Question

[Thạch Nguyễn]

Bài 1:Tính

a)\(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2\) b)\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)c) \(\dfrac{5^4.20^4}{25^5.4^5}\)

Bài 2:Tìm x

a)\(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)

b)\(\left(x+5\right)^3=-64\) c)\(\left(2x-3\right)^2=9\)

c)\(x^2+1=82\)

Bài 3:Tính

M=\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}\)

Bài 4:So sánh \(2^{24}\)và\(3^{16}\)

GIÚP MIK DZỚI,MÌNH TIK CHO ><

Answer 1

[Hoàng Thị Ngọc Anh]

Bài 1:

c) \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.4^4.5^4}{5^{10}.4^5}=\dfrac{5^8.4^4}{5^8.5^2.4^4.4}=\dfrac{1}{25.4}=\dfrac{1}{100}\)

Bài 2: a) \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\\\left(y+0,4\right)^{100}\ge0\forall y\\\left(z-3\right)^{678}\ge0\forall z\end{matrix}\right.\) \(\Rightarrow\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{2}{5}\\z=3\end{matrix}\right.\)

Vậy ...

Bài 3: \(M=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}\)

\(=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}=2^8=256.\)

Vậy \(M=256.\)

Mấy bài kia dễ tự làm.

Answer 2

[Mashiro Shiina]

\(3)\)

\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}=2^8=256\)\(4)\)

\(2^{24}=\left(2^6\right)^4=64^4;3^{16}=\left(3^4\right)^4=81^4\)

\(\Leftrightarrow2^{24}< 3^{16}\)