giải phương trình

a,(\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{9\cdot10}\))(x-1)+\(\frac{1}{10}x\)=\(x-\frac{9}{10}\)

b,\(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+\frac{40}{39}\)

c,(x-10)+(x-19)+(x-18)+...+100+101=101

d,(\(\frac{1}{1\cdot51}+\frac{1}{2\cdot52}+\frac{1}{3\cdot53}+...+\frac{1}{10\cdot60}\))x=\(\frac{1}{1\cdot11}+\frac{1}{1\cdot12}+\frac{1}{1\cdot13}\)

Câu 1 :Tính tổng S=\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{8}\)+......+\(\frac{1}{2^{100}}\)

Câu 2 :Tính tổng M=\(\frac{3}{5}+\frac{3}{5^2}+..........+\frac{3}{5^{201}}\)

Câu 3 :Tính tổng N=101³⁰+101³¹+101³²+.......+101¹⁰¹

Câu 4 :Tính tổng A=2³+4³+6³+........+2012³

Câu 5 :Tính tổng B=1³+3³+5³+........+2011³

Câu 6 :Tính tổng C=2*4*6*8+4*6*8*10+.......+100*102*104*106

                              Giúp mik vs m.n !!

1) tính

a) \(\frac{4}{x+2}+\frac{3}{2-x}+\frac{12}{x^2-4}\)

b) \(x+\frac{x-1}{2}+\frac{x-2}{3}\)

c) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{3x-6}{4-9x^2}\)

d) x - 2 - \(\frac{x^2-10}{x+2}\)

e) \(\frac{1}{2x-2y}-\frac{1}{2x+2y}+\frac{y}{y^2-x^2}\)

f) \(\frac{1}{a+1}-\frac{3}{a^3+1}+\frac{3}{a^2-a+1}\)

g) \(\frac{4-2x+x^2}{x+2}-2-x\)

h)\(\frac{1}{x^3-x}-\frac{1}{x^2-x}+\frac{2}{x^2-1}\)

j) \(\frac{1}{2x+3}-\frac{1}{2x-3}+\frac{x-2}{2x^2-x-3}\)

Tính :

a ) \(S=\frac{1}{\sqrt{1}\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+.....+\)\(\frac{1}{\sqrt{2017}+\sqrt{2019}}\)

b ) \(S=\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+....+\frac{1}{\sqrt{100}+\sqrt{102}}\)

c ) \(S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.....+\frac{1}{\sqrt{100}+\sqrt{101}}\)

d  ) \(S=\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}}+....+\frac{1}{\sqrt{2016}+\sqrt{2019}}\)

a) B= \(\frac{x^2-x+1}{x^2+2x+1}\)=\(\frac{x^2-x+1}{\left(x+1\right)^2}\)

Đặt t = x+1 => x= t-1 => B= \(\frac{\left(t-1\right)^2-\left(t-1\right)+1}{t^2}\)=\(\frac{t^2-2t+1-t+1+1}{t^2}\)

B= \(\frac{t^2-3t+3}{t^2}\)=\(\frac{t^2}{t^2}\)\(-\frac{3t}{t^2}\)\(+\frac{3}{t^2}\)

Đặt a = \(\frac{1}{t}\)=> B= 3a²-3a+1= 3(a²-a+\(\frac{1}{3}\))= 3(a²-2a.\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{12}\))= 3(a-\(\frac{1}{2}\))²+\(\frac{1}{4}\)\(\ge\)\(\frac{1}{4}\)

Min của E= \(\frac{1}{4}\)khi a-\(\frac{1}{2}\)= 0 => a= \(\frac{1}{2}\)=> \(\frac{1}{t}\)=\(\frac{1}{2}\)\(\Leftrightarrow\)\(\frac{1}{x+1}\)=\(\frac{1}{2}\)\(\Leftrightarrow\)x+1=2 => x= 1

bài 1:giải các pt sau:

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

b/\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3}\)

c/\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)

d/\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)

e/\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

f\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)