\(A=\sqrt{\dfrac{9}{16}}+2017^0+\left|-0,25\right|.\)

\(A=\dfrac{3}{4}+1+\dfrac{1}{4}.\)

\(A=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+1.\)

\(A=1+1.\)

\(A=2.\)

Vậy.....

A =3/4+1+0,25=2

A=\(\sqrt{\frac{9}{16}}\)+\(2016^0+\left|-0,25\right|\)

A=\(\frac{3}{4}\)+1+0,25

A=2

\(A=\sqrt{\frac{9}{16}+2016^{ }^0+\left|-0,25\right|}\)

\(A=\frac{3}{4}+1+0,25\)

\(A=0,75+1+0,25\)

\(A=2\)

\(\dfrac{6}{5}\sqrt{1\dfrac{9}{16}}-\left(-\dfrac{3}{4}\right)^2:0,25\)

\(=\dfrac{6}{5}\cdot\sqrt{\dfrac{25}{16}}-\dfrac{9}{16}:0,25\)

\(=\dfrac{6}{5}\cdot\sqrt{\left(\dfrac{5}{4}\right)^2}-\dfrac{9}{16}:\dfrac{1}{4}\)

\(=\dfrac{6}{5}\cdot\dfrac{5}{4}-\dfrac{9\cdot4}{16}\)

\(=\dfrac{6}{4}-\dfrac{9}{4}\)

\(=\dfrac{6-9}{4}\)

\(=-\dfrac{3}{4}\)

a/ \(2016\dfrac{1}{6}:\dfrac{-2}{5}-16\dfrac{1}{6}:\dfrac{-2}{5}\)

\(=2016\dfrac{1}{6}.\dfrac{-5}{2}-16\dfrac{1}{6}.\dfrac{-5}{2}\)

\(=\dfrac{-5}{2}\left(2016\dfrac{1}{6}-16\dfrac{1}{6}\right)\)

\(=\dfrac{-5}{2}.2000\)

\(=-5000\)

b/ \(\left(\dfrac{4}{3}-\dfrac{3}{2}\right)^2-2.\left|-\dfrac{1}{9}\right|+\sqrt{\dfrac{4}{81}}\)

\(=\left(\dfrac{8}{6}-\dfrac{9}{6}\right)^2-2.\dfrac{1}{9}+\dfrac{2}{9}\)

\(=\dfrac{1}{4}-\dfrac{2}{9}+\dfrac{2}{9}\)

\(=\dfrac{1}{36}+\dfrac{2}{9}\)

\(=\dfrac{1}{4}\)

a.

\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)

c.

\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)

d.

\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{-a-5b\sqrt{5}}{\left(a+b\sqrt{5}\right)\left(a-b\sqrt{5}\right)}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{a+5b\sqrt{5}}{a^2-5b^2}=9+20\sqrt{5}\\ \Leftrightarrow\left(9+20\sqrt{5}\right)\left(a^2-5b^2\right)=a+5b\sqrt{5}\\ \Leftrightarrow9\left(a^2-5b^2\right)+\sqrt{5}\left(20a^2-100b^2\right)-5b\sqrt{5}=a\\ \Leftrightarrow\sqrt{5}\left(20a^2-100b^2-5b\right)=9a^2-45b^2+a\)

Vì \(\sqrt{5}\) vô tỉ nên để \(\sqrt{5}\left(20a^2-100b^2-5b\right)\) nguyên thì

\(\left\{{}\begin{matrix}20a^2-100b^2-5b=0\\9a^2-45b^2+a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}180a^2-900b^2-45b=0\\180a^2-900b^2+20a=0\end{matrix}\right.\\ \Leftrightarrow20a+45b=0\\ \Leftrightarrow4a+9b=0\Leftrightarrow a=-\dfrac{9}{4}b\\ \Leftrightarrow9a^2-45b^2+a=\dfrac{729}{16}b^2-45b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow\dfrac{9}{16}b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow b\left(\dfrac{9}{16}b-\dfrac{9}{4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}b=0\\b=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=9\end{matrix}\right.\)

Với \(\left(a;b\right)=\left(0;0\right)\left(loại\right)\)

Vậy \(\left(a;b\right)=\left(9;4\right)\)

a)

x^2-16/25=0

x^2-4^2/5^2=0

=>x-4/5=0

x=0+4/5

x=0/5

\(x=\left(1-\dfrac{1}{\sqrt{4}}\right).\left(1-\dfrac{1}{\sqrt{16}}\right).\left(1-\dfrac{1}{\sqrt{36}}\right).\left(1-\dfrac{1}{\sqrt{64}}\right).\left(1-\dfrac{1}{\sqrt{100}}\right)\)

\(x=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{6}\right).\left(1-\dfrac{1}{8}\right).\left(1-\dfrac{1}{10}\right)\)

\(x=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}.\dfrac{9}{10}\)

\(x=\dfrac{63}{256}\)

và \(y=\sqrt{20+0,25}\)

\(y=\sqrt{20,25}\)

\(y=4,5\)

Do 4,5 > \(\dfrac{63}{256}\)

=> x<y

a: \(=7\cdot\dfrac{6}{7}-5+\dfrac{3\sqrt{2}}{2}=1+\dfrac{3}{2}\sqrt{2}\)

b: \(=-\dfrac{8}{7}-\dfrac{3}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{-16+7}{14}-\dfrac{3}{8}=\dfrac{-9}{14}-\dfrac{3}{8}\)

\(=\dfrac{-72-42}{112}=\dfrac{-114}{112}=-\dfrac{57}{56}\)

c: \(=20\sqrt{5}-\dfrac{1}{4}\cdot\dfrac{4}{3}+\dfrac{3}{2}=20\sqrt{5}+\dfrac{3}{2}-\dfrac{1}{3}=20\sqrt{5}+\dfrac{7}{6}\)

\(\dfrac{6}{5}\cdot\sqrt{\dfrac{25}{16}}-\left(\dfrac{3}{4}\right)^2:0,25\\ =\dfrac{6}{5}\cdot\dfrac{5}{4}-\dfrac{9}{16}\cdot4\\ =\dfrac{3}{2}-\dfrac{9}{4}\\ =-\dfrac{3}{4}\)

\(\dfrac{6}{5}\cdot\sqrt{\dfrac{25}{16}}-\left(\dfrac{3}{4}\right)^2:0,25\)

\(=\dfrac{6}{5}\cdot\dfrac{5}{4}-\dfrac{9}{16}:\dfrac{1}{4}\)

\(=\dfrac{6\cdot5}{5\cdot4}-\dfrac{9\cdot4}{16}\)

\(=\dfrac{6}{4}-\dfrac{9}{4}\)

\(=\dfrac{3}{4}\)

\(\dfrac{5}{6}.\sqrt{\dfrac{25}{16}}-\left(\dfrac{3}{4}\right)^2:0,25\\ =\dfrac{5}{6}.\sqrt{\dfrac{5^2}{4^2}}-\dfrac{9}{16}:\dfrac{1}{4}\\ =\dfrac{5}{6}.\dfrac{5}{4}-\dfrac{9}{16}.4\\ =\dfrac{25}{24}-\dfrac{9}{4}=\dfrac{25-9.6}{24}=\dfrac{-39}{24}=-\dfrac{13}{8}\)