ta có 1-x=-(x-1)

1-x+1=x-1

<=>3=2x

<=>x=2/3

vậy x =2/3

Theo bài ra ta có:1-\(\frac{1}{1-x}\)=\(\frac{1}{1-x}\)

Suy ra:\(\frac{1}{1-x}\)=1-

Theo bài ra ta có:1-\(\frac{1}{1-x}\)=\(\frac{1}{1-x}\)

Suy ra:\(\frac{1}{1-x}\)=1-\(\frac{1}{1-x}\)

         \(\frac{1}{1-x}\)=\(\frac{1-x}{1-x}\)- \(\frac{1}{1-x}\) 

         \(\frac{1}{1-x}\)=\(\frac{1-x-1}{1-x}\)

         \(\frac{1}{1-x}\)=\(\frac{1-1-x}{1-x}\)=\(\frac{-x}{1-x}\)

         \(\frac{1}{1-x}\)=\(\frac{-x}{1-x}\)

Suy ra :1=-x

Suy ra:x=-1

      Vậy x=-1

VT = = 4 - 3 = 1 = VP

Vậy:  2 - 3 2 + 3 = 1

B đúng

B bn a!!!

B

Ta thấy \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

 \(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

......

\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)

hay \(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\)

\(D< 1-\dfrac{1}{10}=\dfrac{9}{10}< 1\) ( đpcm )

Ta có \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)

         \(\dfrac{1}{3.3}\)<\(\dfrac{1}{2.3}\)

         \(\dfrac{1}{4.4}\)<\(\dfrac{1}{3.4}\)

  .........................

         \(\dfrac{1}{10.10}\)<\(\dfrac{1}{9.10}\)

=>\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

=> D <  1 - \(\dfrac{1}{10}\)

=>D < \(\dfrac{9}{10}\)

=> D < \(\dfrac{10}{10}\)

 Vậy D < 1

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(.....\)

\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)

\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)

\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}< 1\)

\(\Rightarrow B< 1\left(dpcm\right)\)

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)

 \(B< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{9\times10}\)

 \(B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(B< 1-\dfrac{1}{10}\)

\(B< \dfrac{9}{10}< 1\)

Vậy \(B< 1\)

B= 1/4+1/9+1/16+1/25+1/36+1/49+1/64+1/81+1/100

vì các số kia dần nhỏ dần và số lớn nhất cũng chỉ có 0,25 nên B<1

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

1/5*7+1/7*9+...+1/63*65=1/2(2/5*7+2/7*9+...+2/63*65)=1/2(1/5-1/7+1/7-1/9+...+1/63-1/65)=1/2(1/5-1/65)=1/2*(13/65-1/65)=1/2*12/65=6/65

Vì Số nghịch đảo của 6/65 là 65/6

nên Số nghịch đảo của 1/5*7+1/7*9+...+1/63*65 là 65/6

Cái này là tính nhanh chứ?

= \(\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\cdot\cdot\cdot\frac{10^2-1}{10^2}=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\cdot\cdot\cdot\frac{9\cdot11}{10\cdot10}=\frac{\left(1\cdot2\cdot3\cdot\cdot\cdot\cdot9\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot\cdot10\cdot11\right)}{\left(2\cdot3\cdot..\cdot10\right)\left(2\cdot3\cdot\cdot\cdot\cdot10\right)}=\frac{11}{2.10}=\frac{11}{20}\)

Nghịch đảo của số đó là 20/11

Số nghịch đảo của -1 là -1