Lời giải:

$A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{2017.2019}$

$=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2019-2017}{2017.2019}$

$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}$

$=1-\frac{1}{2019}=\frac{2018}{2019}$

cho gia tri tong a,b,c

a,

13[x-9] = 169

=> x - 9 = 169/13

=> x - 9 = 13

=> x = 13+9

=> x = 22

b,

Viết lại đề:

7^x+3 = 343

<=> 7^x+3 = 7³

=> x + 3 = 3 

=> x = 3-3

=> x = 0

c,

230 + [16 + [x-5]] = 315 . 2³

=> 230 + [16 + x - 5] = 315 . 8

=> 230 + 16 + x - 5 = 2520

=> 230 + 16 + x = 2520 + 5 = 2525

=> x = 2525 - 230 - 16 = 2279

d,

13.x - 3².x = 2017¹ - 1²⁰¹⁸

=> 13x - 9x = 2017 - 1

=> 4x = 2016

=> x = 504

a) 13 ( x-9 )=169

=> x-9 =169 : 13 =13

=> x=13+9 =22

b)\(7^{x+3}=343\)

\(7^x.7^3=343\)

\(7^x=343:7^3\)

\(7^x=1\Rightarrow x=1\)

c)230 + 16 +x -5 =315.8

241 +x =2520

x=2520-241=2279

d) 13x -\(3^2.x\)=2017-1

x(13-9)=2016

x.4=2016

x=2016:4

x=504

\(\frac{3^2-1}{5^2-1}.\frac{7^2-1}{9^2-1}......\frac{2015^2-1}{2017^2-1}.\frac{2017^2-1}{2019^2-1}\)  \(\Rightarrow\frac{1}{3}.\frac{3}{5}......\frac{1007}{1009}.\frac{504}{505}\)=\(\frac{504}{505}\)

Tính gt bt nhé

Giúp vs a

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