\(1,ĐK:x\ge1\\ PT\Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\\ 2,\Leftrightarrow2x-5=x^2-8x+16\left(x\ge4\right)\\ \Leftrightarrow x^2-10x+21=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=7\left(tm\right)\end{matrix}\right.\\ 3,\Leftrightarrow\left[{}\begin{matrix}2x-1=x+2\left(x\ge\dfrac{1}{2}\right)\\1-2x=x+2\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\dfrac{1}{3}\left(tm\right)\end{matrix}\right.\)

\(ĐK:x\ne-1;y\ne2\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y}{2-y}=-1\\\dfrac{x}{x+1}+\dfrac{2y}{2-y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0y=-2\left(vn\right)\\\dfrac{x}{x+1}+\dfrac{2y}{2-y}=2\end{matrix}\right.\Leftrightarrow x,y\in\varnothing\)

\(c,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{3^{61}}{3^{60}}=3\\ d,=\dfrac{9}{16}\left(-16\right)-\left(-1\right)^{2016}-10=-9-1-10=-20\)

\(a,\Leftrightarrow x^2-25+9-3x-10x+11=0\\ \Leftrightarrow x^2-13x-5=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13+3\sqrt{21}}{2}\\x=\dfrac{13-3\sqrt{21}}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-3-2x-3\right)\left(x-3+2x+3\right)=0\\ \Leftrightarrow3x\left(-x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

\(ĐK:x,y\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{2}{y}=4\\\dfrac{2}{x}+\dfrac{3}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+1=2\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\left(tm\right)\)

Vì \(AC=BC\) nên \(C\in\) trung trực AB

Vì \(OA=OB\) nên \(O\in\) trung trực AB

Do đó OC là trung trực AB

\(\Rightarrow OC\bot AB\) tại H và H là trung điểm AB

Do đó \(AH=\dfrac{1}{2}AB=3\)

Áp dụng HTL vào tam giác AHC: \(AH^2=HC\cdot HO=9\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}-y\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=1\end{matrix}\right.\Leftrightarrow x,y\in\varnothing\left(x,y\in Z\right)\)

\(=\dfrac{-2^{19}\cdot2^{28}\cdot5^{14}}{5^{15}\cdot2^{49}}=\dfrac{-1\cdot1}{5\cdot2^2}=-\dfrac{1}{20}\)

VD2:

\(a,A=\dfrac{x^2}{\left(y+1\right)^2}\cdot\dfrac{y+1}{2x}\cdot\dfrac{y+1}{2x}=\dfrac{1}{4}\\ b,B=\dfrac{x^2}{\left(y+1\right)^2}:\left(\dfrac{2y}{y+1}\cdot\dfrac{y+1}{2x}\right)=\dfrac{x^2}{\left(y+1\right)^2}\cdot1=\dfrac{x^2}{\left(y+1\right)^2}\)

VD1:

\(a,\dfrac{6x}{15y^3}\left(-\dfrac{5y^2}{3x^2}\right)=\dfrac{-2}{3xy}\\ b,=\dfrac{x+1}{x-2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x+1\right)^2}=\dfrac{x+2}{x+1}\\ c,=\dfrac{-3\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x-1}=\dfrac{-3}{x+3}\\ d,=\dfrac{2\left(3x+2\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-2\right)}{3x+2}=\dfrac{2x}{x+2}\)

VD1:

\(a,=\dfrac{\left(1-x\right)\left(1+x\right)}{x\left(x-2\right)}\cdot\dfrac{x}{x+1}=\dfrac{1-x}{x-2}\\ b,=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x+2}\cdot\dfrac{1}{x^2+x+1}=\dfrac{x-1}{x+2}\\ c,=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)}\cdot\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)^2}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)^2}\\ d,=\dfrac{x+2y}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x+2y\right)^2}=\dfrac{x-y}{x+2y}\)