tính : a) a^5.a^7:a^11;a thuộc Z b)x^6:x^3.x^2 c)[(x^8)^3]^0 với x khác 0
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PT
0
TT
1
12 tháng 8 2016
(5/7 + 9/7 - 5/7) = 9/7
12/11 + 2/11 - 14/11 =1
9/7 x 1 = 9/7
1 tháng 7 2021
\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)
H
1 tháng 7 2021
1. Tính
\(a,5\times\dfrac{7}{3}=\dfrac{35}{3}\)
\(b,\dfrac{13}{4}:7=\dfrac{13}{4}\times\dfrac{1}{7}=\dfrac{13}{28}\)
2. Tính
\(a,\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}\)
\(=\dfrac{15}{35}+\dfrac{14}{35}+\dfrac{3}{4}\)
\(=\dfrac{29}{35}+\dfrac{3}{4}\)
\(=\dfrac{116}{140}+\dfrac{105}{140}\)
\(=\dfrac{221}{140}\)
\(b,\dfrac{9}{7}-\dfrac{5}{11}\times\dfrac{11}{7}\)
\(=\dfrac{9}{7}-\dfrac{55}{77}\)
\(=\dfrac{99}{77}-\dfrac{55}{77}\)
\(=\dfrac{44}{77}=\dfrac{4}{7}\)
\(c,\dfrac{3}{5}\times\dfrac{5}{7}+\dfrac{4}{7}\)
\(=\dfrac{3}{5}\times\left(\dfrac{5}{7}+\dfrac{4}{7}\right)\)
\(=\dfrac{3}{5}\times\dfrac{9}{7}\)
\(=\dfrac{27}{35}\)
\(d,\dfrac{7}{9}\times\dfrac{2}{5}:\dfrac{3}{11}\)
\(=\dfrac{14}{45}:\dfrac{3}{11}\)
\(=\dfrac{14}{45}\times\dfrac{11}{3}\)
\(=\dfrac{154}{135}\)
\(e,\dfrac{9}{7}+\dfrac{2}{3}-\dfrac{1}{4}\)
\(=\dfrac{27}{21}+\dfrac{14}{21}-\dfrac{1}{4}\)
\(=\dfrac{41}{21}-\dfrac{1}{4}\)
\(=\dfrac{164}{84}-\dfrac{21}{84}\)
\(=\dfrac{143}{84}\)
\(g,\dfrac{4}{9}:\dfrac{3}{5}\times\dfrac{2}{11}\)
\(=\dfrac{4}{9}\times\dfrac{5}{3}\times\dfrac{2}{11}\)
\(=\dfrac{20}{27}\times\dfrac{2}{11}\)
\(=\dfrac{40}{297}\)
\(h,\dfrac{7}{2}-\dfrac{3}{10}:\dfrac{2}{5}\)
\(=\left(\dfrac{7}{2}-\dfrac{3}{10}\right):\dfrac{2}{5}\)
\(=\left(\dfrac{35}{10}-\dfrac{3}{10}\right):\dfrac{2}{5}\)
\(=\dfrac{32}{10}:\dfrac{2}{5}\)
\(=\dfrac{16}{5}\times\dfrac{5}{2}\)
\(=\dfrac{80}{10}=8\)
KM
3
31 tháng 7 2015
A = 5/3.7 + 5/7.11 + 5/11.15 + ... + 5/35.39
A = 5/4 x ( 4/3.7 + 4/7.11 + 4/11.15 + ... + 4/35.39 )
A = 5/4 x ( 1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/35 - 1/39 )
A = 5/4 x ( 1/3 - 1/39 )
A = 5/4 x 4/13
A = 5/13
ND
1
31 tháng 7 2015
\(A=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{35\cdot39}\)
\(=\frac{5}{4}\cdot\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{35\cdot39}\right)\)
\(=\frac{5}{4}\cdot\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...-\frac{1}{35}+\frac{1}{35}-\frac{1}{39}\right)\)
\(=\frac{5}{4}\cdot\left(1-\frac{1}{39}\right)=\frac{5}{4}\cdot\frac{38}{39}=\frac{95}{78}\)
DT
3
2 tháng 2 2022
\(=\left(\dfrac{9}{11}+\dfrac{20}{11}\right)+\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{8}{13}\\ =\dfrac{29}{11}+1+\dfrac{8}{13}\)
Biết đến đó thôi sorry:<
Y
15 tháng 8 2023
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)
\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)
Vậy \(B=3\)
NM
1
12 tháng 11 2018
tìm a nguyên biết (a^2-1)(a^2-4)(a^2-7)(a^2-10)<0
a) \(a^5.a^7:a^{11}\left(a\inℤ\right)\)
\(\Rightarrow a^5.a^7:a^{11}=a^{5+7}:a^{11}=a^{12}:a^{11}=a^1=a\)
b) \(x^6:x^3.x^2\)
\(\Rightarrow x^{6-3}.x^2=x^3.x^2=x^5\)
c) \(\left[\left(x^8\right)^3\right]^0\left(x\ne0\right)\)
\(\Rightarrow\left[\left(x^8\right)^3\right]^0=\left(x^{24}\right)^0=1\)
a5.a7: a11=a5+7-11=a1=a
x6: x3.x2=x6–3+2=x5
[(x8)3]0=1(vì bất cứ số nào mũ 0 cũng =1)