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

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

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

\(\Leftrightarrow\left|x-\dfrac{1}{3}\right|=\dfrac{11}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{-11}{6}\\x-\dfrac{1}{3}=\dfrac{11}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11}{6}+\dfrac{1}{3}\\x=\dfrac{11}{6}+\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{13}{6}\end{matrix}\right.\)

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

\(=>|x-\dfrac{1}{3}|+\dfrac{2}{3}=\dfrac{5}{2}\)

\(=>|x-\dfrac{1}{3}|=\dfrac{5}{2}-\dfrac{2}{3}=\dfrac{11}{6}\)

\(=>\) $\left[\begin{matrix} x-\dfrac{1}{3}=\dfrac{11}{6}\\ x-\dfrac{1}{3}=\dfrac{-11}{6}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=\dfrac{13}{6}\\ x=\dfrac{-3}{2}\end{matrix}\right.$

\(\left(a^5\right)^7=a^{5.7}=\text{a}^{35}\)

\(\left(a^3\right)^4.a^9=a^{12}.a^9=a^{12+9}=a^{21}\)

\(\left(2^3\right)^5.\left(2^3\right)^4=2^{15}.2^{12}=2^{15+12}=2^{27}\)

( a⁵ ) ⁷ = 5^{5 + 7} = 5^{1 2}

( a ³ ) ⁴ . a⁹ = a^{3 + 4} . a⁹ = a⁷ . a⁹ = a ^{7 +9} = a^{1 6} 

(2³ ) ⁵ . ( 2³ ) ⁴ = 2^{3 + 5} . 2 ^{3 +4} = 2⁸ . 2⁷ = 2^{8 + 7} = 2^{1 5}

a) \(1\dfrac{1}{5}+\dfrac{4}{5}\div x=0,75\)

\(\Leftrightarrow\dfrac{6}{5}+\dfrac{4}{5}\div x=0,75\)

\(\Leftrightarrow\dfrac{4}{5}\div x=\dfrac{-9}{20}\)

\(\Leftrightarrow x=\dfrac{4}{5}\div\dfrac{-9}{20}\)

\(\Leftrightarrow x=-\dfrac{16}{9}\)

b) \(\left|x-\dfrac{1}{3}\right|+\dfrac{2}{3}=2\dfrac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{11}{6}\\x-\dfrac{1}{3}=\dfrac{-11}{6}\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

đoạn ống nước mới dài số m là:

    0,8+1,35 -\(\dfrac{2}{25}\)= 0,8+ 1,35- 0,08=2.07

Độ dài đoạn ống nước mới là:

Đáp số: $2,07m$

Ta có:

\(\dfrac{a-b}{a+b}=\dfrac{b-c}{b+c}\)

\(=>(a-b)(b+c)=(b-c)(a+b)\)

\(=>ab+ac-b^2-bc=ab+b^2-ac-bc\)

\(=>ab-ab+ac+ac=b^2+b^2-bc+bc\)

\(=>2ac=2b^2\)

\(=>ac=b^2\)

Hay \(b^2=ac\)

     `->Đpcm`

mong mn giúp đỡ

Lời giải:

$2^2S=1-\frac{1}{2^2}+\frac{1}{2^4}-.......+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}$

$S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-.....+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}$

$4S+S=1-\frac{1}{2^{2004}}<1$

$5S<1$

$S< \frac{1}{5}=0,2$ (đpcm)

\(a)5 \times \dfrac{7}{5}=7\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(b)\dfrac{3}{4} \times \dfrac{7}{9}+\dfrac{1}{4} \times \dfrac{7}{9}\)

\(=\dfrac{7}{9} \times (\dfrac{3}{4}+\dfrac{1}{4})\)

\(=\dfrac{7}{9} \times \dfrac{4}{4}=\dfrac{7}{9}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(c)\dfrac{1}{7} \times \dfrac{5}{9}+\dfrac{5}{9} \times \dfrac{1}{7}+\dfrac{5}{9} \times \dfrac{3}{7}\)

\(=\dfrac{5}{9} \times (\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\)

\(=\dfrac{5}{9} \times \dfrac{5}{7}=\dfrac{25}{63}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(d)\dfrac{4 \times 11 \times 3}{4 \times 9 \times 121}\)

\(=\dfrac{4 \times 3 \times 11}{4 \times 3 \times 3 times 11 \times 11}\)

\(=\dfrac{1}{3 \times 11}=\dfrac{1}{33}\)

a, 5 x 7/5 = 7

b, 3/4 x 7/9 + 1/4 x 7/9

= 7/9 x ( 3/4 + 1/4 )

= 7/9 x 1

= 7/9

c, 1/7 x 5/9 + 5/9 x 1/7 + 5/9 x 3/7

= 1/7x ( 5/9+ 5/9) + 5/9 x 3/7

= 1/7 x 25/81 + 5/21

= 25/567 + 5/21

= 160/567

d, 4x11x3/4 x 9/121

= 44 x 3/4 x 9/121

= 33 x 9/121

= 27/11