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\(\)Đặt \(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{205}}{\frac{204}{1}+\frac{203}{2}+\frac{202}{3}+...+\frac{1}{204}}=\frac{B}{C}\)
Biến đổi C:
\(C=\left(\frac{204}{1}+1\right)+\left(\frac{203}{2}+1\right)+\left(\frac{202}{3}+1\right)+...+\left(\frac{1}{204}+1\right)-204\)
\(=205+\frac{205}{2}+\frac{205}{3}+..+\frac{205}{204}+\frac{205}{205}-205\)
\(=205.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}\right)\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}}{205.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}\right)}=\frac{1}{205}\)
a= 1/2 + 1/4 + 1/8 - 1 x 1 + 8/1 - 4/1 - 2/1=\(1\frac{7}{8}\)=1,875
b=3/1 - 6/3 - 9/6 - 369/1 : 1/3 + 3/6 + 6/9 - 1/963 \(\approx\)186,665628245067
c=1/1 - 1/2 + 3/1 - 1/4 + 5/1 - 1/6 + 7/1 - 1/8 + 9/1 - 1/10=\(\approx\)23,8583333333333
vậy a>b>c
**************************l i k e***********************************8
A = \(\left(-\frac{1}{8}\right)\times\left(-13\right)=\frac{13}{8}\) => 0 < A < 2
B: Tử âm ; mẫu dương => B < 0
C = \(\left(\frac{1}{1}+\frac{3}{1}+\frac{5}{1}+\frac{7}{1}+\frac{9}{1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
= 25 \(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
Dễ có: B < A < C
b; \(3\frac{1}{4}.x-1\frac{1}{6}.x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x.\left(3\frac{1}{4}-1\frac{1}{6}-1\frac{2}{3}\right)=\frac{5}{12}\)
\(\Rightarrow x.\left(\frac{13}{4}-\frac{7}{6}-\frac{5}{3}\right)=\frac{5}{12}\)
\(\Rightarrow x.\left(\frac{39}{12}-\frac{14}{12}-\frac{20}{12}\right)=\frac{5}{12}\)
\(\Rightarrow x.\frac{5}{12}=\frac{5}{12}\)
\(\Rightarrow x=\frac{5}{12}\div\frac{5}{12}=1\)
Vậy x=1
Bài 1:
\(B=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{4}+\frac{3}{8}-\frac{5}{12}}+\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{8}}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)\(=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{2}\left(\frac{1}{2}+\frac{3}{4}-\frac{5}{6}\right)}+\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{8}\right)}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)
\(=\frac{1}{\frac{1}{2}}+3\) \(=2+3\) \(=5\)
Vậy B=5
Bài 2:
a) x3 - 36x = 0
=> x(x2-36)=0
=> x(x2+6x-6x-36)=0
=> x[x(x+6)-6(x+6) ]=0
=> x(x+6)(x-6)=0
\(\Rightarrow\orbr{\begin{cases}^{x=0}x+6=0\\x-6=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}^{x=0}x=-6\\x=6\end{cases}}\)
Vậy x=0; x=-6; x=6
b) (x - y = 4 => x=4+y)
x−3y−2 =32
=>2(x-3) = 3(y-2)
=>2x-6= 3y-6
=>2x-3y=0
=>2(4+y)-3y=0
=>8+2y-3y=0
=>8-y=0
=>y=8 (thỏa mãn)
Do đó x=4+y=4+8=12 (thỏa mãn)
Vậy x=12 và y =8
B= 1/2 + 3/4 - 5/6/1/2(1.2 + 3/4 - 5/6) + 3(1/4+ 1/5 - 1/8)/ 1/4 1/5 - 1/8
B= 1/ 1/2 + 3
B= 2+3
B=5
B2:
a) x^3 - 36x = 0
x(x^2 - 36) = 0
=> x=0 hoặc x^2-36=0
=> x= 0 hoặc x^2=36
=> x=0 hoặc x= +- 6
\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\)
\(9A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\)
\(9A+A=\left(\frac{1}{3}-\frac{1}{3^{^2}}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\right)+\left(\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)
\(10A=\frac{1}{3}-\frac{1}{3^{204}}\)
A = (1/3 - 1/3204) : 10
Vậy A = (1/3 - 1/3204) : 10.
A= \(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\left(1\right)\\ \)
\(\frac{1}{3^2}A=\frac{1}{3^2}\left(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)
\(\frac{1}{3^2}A=\frac{1}{3^4}-\frac{1}{3^6}+\frac{1}{3^8}-\frac{1}{3^{10}}+...+\frac{1}{3^{204}}-\frac{1}{3^{206}}\left(2\right)\)
Từ (1) và (2) vế theo vế ta có :\(A-\frac{1}{3^2}A=\frac{8}{9}A=\frac{1}{3^2}-\frac{1}{3^{206}}\)
\(\Rightarrow A=\left(\frac{1}{3^2}-\frac{1}{3^{206}}\right):\frac{8}{9}\)