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AH
Akai Haruma
Giáo viên
3 tháng 1
Bài 1:
$10+3(x-6)=5^{10}:5^8=5^2=25$
$3(x-6)=25-10=15$
$x-6=15:3=5$
$x=5+6=11$
AH
Akai Haruma
Giáo viên
3 tháng 1
Bài 2:
a. $100-[150-8(7-4)^2]=100-(150-8.3^2)=100-150+8.3^2$
$=-50+72=72-50=22$
b. $=-999-23+999-10-67=(-999+999)-10-(67+23)$
$=0-10-90=-(10+90)=-100$
KL
21 tháng 8 2017
a, A = (\(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\)) - (\(\dfrac{79}{67}-\dfrac{28}{41}\))
= \(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
= \(\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\)
= \(\dfrac{1}{3}-1+1\)
= \(\dfrac{1}{3}\)
@Mai Tran
TC
2
22 tháng 4 2017
1) a) \(\left(-0,75+\frac{1}{2}\right):\frac{4}{3}\)
\(=\left(-0,75+\frac{1}{2}\right)\cdot\frac{3}{4}\)
\(=\left(-0,75+0,5\right)\cdot0,75\)
\(=-0,25\cdot0,75\)
\(=-0,1875\)
Vậy: \(\left(-0,75+\frac{1}{2}\right):\frac{4}{3}=-0,1875\)
b) \(\frac{7}{5}\cdot\frac{7}{4}-\frac{32}{5}\)
\(=\frac{49}{20}-\frac{32}{5}\)
\(=\frac{49}{20}-\frac{128}{20}\)
\(=-\frac{79}{20}=-3\frac{19}{20}\)
Vậy: \(\frac{7}{5}\cdot\frac{7}{4}-\frac{32}{5}=-3\frac{19}{20}\)
2) \(\frac{10}{3}\cdot x+\frac{67}{4}=-13,25\)
<=> \(\frac{10}{3}\cdot x+\frac{67}{4}=-\frac{53}{4}\)
<=> \(\frac{10}{3}\cdot x=-\left(\frac{53}{4}+\frac{67}{4}\right)\)
<=> \(\frac{10}{3}\cdot x=-30\)
<=> \(x=-30:\frac{10}{3}\)
<=> \(x=-30\cdot\frac{3}{10}\)
<=> \(x=-9\)
Vậy: \(x=-9\)
NH
22 tháng 8 2020
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(A=7\cdot\frac{3}{35}=\frac{21}{35}\)
22 tháng 8 2020
\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)
\(A=7\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\cdot\frac{3}{35}=\frac{3}{5}\)
\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)
\(B=\frac{1}{2}\left(\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)
\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)
\(C=\frac{4}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\cdot\frac{502}{1005}=\frac{1004}{1005}\)
H
9