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1.= 574.499+3
=(574)499.573
= .....1499.B3
= A1.B3
=......3
Cau kia van dung ma lam
a, ( 1999 + 313 ) - 1999
= 1999 - 1999 + 313
= 313
b, 364 - ( 364 - 111 )
= 364 - 364 - 111
= -111
c, 249 - ( 75- 51 )
= 249 - 24
= 225
giúp mik nha chiều này 6:00 mik nộp rồi
ai nhanh mik sẽ k cho 3 k
\(2\frac{3}{5}x-\frac{1}{7}=1\frac{9}{35}\)
\(\frac{13}{5}x=\frac{44}{35}+\frac{1}{7}\)
\(\frac{13}{5}x=\frac{7}{5}\)
\(x=\frac{7}{5}:\frac{13}{5}\\ x=\frac{7}{13}\)
1) Ta có : \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)
=> x + 1 = 0
=> x = - 1
b) \(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)
=> \(\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+1}{2009}+1\right)\)
=> \(\frac{x+2010}{2006}+\frac{x+2010}{2007}=\frac{x+2010}{2008}+\frac{x+2010}{2009}\)
=> \(\left(x+2010\right)\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
Vì \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)
=> x + 2010 = 0
=> x = -2010
c) \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)
\(\Rightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)=\left(\frac{x+1975}{75}-1\right)+\left(\frac{x+1969}{69}-1\right)\)
=> \(\frac{x+1900}{45}+\frac{x+1900}{54}=\frac{x+1900}{75}+\frac{x+1900}{69}\)
=> \(\left(x+1900\right)\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)
=> \(x+1900=0\left(\text{Vì }\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\ne0\right)\)
=> x = -1900
d) \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)
=> \(\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)=\left(\frac{x+2012}{8}+2\right)+\left(\frac{x+2014}{7}+2\right)\)
=> \(\frac{x+2028}{10}+\frac{x+2028}{9}=\frac{x+2028}{8}+\frac{x+2028}{7}\)
=> \(\left(x+2028\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)
=> x + 2028 = 0 \(\left(\text{Vì }\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\ne0\right)\)
=> x = -2028
1) Ta có: \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
+ TH1: \(x+1=0\)\(\Leftrightarrow\)\(x=-1\)
+ TH2: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)
Vì \(\hept{\begin{cases}\frac{1}{10}>\frac{1}{13}\\\frac{1}{11}>\frac{1}{14}\\\frac{1}{12}>0\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)
\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)
mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)
\(\Rightarrow\)Phương trình trên vô nghiệm
Vậy \(x=-1\)
2) Ta có: \(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)
\(\Leftrightarrow\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+2}{2008}+1\right)-\left(\frac{x+1}{2009}+1\right)=0\)
\(\Leftrightarrow\frac{x+2010}{2006}+\frac{x+2010}{2007}-\frac{x+2010}{2008}-\frac{x+2010}{2009}=0\)
\(\Leftrightarrow\left(x+2010\right).\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
+ TH1: \(x+2010=0\)\(\Leftrightarrow\)\(x=-2010\)
+ TH2: \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)
Vì \(\hept{\begin{cases}\frac{1}{2006}>\frac{1}{2008}\\\frac{1}{2007}>\frac{1}{2009}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}>\frac{1}{2008}+\frac{1}{2009}\)
\(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}>0\)
mà \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)
\(\Rightarrow\)Phương trình trên vô nghiệm
Vậy \(x=-2010\)
3) Ta có: \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)
\(\Leftrightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)-\left(\frac{x+1975}{75}-1\right)-\left(\frac{x+1969}{69}-1\right)=0\)
\(\Leftrightarrow\frac{x+1900}{45}+\frac{x+1900}{54}-\frac{x+1900}{75}-\frac{x+1900}{69}=0\)
\(\Leftrightarrow\left(x+1900\right).\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)
+ TH1: \(x+1900=0\)\(\Leftrightarrow\)\(x=-1900\)
+ TH2: \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)
Vì \(\hept{\begin{cases}\frac{1}{45}>\frac{1}{75}\\\frac{1}{54}>\frac{1}{69}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}>\frac{1}{75}+\frac{1}{69}\)
\(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}>0\)
mà \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)
\(\Rightarrow\)Phương trình trên vô nghiệm
Vậy \(x=-1900\)
4) Ta có: \(\frac{x-99}{5}+\frac{x-97}{7}=\frac{x-95}{9}+\frac{x-93}{11}\)
\(\Leftrightarrow\left(\frac{x-99}{5}-1\right)+\left(\frac{x-97}{7}-1\right)-\left(\frac{x-95}{9}-1\right)-\left(\frac{x-93}{11}-1\right)=0\)
\(\Leftrightarrow\frac{x-104}{5}+\frac{x-104}{7}-\frac{x-104}{9}-\frac{x-104}{11}=0\)
\(\Leftrightarrow\left(x-104\right).\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)=0\)
+ TH1: \(x-104=0\)\(\Leftrightarrow\)\(x=104\)
+ TH2: \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)
Vì \(\hept{\begin{cases}\frac{1}{5}>\frac{1}{7}\\\frac{1}{9}>\frac{1}{11}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}>\frac{1}{9}+\frac{1}{11}\)
\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}>0\)
mà \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)
\(\Rightarrow\)Phương trình trên vô nghiệm
Vậy \(x=104\)
5) Ta có: \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)
\(\Leftrightarrow\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)-\left(\frac{x+2012}{8}+2\right)-\left(\frac{x+2014}{7}+2\right)=0\)
\(\Leftrightarrow\frac{x+2028}{10}+\frac{x+2028}{9}-\frac{x+2028}{8}-\frac{x+2028}{7}=0\)
\(\Leftrightarrow\left(x+2028\right).\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)
+ TH1: \(x+2028=0\)\(\Leftrightarrow\)\(x=-2028\)
+ TH2: \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)
Vì \(\hept{\begin{cases}\frac{1}{10}< \frac{1}{8}\\\frac{1}{9}< \frac{1}{7}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}< \frac{1}{8}+\frac{1}{7}\)
\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}< 0\)
mà \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)
\(\Rightarrow\)Phương trình trên vô nghiệm
Vậy \(x=-2028\)
Chúc bn hok tốt nha
1. \(\frac{8^{10}}{4^{14}}=\frac{\left(2^3\right)^{10}}{\left(2^2\right)^{14}}=\frac{2^{30}}{2^{28}}=2^2=4\)
2. \(\frac{6^5.5^3}{10^3}=\frac{6^5}{2^3}=\frac{6^3.6^2}{2^3}=3^3.6^2=27.36=972\)
1) \(\frac{8^{10}}{4^{14}}=\frac{\left(2^3\right)^{10}}{\left(2^2\right)^{14}}=\frac{2^{30}}{2^{28}}=2^2=4\)
2) \(\frac{6^5.5^3}{10^3}=\frac{2^5.3^5.5^3}{2^3.5^3}=2^2.3^5=972\)
Học tốt!!!!
100.2.5.10 = 10.10.10.10 = 104
A = 103103 = 7.11.13.103
B = 102.102 = 22.32.17
C = 102.10.10102.10.10 = 22.3.17.103.5051
D = 104104 = 23.7.11.132
Giải :
( 1435 + 213 ) – 13 = 1435 + ( 213 – 13 ) = 1435 + 200 = 1635
1972 – ( 364 + 972 ) = 1972 – 972 – 368 = 1000 – 368 = 632
c. 364 – ( 364 – 111 ) = 364 – 364 + 111 = 111
249 – ( 75-51 ) = ( 249 + 51 ) – 75 = 300 – 75 = 225