E=1/5^2 + 1/9^2+ ....+1/2017^2
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Lời giải:
$A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{2017.2019}$
$=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2019-2017}{2017.2019}$
$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}$
$=1-\frac{1}{2019}=\frac{2018}{2019}$
a,
13[x-9] = 169
=> x - 9 = 169/13
=> x - 9 = 13
=> x = 13+9
=> x = 22
b,
Viết lại đề:
7x+3 = 343
<=> 7x+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
\(\frac{3^2-1}{5^2-1}.\frac{7^2-1}{9^2-1}......\frac{2015^2-1}{2017^2-1}.\frac{2017^2-1}{2019^2-1}\) \(\Rightarrow\frac{1}{3}.\frac{3}{5}......\frac{1007}{1009}.\frac{504}{505}\)=\(\frac{504}{505}\)
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.
\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)