B1. Tìm x
a) 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... +1/x(x+1) =98/99
b) x - 20/11.13 - 20/13.15 - 20/15.17 - ... - 20/53.55 = 3/11
c) 5^x + 2.5^x + 2^3=83
B2. So sánh 2 biểu thức sau
A =2016^2016 + 2/2016^2016 - 1 và B = 2016^2016/2016^2016 - 3
giúp mình với các bạn ơi sắp thi rồi thầy cho nhiều bài lắm nhớ giải chi tiết giúp mình nhé
1.
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{99}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x-1}=\frac{98}{99}\)
\(1-\frac{1}{x-1}=\frac{98}{99}\)
\(\frac{1}{x-1}=1-\frac{98}{99}\)
\(\frac{1}{x-1}=\frac{1}{99}\)
\(\Rightarrow x-1=99\)
\(\Rightarrow x=99+1=100\)
b) \(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}\right)+10.\left(\frac{1}{13}-\frac{1}{15}\right)+10.\left(\frac{1}{15}-\frac{1}{17}\right)+...+10.\left(\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left[10.\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-10.\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
\(\Rightarrow x=\frac{3}{11}+\frac{8}{11}=1\)
c) 5x + 2 . 5x + 23 = 83
5x . ( 1 + 2 ) + 8 = 83
5x . 3 = 83 - 8
5x . 3 = 75
5x = 75 : 3
5x = 25
\(\Rightarrow\)5x = 52
\(\Rightarrow\)x = 2
2.
Ta thấy \(2016^{2016}>2016^{2016}-3\)
\(\Rightarrow B=\frac{2016^{2016}}{2016^{2016}-3}>\frac{2016^{2016}+2}{2016^{2016}-3+2}=\frac{2016^{2016}+2}{2016^{2016}-1}=A\)
\(\Rightarrow A< B\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{99}\)
Ta có \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{99}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{98}{99}\)(áp dụng công thức)
= \(1-\frac{1}{x+1}=\frac{98}{99}\)
= \(\frac{1}{x+1}=1-\frac{98}{99}\)(quy tắc tìm số trừ)
= \(\frac{1}{x+1}=\frac{1}{99}\Rightarrow\frac{1}{x+1}=\frac{1}{98+1}\Rightarrow x=98\)
Vậy x = 98 :)
Còn nữa, công thức mà mình áp dụng là: \(\frac{a}{b.c}=\frac{1}{b}-\frac{1}{c}\)nếu \(a=c-b\)
\(a,\frac{1}{1\cdot2}+.........+\frac{1}{x\cdot\left(x+1\right)}=\frac{98}{99}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{98}{99}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{98}{99}\Rightarrow\frac{1}{x+1}=1-\frac{98}{99}=\frac{1}{99}\Rightarrow x+1=99\Rightarrow x=98\)
Bai hoc a/b*c luon luon tach ra duoc la 1/b-1/c khi c-b=a hoac b-c=a
\(b,x-\frac{20}{11\cdot13}-......-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(\Rightarrow x-\left(\frac{20}{11\cdot13}+.....+\frac{20}{53\cdot55}\right)=\frac{3}{11}\Rightarrow x-2\left(\frac{2}{11\cdot13}+...+\frac{2}{53\cdot55}\right)=\frac{3}{11}\)
\(\Rightarrow x-2\cdot\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\Rightarrow x-2\cdot\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Rightarrow x-2\cdot\left(\frac{5}{55}-\frac{1}{55}\right)=\frac{3}{11}\Rightarrow x-2\cdot\frac{4}{55}=\frac{3}{11}\Rightarrow x-\frac{8}{55}=\frac{15}{55}\Rightarrow x=\frac{15}{55}+\frac{8}{55}=\frac{23}{55}\)
\(c,5^x+2\cdot5^x+2^3=83\)
\(\Rightarrow5^x\left(1+2\right)+8=83\Rightarrow3\cdot5^x=75\Rightarrow5^x=75\div3=25=5^2\Rightarrow x=2\)
Bai2
Ban co the ghi de ro hon dc ko
\(A=2016^{2016}+\frac{2}{2016^{2016}-1};B=\frac{2016^{2016}}{2016^{2016}-3}hayA=2016^{2016}+\frac{2}{2016^{2016}}-1;B=\frac{2016^{2016}}{2016^{2016}}-3\)
