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6\(^2\)+ 64 : ( x - 1 ) = 52
36 + 64 : ( x - 1 ) =52
64 ; ( x - 1 ) =64 : 52
x - 1 = \(\frac{16}{13}\)
x = \(\frac{16}{13}\)+1
x = \(\frac{29}{13}\)
HT
b)\(\left(x-8\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-8=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=2\end{cases}}\)
c) \(\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)=9x+200\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+...+10\right)=9x+200\) (10 số hạng x)
\(\Leftrightarrow10x+55=9x+200\Leftrightarrow x+55=200\)
\(\Leftrightarrow x=145\)
Lời giải:
a.
$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$
b.
$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$
$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$
c.
$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$
d.
$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$
$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$
e.
$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$
$\frac{-19}{15}: x=1$
$x=\frac{-19}{15}:1 =\frac{-19}{15}$
f.
$(-\frac{3}{4}+x).2\frac{2}{3}=1$
$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$
$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$
3x+3x-1+3x-2=1053
=> 3x-2.32+3x-2.3+3x-2=1053
3x-2.9+3x-2.3+3x-2=1053
=>3x-2.(9+3+1)=1053
3x-2.13=1053
3x-2=1053:13=81
3x-2=34
=>x-2=4
x=4+2
x=6
1. Tìm x
a) 1+2+3+...+x = 210
=> \(\frac{x\left(x+1\right)}{2}=210\)
=> x = 20
b) \(32.3^x=9.3^{10}+5.27^3\)
=>\(32.3^x=9.3^{10}+5.3^9\)(\(27^3=\left(3^3\right)^3=3^9\))
=>\(32.3^x=9.3.3^9+5.3^9\)
=>\(32.3^x=3^9\left(9.3+5\right)\)
=>\(32.3^x=3^9.32\)
=>x = 9
2.
Ta có 2A = 3A - A
=> 2A = \(3\left(1+3+3^2+3^3+....+3^{10}\right)\)\(-\)\(1-3-3^2-3^3-....-3^{10}\)
=> 2A = \(3+3^2+3^3+.....+3^{11}-\)\(1-3-3^2-3^3-...-3^{10}\)
=> 2A = \(3^{11}-1\)
=> 2A+1 = \(3^{11}-1+1\)=\(3^{11}\)
=> n = 11
Ta có : a)1 + 2 + 3 + ... + x = 210
=> \(\frac{x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 420
=> x(x + 1) = 20.21
=> x = 20
a, (sửa đề )
\(1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{x.\left(x+1\right)}=\frac{1999}{2000}\)
=\(1+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}\right)=\frac{1999}{2000}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x+\left(x+1\right)}=1-\frac{1999}{2000}=\frac{1}{2000}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2000}\)
=\(\frac{1}{1}-\frac{1}{x+1}=\frac{1}{2000}\)
=\(\frac{1}{x+1}=\frac{1}{1}-\frac{1}{2000}=\frac{1999}{2000}\)
=> \(x+1=1:\frac{1999}{2000}=\frac{2000}{1999}\)
=>\(x=\frac{2000}{1999}-1=\frac{1}{1999}\)
Vậy x ∈{ \(\frac{1}{1999}\)}
b, \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
=>\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
=>2.(\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x.\left(x+1\right)}\))=\(\frac{2}{9}\)
=>\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x+\left(x+1\right)}=\frac{2}{9}:2=\frac{1}{9}\)
=>\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
=>\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
=>\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
=>\(x+1=18\)
=>\(x=18-1=17\)
=>x∈{17}