Bài 2: Giải:

Ta có:

\(2a^2+a=3b^2+b\Leftrightarrow2a^2-2b^2+a-b=b^2\)

\(\Leftrightarrow\left(a-b\right)\left(2a+b+1\right)=b^2\left(1\right)\)

Đặt \(ƯCLN\left(a-b;2a+2b+1\right)=d\)

\(\Rightarrow\) \(\begin{cases}a-b\vdots d\\2a+2b+1\vdots d\end{cases}\) \(\Rightarrow b^2=\left(a-b\right)\left(2a+2b+1\right)⋮d^2\)

\(\Rightarrow b⋮d.\) Lại có: \(2\left(a-b\right)-\left(2a+2b+1\right)⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\Rightarrow\left(a-b;2a+2b+1\right)=1\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\) \(\Rightarrow\) Đpcm

a: \(=-\dfrac{3^{10}\cdot5^{21}}{5^{20}\cdot3^{12}}=-5\cdot\dfrac{1}{9}=-\dfrac{5}{9}\)

b: \(=-\dfrac{11^5}{11^5}\cdot\dfrac{13^7}{13^8}=-\dfrac{1}{13}\)

c: \(=\dfrac{2^{10}\cdot3^9\cdot\left(3-1\right)}{2^9\cdot3^{10}}=\dfrac{2^{11}\cdot3^9}{2^9\cdot3^{10}}=\dfrac{4}{3}\)

d: \(=\dfrac{5^{11}\cdot7^{11}\cdot8}{5^{11}\cdot7^{11}\cdot\left(5\cdot7+9\right)}=\dfrac{8}{35+9}=\dfrac{8}{44}=\dfrac{2}{11}\)

3³⁸⁵   >   3¹⁰⁰

^{Vì số mũ lớn hơn}

\(a,3x+17=-5\\ \Rightarrow3x=-22\\ \Rightarrow x=-\dfrac{22}{3}\\ b,\dfrac{-5}{8}=\dfrac{x}{16}\\ \Rightarrow x=\dfrac{-5}{8}.16\\ \Rightarrow x=-10\\ c,\dfrac{4}{9}+\dfrac{7}{3}:x=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{3}:x=-\dfrac{11}{45}\\ \Rightarrow x=-\dfrac{105}{11}\)

\(a)3.x+17=-5\)

\(3.x\)         \(=\left(-5\right)-170=-22\)

  \(x\)          \(=\left(-22\right)\div3\)

  \(x\)          \(=\dfrac{-22}{3}\)

\(b)\dfrac{-5}{8}=\dfrac{x}{16}\)

\(\Rightarrow x=\dfrac{\left(-5\right)\times16}{8}=\dfrac{-80}{8}=-10\)

\(c)\dfrac{4}{9}+\dfrac{7}{3}:x=\dfrac{1}{5}\)

           \(\dfrac{7}{3}:x=\dfrac{1}{5}-\dfrac{4}{9}\)

           \(\dfrac{7}{3}:x=\dfrac{1}{5}+\left(\dfrac{-4}{9}\right)=\dfrac{9}{45}+\left(\dfrac{-20}{45}\right)=\dfrac{-11}{45}\)

                 \(x=\dfrac{7}{3}:\left(\dfrac{-11}{45}\right)=\dfrac{7}{3}\times\left(\dfrac{-45}{11}\right)\)

                 \(x=\dfrac{-315}{33}=\dfrac{-105}{11}\)