sao nhieu qua zay

a: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)

=>\(3A=2^{101}-2\)

=>\(A=\dfrac{2^{101}-2}{3}\)

b: Sửa đề: \(A=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)

\(A=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)

\(=\dfrac{2^{11}\cdot3^6\left(2^3+3^3\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)

\(=\dfrac{2}{3}\cdot\dfrac{4+27}{16+15}=\dfrac{2}{3}\)

c: \(B=\dfrac{4^5\cdot9^4-2\cdot6^4}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2\cdot2^4\cdot3^4}{2^{10}\cdot3^8+2^8\cdot2^2\cdot5\cdot3^8}\)

\(=\dfrac{2^5\cdot3^4\left(2^5\cdot3^4-1\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1}{2^5\cdot3^4}\cdot\dfrac{32\cdot81-1}{6}\)

\(=\dfrac{2591}{2^6\cdot3^5}\)

help

a) x+(-30)-(95-40-30+x)

=x+(-30)-95+40+30-x

=(x-x) +[(-30)-95+40+30]

=-55

b) a+(273-a-120)-(270-120)

=a+273-a-120-150

=(a-a)+(273-120-150)

=3

c)b-(294+130)+(94+130)

=b-424+224

=b-(424-224)

=b-200

\(\dfrac{63}{27}=\dfrac{7}{3}\)

\(\dfrac{3\times4\times6}{12\times8\times5}=\dfrac{3\times1\times1}{1\times1\times5}=\dfrac{3}{5}\)

a)\(\frac{20}{-160}=-\frac{1}{8}\)

b)\(\frac{-24}{-72}=\frac{1}{3}\)

c)\(-\frac{270}{450}=-\frac{3}{5}\)

d)\(\frac{3.7.11}{33.14}=\frac{33.7}{33.14}=\frac{7}{14}=\frac{1}{2}\)

e)\(\frac{49+7.49}{-49}=\frac{49.\left(1+7\right)}{-49}=\frac{49.8}{-49}=-8\)

\(a,\frac{20}{-160}=\frac{-20:20}{160:20}=\frac{-1}{8}\)

\(b,\frac{-24}{-72}=\frac{-24:24}{-72:24}=\frac{-1}{-3}=\frac{1}{3}\)

\(c,\frac{-270}{450}=\frac{-270:90}{450:90}=\frac{-3}{5}\)

\(d,\frac{3.7.11}{33.14}=\frac{3.7.11}{3.11.7.2}=\frac{1}{2}\)

\(e,\frac{49+7.49}{-49}=-\frac{49.\left(7+1\right)}{49}=-8\)

Học tốt

a) 20/-160 = 1/-8             b) -24/-72 = 1/3              c) -270/450 = -3/5     

d) 3.7.11/33.14 = 3.7.11/3.11.7.2 = 1/2     

 e) 49+7.49/ -49 = 7.7+7.49/-49 = 7.(7+49) /-49 = 7.56/-49 = 392/-49 = -8

49/69 Phân.số.tối.giản

50/ 80 =5/ 8

35/ 40=7/ 8

125/ 1000=1/ 8

\(\dfrac{49}{69}Phân.số.tối.giản\\ \dfrac{50}{80}=\dfrac{50: 10}{80:10}=\dfrac{5}{8}\\ \dfrac{35}{40}=\dfrac{35:5}{40:5}=\dfrac{7}{8}\\ \dfrac{125}{1000}=\dfrac{125:125}{1000:125}=\dfrac{1}{8}\)

\(P=\left(\dfrac{x-1}{\sqrt{x}+1}-\dfrac{x-2\sqrt{x}+1}{x-\sqrt{x}}+1\right).\dfrac{1}{x\sqrt{x}+1}\)

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

\(=\left(\sqrt{x}-1-\dfrac{\sqrt{x}-1}{\sqrt{x}}+1\right).\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)+\sqrt{x}}{\sqrt{x}}.\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}.\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

Bài 2: 

Ta có: \(P=\left(\dfrac{x-1}{\sqrt{x}+1}-\dfrac{x-2\sqrt{x}+1}{x-\sqrt{x}}+1\right)\cdot\dfrac{1}{x\sqrt{x}+1}\)

\(=\left(\sqrt{x}-1-\dfrac{\sqrt{x}-1}{\sqrt{x}}+1\right)\cdot\dfrac{1}{x\sqrt{x}+1}\)

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{1}{x+\sqrt{x}}\)