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\(\dfrac{75^{50}.2^{100}}{100^{50}.9^{25}}=\dfrac{\left(25.3\right)^{50}.2^{100}}{\left(25.4\right)^{50}.\left(3^2\right)^{25}}=\dfrac{25^{50}.3^{50}.2^{100}}{25^{50}.4^{50}.3^{50}}=\dfrac{25^{50}.3^{50}.2^{100}}{25^{50}.\left(2^{2^{ }}\right)^{50}.3^{50}}=\dfrac{25^{50}.3^{50}.2^{100}}{25^{50}.2^{100}.3^{50}}=1\)
\(\dfrac{75^{50}\cdot2^{100}}{100^{50}\cdot9^{25}}=\dfrac{25^{50}\cdot3^{50}\cdot4^{50}}{25^{50}\cdot4^{50}\cdot3^{50}}=1\)
\(\dfrac{75^{50}.2^{100}}{100^{50}.9^{25}}=\dfrac{\left(25.3\right)^{50}.2^{100}}{\left(25.4\right)^{50}.\left(3^2\right)^{25}}=\dfrac{25^{50}.3^{50}.2^{100}}{\left(25^{50}\right).\left(2^2\right)^{50}.\left(3^2\right)^{25}}=\dfrac{25^{50}.3^{50}.2^{100}}{25^{50}.2^{100}.3^{50}}=1\)
\(\dfrac{75^{50}.2^{100}}{100^{50}.9^{25}}\)
\(=\dfrac{\left(3.5^2\right)^{50}.\left(2^2\right)^{50}}{\left(2^2.5^2\right)+\left(3^2\right)^{50}}\)
\(=\dfrac{ \left(3.25\right)^{50}.4^{50}}{\left(4.25\right)^{50}.3^{50}}\)
\(=\dfrac{3^{50}.25^{50}.4^{50}}{4^{50}.25^{50}.3^{50}}=1\)
n=ghi lộn nhé !!
a)\(10.\sqrt{0,01.\sqrt{ }\frac{16}{9}}+3\sqrt{49-\frac{1}{6}}\sqrt{4}\)
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 ( -_- )
=1 chúc bạn học tốt
bạn giải chi tiết ra cho mk với