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Bài 3:

a: \(S=1+5^2+5^4+\cdots+5^{200}\)

=>25S=\(5^2+5^4+5^6+\cdots+5^{202}\)

=>25S-S=\(5^2+5^4+\cdots+5^{202}-1-5^2-\cdots-5^{200}\)

=>24S=\(5^{202}-1\)

=>\(S=\frac{5^{202}-1}{24}\)

b: \(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}\cdot2^{30}=8^{10}\cdot4^{15}\)

\(3\cdot24^{10}=3\cdot3^{10}\cdot8^{10}=8^{10}\cdot3^{11}\)

mà \(4^{15}>3^{11}\)

nên \(4^{30}>3\cdot24^{10}\)

=>\(2^{30}+3^{30}+4^{30}>3\cdot24^{10}\)

Bài 2:

a: |2x-3|>5

=>\(\left[\begin{array}{l}2x-3>5\\ 2x-3<-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x>8\\ 2x<-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ x<-1\end{array}\right.\)

c: |3x-1|<=7

=>-7<=3x-1<=7

=>-6<=3x<=8

=>\(-2\le x\le\frac83\)

d: \(\left|3x-5\right|+\left|2x+3\right|=7\) (1)

TH1: \(x<-\frac32\)

=>2x+3<0; 3x-5<0

(1) sẽ trở thành: -2x-3-3x+5=7

=>-5x+2=7

=>-5x=5

=>x=-1(loại)

TH2: -3/2<=x<5/3

=>2x+3>=0; 3x-5<0

(1) sẽ trở thành: 2x+3-3x+5=7

=>-x+8=7

=>-x=-1

=>x=-1(nhận)

TH3: x>=5/3

=>2x+3>0; 3x-5>=0

(1) sẽ trở thành: 2x+3+3x-5=7

=>5x-2=7

=>5x=9

=>x=9/5(nhận)

a)\(\frac{X}{5}=\frac{5}{6}+\frac{-19}{30}\)

   \(\frac{X}{5}=\frac{1}{5}\)

Vậy \(X=1\)

b)\(X-\frac{41}{5}=\frac{-2}{3}\)

                   \(X=\frac{-2}{3}+\frac{41}{5}\)

                   \(X=\frac{113}{15}\)

Vậy \(X=\frac{113}{15}\)

c)\(\frac{31}{5}-X=\frac{11}{3}+\frac{7}{10}\)

    \(\frac{31}{5}-X=\frac{131}{30}\)

                   \(X=\frac{31}{5}-\frac{131}{30}\)

                    \(X=\frac{11}{6}\)

Vậy \(X=\frac{11}{6}\)

d)\(\frac{9}{X}=\frac{2}{5}+\frac{-7}{20}\)

   \(\frac{9}{X}=\frac{1}{20}\)

     \(X=9:\frac{1}{20}\)

     \(X=180\)

Vậy \(X=180\)

Hc tốt

a,x=9/10

b,x=16/3

c,1/7

d,1205/741

e,313/175

bạn ơi bạn bị sai mất 2 câu cuối rồi đó làm ơn tính lại giùm mình nha

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)

\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{1}{9}\)

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

\(\Leftrightarrow\dfrac{4}{3}x=2\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...

7/5 x 5/8 x30/7x16x1/1000 = 3/50

1/2 + 3/5 +1/9 +1/217+7/18+4/35-2/7= 311/21

các bạn nhớ diễn giải ra nhá

\(a,\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=\dfrac{68}{7}-\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=9\)

\(\Leftrightarrow x-\dfrac{1}{3}=3\)

\(\Leftrightarrow x=3+\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{9}{3}+\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

\(b,x+30\%x=-1,31\)

\(\Leftrightarrow x+\dfrac{3}{10}.x=-\dfrac{131}{100}\)

\(\Leftrightarrow x.\left(1+\dfrac{3}{10}\right)=-\dfrac{131}{100}\)

\(\Leftrightarrow x.\dfrac{13}{10}=-\dfrac{131}{100}\)

\(\Leftrightarrow x=-\dfrac{131}{100}.\dfrac{10}{13}\)

\(\Leftrightarrow x=-\dfrac{131}{130}\)

\(c,-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{2}{10}\)

\(\Leftrightarrow-\dfrac{2}{3}x=-\dfrac{1}{10}\)

\(\Leftrightarrow x=-\dfrac{1}{10}.\left(-\dfrac{3}{2}\right)\)

\(\Leftrightarrow x=\dfrac{3}{20}\)