Tìm x, biết
\(\left(\frac{1}{2}-\frac{1}{3}\right)\cdot6^x+1+6^x+1=7\cdot6^9\)
K
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NV
27 tháng 7 2016
\(\left(\frac{1}{2}-\frac{1}{3}\right).6^{x+1}+6^{x+1}=7.6^9\)
\(\Rightarrow\frac{1}{6}.6.6^x+6.6^x=7.6^9\)
\(\Rightarrow6^x+6.6^x=7.6^9\)
\(\Rightarrow6^x.\left(1+6\right)=7.6^9\)
\(\Rightarrow6^x=\frac{7.6^9}{7}=6^9\)
\(\Rightarrow x=9\)
27 tháng 7 2016
\(\left(\frac{1}{2}-\frac{1}{3}\right).6^{x+1}+6^{x+1}=7.6^9\)
\(\Leftrightarrow\frac{1}{6}.6^{x+1}+6^{x+1}=7.6^9\)
\(\Leftrightarrow6^{x+1}.\left(\frac{1}{6}+1\right)=7.6^9\)
\(\Leftrightarrow6^{x+1}.\frac{7}{6}=7.6^9\)
\(\Leftrightarrow6^{x+1}=7.6^9:\frac{7}{6}\)
\(\Leftrightarrow6^{x+1}=7.6^9.\frac{6}{7}\)
\(\Leftrightarrow6^{x+1}=\left(7.\frac{6}{7}\right).6^9\)
\(\Leftrightarrow6^{x+1}=6.6^9\)
\(\Leftrightarrow6^{x+1}=6^{10}\)
\(\Leftrightarrow x+1=10\)
\(\Leftrightarrow x=9\)
11 tháng 2 2020
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
TB
1
11 tháng 5 2017
= ( x. 47/7 ) . 92/35 = -2 ( lưu ý (/ ) là dấu gạch ngang phân số )
=(x. 47 /7 )= 35/46
x=35/46 x 47 /7
x= 235/46
DP
16 tháng 7 2017
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{2.6}+\frac{2}{2.10}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\Leftrightarrow2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)
\(\Leftrightarrow2\left(1-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)
\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=1\frac{1991}{1993}\div2\)
\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=\frac{1992}{1993}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=1-\frac{1992}{1993}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{1993}\)
\(\Leftrightarrow x=1992\)
\(\text{Vậy x = 1992 }\)
AH
Akai Haruma
Giáo viên
6 tháng 7
Lời giải:
$x(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7})< 1\frac{6}{7}$
$x(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7})< \frac{13}{7}$
$x(1-\frac{1}{7})< \frac{13}{7}$
$x.\frac{6}{7}< \frac{13}{7}$
$x< \frac{13}{7}: \frac{6}{7}=\frac{13}{6}$
Vì $x$ là số nguyên nên $x\leq 2$
Vậy $x$ là các số nguyên sao cho $x\leq 2$.
DP
21 tháng 7 2017
Bài 1 :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)
\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)
\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)
Bài 2 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)