2 . \(\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)
= 2 . \(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)
= 2 . \(\left(\frac{1}{2}-\frac{1}{32}\right)\)= ....
a) x + 10 = 20
<=> x = 20 - 10 = 10
Vậy x = 10
b) 2x + 15 = 35
<=> 2x = 35 - 15 = 20
<=> x = 10
Vậy x = 10
c) 3(x + 2) = 15
<=> x + 2 = 15 : 3 = 5
<=> x = 5 - 2 = 3
Vậy x = 3
d) 10x + 15.11 = 20.10
<=> 10x + 165 = 200
<=> 10x = 200 - 165 = 35
<=> x = 35 : 10 = 3,5
Vậy x = 3,5
e) 4(x + 2) = 3.4
<=> x + 2 = 3
<=> x = 3 - 2 = 1
Vậy x = 1
f) 33x + 135 = 26.9
<=> 33x + 135 = 234
<=> 33x = 234 - 135 = 99
<=> x = 99 : 33 = 3
Vậy x = 3
g) 2x + 15 + 16 + 17 = 100
<=> 2x + 48 = 100
<=> 2x = 100 - 48 = 52
<=> x = 52 : 2 = 26
Vậy x = 26
h) 2(x + 9 + 10 + 11) = 4.12.5
<=> x + 30 = 120
<=> x = 120 - 30 = 90
Vậy x = 90
4:
a: =4/15-2,9+11/15=1-2,9=-1,9
b: \(=-36,75+3,7-63,25+6,3=10-100=-90\)
c: \(=6,5+3,5-\dfrac{10}{17}-\dfrac{7}{17}=10-1=9\)
d: \(=\dfrac{13}{25}\left(-39,1-60,9\right)=\dfrac{13}{25}\left(-100\right)=-52\)
e: =-5/12-7/12-3,7-6,3=-1-10=-11
f: =2,8(-6/13-7/13)-7,2=-2,8-7,2=-10
Bài 1:
\(A=\frac{8}{7}+\frac{4}{11}(\frac{-6}{7}-\frac{5}{11})=\frac{8}{7}+\frac{-404}{847}=\frac{564}{847}\)
\(B=\frac{1}{5}.10-\frac{1}{3}.\frac{-21}{20}-\frac{1}{8}=2+\frac{7}{20}-\frac{1}{8}=\frac{89}{40}\)
Bài 2:
a.
$\frac{3}{4}+\frac{1}{4}:x=-3$
$\frac{1}{4}:x =-3-\frac{3}{4}=\frac{-15}{4}$
$x=\frac{1}{4}: \frac{-15}{4}=\frac{-1}{15}$
b.
$(x-\frac{1}{3})^2=1-\frac{5}{9}=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2$
$\Rightarrow x-\frac{1}{3}=\frac{2}{3}$ hoặc $x-\frac{1}{3}=\frac{-2}{3}$
$\Rightarrow x=1$ hoặc $x=\frac{-1}{3}$
\(x:\left(9+\frac{1}{2}-\frac{3}{2}\right)=\frac{2\left(\frac{1}{5}+\frac{1}{9}+\frac{1}{11}\right)}{8\left(\frac{1}{5}+\frac{1}{9}+\frac{1}{11}\right)}\)
\(x:8=\frac{1}{4}\)
x=2
1b)\(\frac{7}{19}x\frac{8}{11}+\frac{3}{11}:\frac{19}{7}-\frac{2}{-19}=\frac{7}{19}x\frac{8}{11}+\frac{3}{11}x\frac{7}{19}+\frac{2}{19}=\left(\frac{8}{11}+\frac{3}{11}\right)\frac{7}{19}+\frac{2}{19}=\frac{7}{19}+\frac{2}{19}=\frac{9}{19}\)
c)\(4\left(\frac{4}{9}+\frac{7}{11}-\frac{4}{9}\right)=4\frac{7}{11}\)
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