= 45*95*95

= 4275*95

= 406125

**** mình nha

406125

tick de anh em

D ban nhe

a) Ta có: -12<-10

\(\Leftrightarrow30\cdot\left(-12\right)< 30\cdot\left(-10\right)\)

\(\Leftrightarrow30\cdot\left(-12\right)+2020< 30\cdot\left(-10\right)+2020\)(đpcm)

b) Ta có: 5>-5

\(\Leftrightarrow\left(-45\right)\cdot5< \left(-45\right)\cdot\left(-5\right)\)

\(\Leftrightarrow\left(-45\right)\cdot5+95< \left(-45\right)\cdot\left(-5\right)+95\)(đpcm)

=1/3(3/2*5+3/5*8+...+3/95*98)

=1/3(1/2-1/5+1/5-1/8+...+1/95-1/98)

=1/3*96/196

=32/196

=8/49

Ta có: \(\dfrac{95}{x}+\dfrac{95}{x+1}=5\)

\(\Leftrightarrow5x\left(x+1\right)=95x+95+95x\)

\(\Leftrightarrow5x^2+5x-190x-95=0\)

\(\Leftrightarrow5x^2-185x-95=0\)

\(\Leftrightarrow x^2-37x-19=0\)

\(\Delta=\left(-37\right)^2-4\cdot1\cdot\left(-19\right)=1445\)

Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{37-17\sqrt{5}}{2}\\x_2=\dfrac{37+17\sqrt{5}}{2}\end{matrix}\right.\)

\(\dfrac{95}{x}+\dfrac{95}{x+1}=5\)

<=> \(\dfrac{95\left(x+1\right)}{x\left(x+1\right)}+\dfrac{95x}{x\left(x+1\right)}=\dfrac{5x\left(x+1\right)}{x\left(x+1\right)}\)

=> 95x + 95 + 95x = 5x² + 5x

<=> 190x - 5x - 5x² = -95

<=> -5x² + 185x + 95 = 0

<=> x = 37,5

Ta có:\(A=\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{11}+...+\dfrac{31}{92}-\dfrac{32}{95}+\dfrac{32}{95}-\dfrac{33}{98}\)

                \(=\dfrac{1}{2}+\dfrac{33}{98}=\dfrac{82}{98}=\dfrac{41}{49}\)

Ta có: \(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot98}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)

\(=\dfrac{8}{49}\)

\(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-99}{1}-1\right)+\left(\dfrac{x-3}{97}-1\right)+\left(\dfrac{x-97}{3}-1\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-95}{5}-1\right)=0\)

=>x-100=0

=>x=100

\(A=\frac{2}{2\cdot5}+\frac{2}{5\cdot8}+\frac{2}{8\cdot11}+...+\frac{2}{92\cdot95}+\frac{2}{95\cdot98}\)

\(A=\frac{2}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\right]\)

\(A=\frac{2}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right]\)

\(A=\frac{2}{3}\left[\frac{1}{2}-\frac{1}{98}\right]=\frac{2}{3}\left[\frac{49}{98}-\frac{1}{98}\right]=\frac{2}{3}\cdot\frac{48}{98}=\frac{2}{3}\cdot\frac{24}{49}=\frac{2}{1}\cdot\frac{8}{49}=\frac{16}{49}\)

\(A=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{92.95}+\frac{2}{95.98}\)

\(=\frac{2}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)

\(=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)

\(=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(=\frac{2}{3}.\frac{24}{49}\)

\(=\frac{16}{49}\)

#)Giải :

\(A=2-\frac{2}{5}+\frac{2}{5}-\frac{2}{8}+\frac{2}{8}-\frac{2}{11}+...+\frac{2}{95}-\frac{2}{98}\)

\(A=2-\frac{2}{98}\)

\(A=1\frac{48}{49}=\frac{97}{49}\)

         #~Will~be~Pens~#