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Bài 1 :
\(a+b=3.\left(a-b\right)=\)\(2\frac{a}{b}\)
\(\Rightarrow a+b=3.\left(a-b\right)\)
\(\Rightarrow a+b=3a-3b\)
\(\Rightarrow3a-3b-a-b=0\)
\(\Rightarrow2a-4b=0\)
\(\Rightarrow2.\left(a-2b\right)=0\)
\(\Rightarrow\hept{\begin{cases}a-2b=0\\a=2b\end{cases}}\)
Ta có : \(a+b=\frac{2a}{b}\)
Thay \(a=2b\) vào
\(\Rightarrow2b+b=\frac{2.23}{b}\)
\(\Rightarrow3b=\frac{4b}{b}\Rightarrow3b=4\)
\(\Rightarrow b=\frac{4}{3}\Rightarrow a=2.\frac{4}{3}=\frac{8}{3}\)
Vậy \(a=\frac{8}{3}\) và \(b=\frac{4}{3}\)
Chúc bạn học tốt ( -_- )
Bài 2 :
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{5}+\frac{100}{6.7}+...+\)\(\frac{100}{98.99}+\frac{1}{99}\)
\(B=\frac{100}{2}+\frac{100}{6}+\frac{100}{12}+\frac{100}{20}+\frac{100}{30}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{9900}\)
\(B=\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+\frac{100}{4.5}+\frac{100}{5.6}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{99.100}\)
\(B=100.\frac{100}{2}+\frac{100}{2}-\frac{1}{3}+\frac{100}{3}-\frac{100}{4}+\frac{100}{4}-\frac{100}{5}+\frac{100}{5}-\frac{100}{6}+\frac{100}{6}\)\(-\frac{100}{7}+...+\frac{100}{98}+\frac{100}{99}+\frac{100}{99}-1\)
\(B=100-1\)
\(B=99\)
Chúc bạn học tốt ( -_- )
n=ghi lộn nhé !!
a)\(10.\sqrt{0,01.\sqrt{ }\frac{16}{9}}+3\sqrt{49-\frac{1}{6}}\sqrt{4}\)
a) \(\left(\frac{2}{3}\right)^x=\left(\frac{4}{9}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2^2}{3^2}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^{100}\)
\(\Rightarrow x=100\)
Vậy x = 100
b) \(\left(\frac{2}{3}-x\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(\frac{2}{3}-x\right)^2=\left(\frac{1}{6}\right)^2\)
\(\Rightarrow\frac{2}{3}-x=\frac{1}{6}\)
\(\Rightarrow x=\frac{2}{3}-\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
2)
Ta có:
\(74^{m+1}+74^m=74^m.74^1+74^m=74^m.\left(74+1\right)=74^m.75⋮25\)
( vì \(75⋮25\) )
\(\Rightarrowđpcm\)
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
3) C thiếu đề
4) \(D=\frac{1}{9}-\left|\frac{-5}{23}\right|-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{5}{23}+\frac{5}{23}-\frac{1}{9}-\frac{25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{1}{9}-\frac{5}{23}+\frac{5}{23}+\frac{-25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=0+0+\frac{125}{14}-\frac{7}{30}\)
\(D=\frac{913}{105}\)
Bài 1 và Bài 2 dễ, bn có thể tự làm được!
Bài 3:
a) ta có: 1020 = (102)10 = 10010
=> 10010>910
=> 1020>910
b) ta có: (-5)30 = 530 =( 53)10 = 12510 ( vì là lũy thừa bậc chẵn)
(-3)50 = 350 = (35)10= 24310
=> 12510 < 24310
=> (-5)30 < (-3)50
c) ta có: 648 = (26)8= 248
1612 = ( 24)12 = 248
=> 648 = 1612
d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)