\(1,M=\dfrac{x-5+2x-x-5}{x\left(x-5\right)}=\dfrac{2\left(x-5\right)}{x\left(x-5\right)}=\dfrac{2}{x}\\ 2,\\ a,D=\dfrac{5+4+1}{1}=10\\ b,D=2\Leftrightarrow5x^2+4x+1=2x^2\\ \Leftrightarrow3x^2+3x+x+1=0\\ \Leftrightarrow\left(x+1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{3}\end{matrix}\right.\left(tm\right)\\ c,D=\dfrac{5x^2+4x+1}{x^2}\\ \Leftrightarrow Dx^2-5x^2-4x-1=0\\ \Leftrightarrow x^2\left(D-5\right)-4x-1=0\\ \Leftrightarrow\Delta'\ge0\\ \Leftrightarrow4+\left(D-5\right)\ge0\\ \Leftrightarrow D-1\ge0\Leftrightarrow D\ge1\\ \Leftrightarrow D_{max}=1\Leftrightarrow5x^2+4x+1=x^2\Leftrightarrow x=-\dfrac{1}{2}\)

\(\text{Đặt }\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow a=bk;c=dk\\ \Leftrightarrow\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\left(1\right)\\ \dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\left(2\right)\\ \left(1\right)\left(2\right)\LeftrightarrowĐpcm\)

\(a,ĐK:a\ne0;a\ne-5\\ b,A=\dfrac{a^3+2a^2+2a^2-50+50-5a}{2a\left(a+5\right)}=\dfrac{a^3+4a^2-5a}{2a\left(a+5\right)}\\ A=\dfrac{a\left(a^2+4a-5\right)}{2a\left(a+5\right)}=\dfrac{a\left(a+5\right)\left(a-1\right)}{2a\left(a+5\right)}=\dfrac{a-1}{2}\\ c,a=-1\Leftrightarrow A=\dfrac{-1-1}{2}=-1\\ d,A=0\Leftrightarrow a-1=0\Leftrightarrow a=1\left(tm\right)\)

Gọi số hs 7A,7B,7C lần lượt là \(a,b,c\in \mathbb{N^*},hs\)

Áp dụng tc dtsbn:

\(\dfrac{a}{9}=\dfrac{b}{10}=\dfrac{c}{8}=\dfrac{b-a}{10-9}=\dfrac{5}{1}=5\\ \Leftrightarrow\left\{{}\begin{matrix}a=45\\b=50\\c=40\end{matrix}\right.\)

Vậy ...

GỌi số cây 7A,7B,7C lần lượt là \(a,b,c\in \mathbb{N^*},cây\)

Áp dụng tc dtsbn:

\(\dfrac{a}{5}=\dfrac{b}{8}=\dfrac{c}{9}=\dfrac{c-a}{9-5}=\dfrac{8}{4}=2\\ \Leftrightarrow\left\{{}\begin{matrix}a=10\\b=16\\c=18\end{matrix}\right.\)

Vậy ...

\(\Leftrightarrow2\left(x-5\right)-5x=3-x\\ \Leftrightarrow2x-10-5x+x=3\\ \Leftrightarrow-2x=13\Leftrightarrow x=-\dfrac{13}{2}\)

\(4^{20}+4^{21}+4^{22}+4^{23}=4^{20}\left(1+4+4^2+4^3\right)=4^{20}\cdot85⋮5\left(85⋮5\right)\)

\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(2+2^2+2^3\right)+2^3\left(2+2^2+2^3\right)+...+2^{2007}\left(2+2^2+2^3\right)\\ A=\left(2+2^2+2^3\right)\left(1+2^3+...+2^{2007}\right)\\ A=14\left(1+2^3+...+2^{2007}\right)⋮2\text{ và }7\left(14⋮2\text{ và }7\right)\)

\(ĐK:x^2-4=\left(x-2\right)\left(x+2\right)\ne0\Leftrightarrow x\ne\pm2\)