\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}=\frac{50-1}{100}=\frac{49}{100}\)

1/2*3+1/3*4+.....+1/99*100

=1/2-1/3+1/3-1/4+........1/99*100

=1/2+(-1/3+1/3)+(-1/4+1/4)+.........+(-1/99+1/99)-1/100

=1/2-1/100

=50/100-1/100

=49/100

\(\frac{2015^{35}+1}{2015^{34}+1}=\frac{2015^{35}+2015-2014}{2015^{34}+1}=\frac{2015\left(2015^{34}+1\right)-2014}{2015^{34}+1}=\frac{2015\left(2015^{34}+1\right)}{2015^{34}+1}-\frac{2014}{2015^{34}+1}=2015-\frac{2014}{2015^{34}+1}\)

\(\frac{2015^{34}+1}{2015^{33}+1}=\frac{2015^{34}+2015-2014}{2015^{33}+1}=\frac{2015\left(2015^{33}+1\right)-2014}{2015^{33}+1}=\frac{2015\left(2015^{33}+1\right)}{2015^{33}+1}-\frac{2014}{2015^{33}+1}=2015-\frac{2014}{2015^{33}+1}\)

Mà \(2015-\frac{2014}{2015^{34}+1}>2015-\frac{2014}{2015^{33}+1}\)

Vậy\(\frac{2015^{35}+1}{2015^{34}+1}>\frac{2015^{34}+1}{2015^{33}+1}\)

CẢM ƠN BẠN RẤT ẤT ẤT T T T .... NHIỀU

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\Leftrightarrow\left(\frac{1}{3}+\frac{2}{5}\right)x=\frac{2}{5}\)

\(\Leftrightarrow\frac{11}{15}x=\frac{2}{5}\)

\(\Leftrightarrow x=\frac{6}{11}\)

1/3 x + 2/5 x - 2/5 = 0

1/3x + 2/5 x = 2/5

11/15 x   = 2/5

x= 2/5 : 11/15 = 6/11

B < 1+1+1/2.3+1/3.4+...+1/62.63

B < 2+(1/2-1/3+1/3-1/4+...+1/62-1/63)

B < 2+(1/2-1/63)

B < 2+61/126 suy ra B < 2 và 6/126

Mà 2 + 61/126 <6

Suy ra B< 2+6/126<6 suy tiếp B < 6

\(2\frac{2}{3}x+1\frac{1}{3}x=\frac{2}{3}\)

\(\Rightarrow\) \(\frac{8}{3}x+\frac{4}{3}x=\frac{2}{3}\)

\(\Rightarrow\) \(\left(\frac{8}{3}+\frac{4}{3}\right)x=\frac{2}{3}\)

\(\Rightarrow\)        \(4x=\frac{2}{3}\)

\(\Rightarrow\)\(x=\frac{2}{3}:4\)

\(\Rightarrow x=\frac{1}{6}\)

mọng mina giúp đỡ

\(1,3x+3\frac{1}{2}=x\)

\(1,3x+\frac{7}{2}=x\)

\(1,3x-x+\frac{7}{2}=0\)

\(x.\left(1,3-1\right)=-\frac{7}{2}\)

\(x\cdot0,3=-\frac{7}{2}\)

\(x\cdot\frac{3}{10}=-\frac{7}{2}\)

\(x=-\frac{7}{2}:\frac{3}{10}=-\frac{7}{2}\cdot\frac{10}{3}=-\frac{35}{3}\)

Trả lời

a) Đặt \(H=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(\Rightarrow H< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(\Leftrightarrow H< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow H< 1-\frac{1}{100}\)

\(\Leftrightarrow H< \frac{99}{100}\)

\(\Leftrightarrow A< 1+\frac{99}{100}\)

Ta thấy \(\frac{99}{100}< 1\Rightarrow A< 2\)

Vậy A<2 (đpcm)

b) Ta có: 1=1

             \(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)

               \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< \frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)

               \(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+...+\frac{1}{15}< \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}=1\)

                \(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}< \frac{1}{16}+\frac{1}{16}+...+\frac{1}{16}=1\)

                \(\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}< \frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}=1\)

                 \(\Rightarrow B< 1+1+1+1+1+1\)

                 \(\Rightarrow B< 6\)

   Vậy B<6 (đpcm)

(2x-3)(6-2x) =0

=> 2 x -3=0 hoặc 6-2x=0

      2x=0+3              2x=6-0

       2x =3                    2x=6

x=2/3                              x=6:2

                                       x=3

Ta có: \(\frac{1}{3x}+\frac{2}{5\left(x-1\right)}=0\)

\(\Rightarrow\frac{5\left(x-1\right)}{3x.5\left(x-1\right)}+\frac{2.3x}{3x.5\left(x-1\right)}=0\)

\(\Rightarrow\frac{5\left(x-1\right)+6x}{15x.\left(x-1\right)}=0\)

\(\Rightarrow\frac{11x-5}{15x\left(x-1\right)}=0\)(với điều kiện \(15x\left(x-1\right)>0\forall x\))

\(\Rightarrow11x-5=0\Rightarrow11x=5\Rightarrow x=\frac{5}{11}\)

Tương tác với mình nha mn!