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

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

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

Ta có:

\(B=\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)

\(9B=3+\dfrac{1}{3}+...+\dfrac{1}{3^{97}}\)

\(9B-B=3-\dfrac{1}{3^{99}}\)

\(B=\dfrac{3-\dfrac{1}{3^{99}}}{8}\)

\(A=-\dfrac{2}{3}B=\dfrac{-2}{3}.\dfrac{3-\dfrac{1}{99}}{8}=\dfrac{\dfrac{1}{3^{100}}-1}{4}\)

1,

\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)

\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)

2.

\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)

3.

Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)

4.

\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)

\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)

5.

\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)

\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)

\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)

\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)

Đề sai rồi bạn

bài 2:

\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)

bài 3:

\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)

\(=>x=3\)

\(P=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)

\(\Rightarrow\dfrac{1}{2}P=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}\)

\(\Rightarrow\dfrac{1}{2}P-P=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{100}}\)

\(\Rightarrow-\dfrac{1}{2}P=\dfrac{1}{2^{101}}-\dfrac{1}{2^2}\)

\(\Rightarrow P=\left(\dfrac{1}{2^{101}}-\dfrac{1}{2^2}\right):\left(-\dfrac{1}{2}\right)\)

\(E=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{2}+\dfrac{1}{12}\)

\(E=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}\right)\)

\(E=\dfrac{2}{2}+\dfrac{3}{6}+\left(\dfrac{1}{8}+\dfrac{3}{24}\right)\)

\(E=1+\dfrac{1}{2}+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\)

\(E=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\dfrac{2}{8}\)

\(E=\dfrac{3}{2}+\dfrac{1}{4}\)

\(E=\dfrac{6}{4}+\dfrac{1}{4}\)

\(E=\dfrac{7}{4}\)

a: Ta có: |x+4|=1

=>x+4=1 hoặc x+4=-1

=>x=-3(loại) hoặc x=-5

Khi x=-5 thì \(A=\dfrac{\left(-5\right)^2-5}{3\left(-5+3\right)}=\dfrac{20}{3\cdot\left(-2\right)}=\dfrac{-10}{3}\)

b: \(B=\dfrac{x-1+x+1-3+x}{\left(x-1\right)\left(x+1\right)}=\dfrac{3x-3}{\left(x-1\right)\left(x+1\right)}=\dfrac{3}{x+1}\)

\(\sqrt{1+\dfrac{1}{n}+\dfrac{1}{\left(n+1\right)^2}}\\ =\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}+\dfrac{2}{n}-\dfrac{2}{n+1}-\dfrac{2}{n\left(n+1\right)}}\\ =\sqrt{\left[1+\dfrac{1}{n}-\dfrac{1}{\left(n+1\right)}\right]^2}=\left|1+\dfrac{1}{n}-\dfrac{1}{\left(n+1\right)}\right|\)

\(\Leftrightarrow P=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{99}-\dfrac{1}{100}=98+\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{9849}{100}\)

a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)

=1-2/4=1/2

b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)

c: x-y=0 nên x=y

\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)

=2019

a: \(A=2^{\dfrac{1}{3}}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{3}+\dfrac{2}{3}}=2^{\dfrac{3}{3}}=2^1=2\)

b: \(B=36^{\dfrac{3}{2}}=\left(6^2\right)^{\dfrac{3}{2}}=6^{2\cdot\dfrac{3}{2}}=6^3=216\)

c: \(C=36^{\dfrac{3}{2}}\cdot\left(\dfrac{1}{6}\right)^2=\left(6^2\right)^{\dfrac{3}{2}}\cdot\dfrac{1}{6^2}=\dfrac{6^{2\cdot\dfrac{3}{2}}}{6^2}=\dfrac{6^3}{6^2}=6\)

d: \(D=\sqrt{81}\cdot\left(\dfrac{1}{3}\right)^2=9\cdot\dfrac{1}{3^2}=9\cdot\dfrac{1}{9}=1\)

e: \(E=\left(3+2\sqrt{2}\right)^{50}\cdot\left(3-2\sqrt{2}\right)^{50}\)

\(=\left[\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)\right]^{50}\)

\(=\left(9-8\right)^{50}=1^{50}=1\)

f: \(F=120^{\sqrt{5}+1}\cdot120^{3-\sqrt{5}}\)

\(=120^{\sqrt{5}+1+3-\sqrt{5}}=120^4\)

g: \(G=\left(3+2\sqrt{2}\right)^{2019}\cdot\left(3\sqrt{2}-4\right)^{2018}\)

\(=\left(3+2\sqrt{2}\right)^{2018}\cdot\left(3\sqrt{2}-4\right)^{2018}\cdot\left(3+2\sqrt{2}\right)\)

\(=\left[\left(3+2\sqrt{2}\right)\left(3\sqrt{2}-4\right)\right]^{2018}\left(3+2\sqrt{2}\right)\)

\(=\left(9\sqrt{2}-12+12-8\sqrt{2}\right)^{2018}\cdot\left(3+2\sqrt{2}\right)\)

\(=\left(\sqrt{2}\right)^{2018}\cdot\left(3+2\sqrt{2}\right)=2^{\dfrac{1}{2}\cdot2018}\cdot\left(3+2\sqrt{2}\right)\)

\(=2^{1009}\cdot\left(3+2\sqrt{2}\right)\)