10.42 + 58.68 (21+x) : = 95 : 94
-x-94+62-99=95+62-99
-x - 94 + 62 - 99 = 95 + 62 - 99.
=> -x - 94 + 62 - 99 = 157 - 99.
=> -x - 94 + 62 - 99 = 58.
=> -x - 94 + 62 = 58 + 99
=> -x - 94 + 62 = 157.
=> -x - 94 = 157 - 62.
=> -x - 94 = 95.
=> -x = 95 + 94
=> -x = 189.
Vậy -x = 189.
#Học tốt.
Cho \(3,2.x+\)\(\dfrac{3}{2}\).\(x\)\(=\)\(-0,15\)
Giá trị của \(x\) là:
\(A.\dfrac{3}{94}\)\(B.-\dfrac{3}{94}\)\(C.-\dfrac{95}{3}\)\(D\). Tất cả đều sai
\(:>\)
Lạy ông đi quá lạy bà đi lại, ai có thể cho con xin cái đáp án k:>
\(\frac{X^{95}+x^{94}+x^{93}+...+x+1}{x^{31}+x^{30}+x^{29}+...x+1}\)
\(\frac{\left(x^{95}+x^{94}\right)+.....+\left(x+1\right)}{\left(x^{31}+x^{30}\right)+.....+\left(x+1\right)}=\frac{x^{94}\left(x+1\right)+......+\left(x+1\right)}{x^{30}\left(x+1\right)+.....+\left(x+1\right)}=\frac{x^{94}+x^{92}+....+x^2+1}{x^{30}+x^{28}+....+x^2+1}=\frac{\left(x^2+1\right)x^{92}+x^{88}\left(x^2+1\right).....+\left(x^2+1\right)}{\left(x^2+1\right)x^{28}+\left(x^2+1\right)x^{24}+....+\left(x^2+1\right)}=\frac{x^{92}+x^{88}+......+x^4+1}{x^{28}+x^{24}+.....+x^4+1}=\frac{x^{88}\left(x^4+1\right)+x^{80}\left(x^4+1\right)+....+\left(x^4+1\right)}{x^{24}\left(x^4+1\right)+x^{16}\left(x^4+1\right)+.....+\left(x^4+1\right)}=\frac{x^{88}+x^{80}+....+1}{x^{24}+x^{16}+...+1}\)
\(=\frac{x^{80}\left(x^8+1\right)+x^{64}\left(x^8+1\right)+.....+\left(x^8+1\right)}{x^{16}\left(x^8+1\right)+\left(x^8+1\right)}=\frac{x^{80}+x^{64}+.....+1}{x^{16}+1}=\frac{x^{64}\left(x^{16}+1\right)+.....+x^{16}+1}{x^{16}+1}=x^{64}+x^{32}+1\)
giải phương trình sau
\(\dfrac{2x+5}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}=\dfrac{2x+93}{7}+\dfrac{2x+94}{6}\dfrac{2x+95}{5}\)
Ta có : \(\dfrac{2x+5}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}=\dfrac{2x+93}{7}+\dfrac{2x+94}{6}+\dfrac{2x+95}{5}\)
\(\Leftrightarrow\dfrac{2x+5}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}-\dfrac{2x+93}{7}-\dfrac{2x+94}{6}-\dfrac{2x+95}{5}=0\)
\(\Leftrightarrow\dfrac{2x+5}{95}+1+\dfrac{2x+6}{94}+1+\dfrac{2x+7}{93}+1-\dfrac{2x+93}{7}-1-\dfrac{2x+94}{6}-1-\dfrac{2x+95}{5}-1=0\)
\(\Leftrightarrow\dfrac{2x+100}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}-\dfrac{2x+100}{7}-\dfrac{2x+100}{6}-\dfrac{2x+100}{5}=0\)
\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\right)=0\)
Thấy : \(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\ne0\)
\(\Rightarrow2x+100=0\)
\(\Leftrightarrow x=-50\)
Vậy ...
Ta có: \(\dfrac{2x+5}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}=\dfrac{2x+93}{7}+\dfrac{2x+94}{6}+\dfrac{2x+95}{5}\)
\(\Leftrightarrow\dfrac{2x+5}{95}+1+\dfrac{2x+6}{94}+1+\dfrac{2x+7}{93}+1=\dfrac{2x+93}{7}+1+\dfrac{2x+94}{6}+1+\dfrac{2x+95}{5}+1\)
\(\Leftrightarrow\dfrac{2x+100}{95}+\dfrac{2x+100}{94}+\dfrac{2x+100}{93}=\dfrac{2x+100}{7}+\dfrac{2x+100}{6}+\dfrac{2x+100}{5}\)
\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}\right)=\left(2x+100\right)\left(\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{5}\right)\)
\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}\right)-\left(2x+100\right)\left(\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{5}\right)=0\)
\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\right)=0\)
mà \(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\ne0\)
nên 2x+100=0
\(\Leftrightarrow2x=-100\)
hay x=-50
Vậy: S={-50}
a, 3.(37 - \(x\)) - 56 = 1
3.(37 - \(x\)) = 1 + 56
3.(37 - \(x\)) = 57
37 - \(x\) = 57 : 3
37 - \(x\) = 19
\(x\) = 37 - 19
\(x\) = 18
b, 95 - (2\(x\) - 1).4 = 14
(2\(x\) - 1).4 = 95 - 14
(2\(x\) - 1).4 = 81
2\(x\) - 1 = 81 : 4
2\(x\) = \(\dfrac{81}{4}\) - 1
2\(x\) = \(\dfrac{85}{4}\)
\(x\) = \(\dfrac{85}{4}\) : 2
\(x\) = \(\dfrac{85}{8}\)
c,
-45 + (\(x\) - 15) = 23
- 45 + \(x\) - 15 = 23
- (45 + 15) + \(x\) = 23
- 60 + \(x\) = 23
\(x\) = 23 + 60
\(x\) = 83
Rút gọn A=\(\frac{x^{95}+x^{94}+x^{93}+....+x+1}{x^{31}+x^{30}+x^{29}+...+x+1}\)
\(\frac{x-1}{95}+\frac{x-2}{94}=\frac{x-3}{93}+\frac{x-4}{92}\)
\(\frac{x-1}{95}+\frac{x-2}{94}=\frac{x-3}{93}+\frac{x-4}{92}\)
\(\Rightarrow\frac{x-1}{95}+\frac{x-2}{94}-\frac{x-3}{93}-\frac{x-4}{92}=0\)
\(\Rightarrow\left(\frac{x-1}{95}-1\right)+\left(\frac{x-2}{94}-1\right)-\left(\frac{x-3}{93}-1\right)-\left(\frac{x-4}{92}-1\right)=0\)
\(\Rightarrow\frac{x-96}{95}+\frac{x-96}{94}-\frac{x-96}{93}-\frac{x-96}{92}=0\)
\(\Rightarrow\left(x-96\right)\left(\frac{1}{95}+\frac{1}{94}-\frac{1}{93}-\frac{1}{92}\right)=0\)
\(\Rightarrow x-96=0\left(vì\frac{1}{95}+\frac{1}{94}-\frac{1}{93}-\frac{1}{92}\ne0\right)\)
\(=>x=96\)
cho mình hỏi là viết phân số làm sao vậy
tính nhanh A | 38 + 125 + 29 +17
B | 312 X 45 : 15
C | 100 - 99+ 98 - 97 + 96 - 95 + 94 - 93 + 92 -91
D |(125 x 36 ) : ( 5 x 9 )
A=[(38+17)+125]+29
A=175+29
A=204
B=312.3
B=936
C=1+1+1+1+1
C=5
D=(125:5).(36:9)
D=25.4
D=100
Tìm x:
\(\frac{x+1}{96}+\frac{x+2}{95}=\frac{x+3}{94}+\frac{x+4}{93}\)
\(\frac{x+1}{96}+\frac{x+2}{95}=\frac{x+3}{94}+\frac{x+4}{93}\)
\(\Rightarrow\left(\frac{x+1}{96}+1\right)+\left(\frac{x+2}{95}+1\right)=\left(\frac{x+3}{94}+1\right)+\left(\frac{x+4}{93}+1\right)\)
\(\Rightarrow\frac{x+97}{96}+\frac{x+97}{95}=\frac{x+97}{94}+\frac{x+97}{93}\)
\(\Rightarrow\frac{x+97}{96}+\frac{x+97}{95}-\frac{x+97}{94}-\frac{x+97}{93}=0\)
\(\Rightarrow\left(x+97\right)\left(\frac{1}{96}+\frac{1}{95}-\frac{1}{94}-\frac{1}{93}\right)=0\)
Mà \(\frac{1}{96}+\frac{1}{95}-\frac{1}{94}-\frac{1}{93}\ne0\)
\(\Rightarrow x+97=0\)
\(\Rightarrow x=-97\)
Vậy x = -97
Có : \(\frac{x+1}{96}+\frac{x+2}{95}=\frac{x+3}{94}+\frac{x+4}{93}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{96}+1\right)+\left(\frac{x+2}{95}+1\right)\)= \(\left(\frac{x+3}{94}+1\right)+\left(\frac{x+4}{93}+1\right)\)
\(\Leftrightarrow\) \(\frac{x+97}{96}+\frac{x+97}{95}=\frac{x+97}{94}+\frac{x+97}{93}\)
\(\Leftrightarrow\) \(\frac{x+97}{96}+\frac{x+97}{95}-\frac{x+97}{94}-\frac{x+97}{93}=0\)
\(\Leftrightarrow\) \(\left(x+97\right)\left(\frac{1}{96}+\frac{1}{95}-\frac{1}{94}-\frac{1}{93}\right)=0\)
\(\Leftrightarrow\) \(\left(x+97\right)=0\) ( \(\frac{1}{96}+\frac{1}{95}-\frac{1}{94}-\frac{1}{93}\)) \(\ne0\)
\(\Leftrightarrow\)\(x=-97\)
Vậy \(x=-97\)