Question

1) So sánh :

a) \(3^{2^3}\) và (3²)³                  b) (-8)⁹ và (-32)⁵                   c) 2²¹ và 3¹⁴

2) Cho \(\dfrac{a}{b}=\dfrac{c}{d}.\) Chứng minh rằng :

a)\(\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)                            b) \(\dfrac{ab}{cd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

Answer 1

 Mk săpp thi rồi nên hơi nhiều bài mong mn giúp mk !!!

Answer 2

\(1,\\ a,3^{2^3}=3^8>3^6=\left(3^2\right)^3\\ b,\left(-8\right)^9=\left(-2\right)^{27}< \left(-2\right)^{25}=\left(-32\right)^5\\ c,2^{21}=8^7< 9^7=3^{14}\\ 2,\)

\(a,\) Áp dụng tcdtsbn:

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

\(b,\) Sửa: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow a=bk;c=dk\)

\(\Leftrightarrow\dfrac{ab}{cd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2};\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2}{d^2}\\ \LeftrightarrowĐpcm\)