2, \(=>9A=3^3+3^5+3^7+......+3^{39}+3^{41}\)

\(=>9A-A=3^{41}-3\)

\(=>A=\dfrac{3^{41}-3}{8}\)

CHÚC BẠN HỌC TỐT........

Vì \(a^2+b^2+c^2=1\)

\(\Rightarrow-1\le a,b,c\le1\)

\(\Rightarrow a-1\le0;b-1\le0;c-1\le0\)

Lây cai xau trừ cai trươc được

\(\left(a^3+b^3+c^3\right)-\left(a^2+b^2+c^2\right)=0\)

\(\Leftrightarrow a^2\left(a-1\right)+b^2\left(b-1\right)+c^2\left(c-1\right)=0\)

Ta co \(VT\le0\)

Dâu = xảy ra khi: \(\left(a,b,c\right)=\left\{0,0,1;0,1,0;1,0,0\right\}\)

\(\Rightarrow S=1\) 

a. x + \(\dfrac{3}{7}\)= \(\dfrac{2}{5}:\dfrac{18}{25}=>x+\dfrac{3}{7}=\dfrac{2}{5}\)x\(\dfrac{35}{18}=>x+\dfrac{3}{7}=\dfrac{7}{9}\)

=> x = \(\dfrac{7}{9}-\dfrac{3}{7}=\dfrac{49}{63}-\dfrac{27}{63}=\dfrac{22}{63}\)

b. \(x\) x \(\dfrac{5}{9}\)= \(\dfrac{4}{5}-\dfrac{1}{3}\)

=> \(x\) x \(\dfrac{5}{9}\)= \(\dfrac{12}{15}-\dfrac{5}{15}=>x\) x \(\dfrac{5}{9}\)= \(\dfrac{7}{15}\)

=> x = \(\dfrac{7}{15}:\dfrac{5}{9}\)

=> x = \(\dfrac{21}{25}\)

a,x + 3/7 = 2/5 : 18/35

   x + 3/7 = 7/9

   x          = 7/9 - 3/7

   x          = 22/63

vậy x = ...

b, X x 5/9 = 4/5 - 1/3

    X x 5/9 = 7/15

            X  = 7/15 : 5/9

            X  = 21/25

vậy X = ...

\(a.x+\dfrac{3}{7}=\dfrac{2}{5}:\dfrac{18}{35}\\x+\dfrac{3}{7}=\dfrac{2}{5}\times\dfrac{35}{18} \\ x+\dfrac{3}{7}=\dfrac{7}{9}\\ x=\dfrac{7}{9}-\dfrac{3}{7}\\ x=\dfrac{22}{63}\)

\(b.x\times\dfrac{5}{9}=\dfrac{4}{5}-\dfrac{1}{3}\\x\times\dfrac{5}{9}=\dfrac{7}{15}\\ x=\dfrac{7}{15}:\dfrac{5}{9}\\ x= \dfrac{21}{25}\)

1) 52-|x-3|=80

<=> |x-3|=28

<=> x-3=28 hoặc x-3=-28

<=> x=31 hoặc x=-25

Đáp số x= 31 hoặc x=-25

2)  x*(x+2)=0

<=> x=0 hoặc x+2=0

<=> x=0 hoặc x=-2

vậy .......

b: \(\dfrac{x^2}{6}=\dfrac{24}{25}\)

nên \(x^2=\dfrac{144}{25}\)

=>x=12/5 hoặc x=-12/5

c: \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\)

\(\Leftrightarrow x^8-x^7=0\)

hay \(x\in\left\{0;1\right\}\)

d: \(x^{12}=25x^{10}\)

\(\Leftrightarrow x^{10}\left(x^2-25\right)=0\)

hay \(x\in\left\{0;5;-5\right\}\)