Bài 1 :

Đặt :

\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=3k\\3y=4k\\4z=5k\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3k}{2}\\y=\dfrac{4k}{3}\\z=\dfrac{5k}{4}\end{matrix}\right.\)

Thay vào \(x+y+z=49\) ta được :

\(\dfrac{3k}{2}=\dfrac{4k}{3}=\dfrac{5k}{4}=49\)

\(\Leftrightarrow\dfrac{18k+16k+15k}{12}=\dfrac{588}{12}\)

\(\Leftrightarrow49k=588\)

\(\Leftrightarrow k=12\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3.12}{2}=18\\y=\dfrac{4.12}{3}=16\\z=\dfrac{5.12}{4}=15\end{matrix}\right.\)

Vậy ....

Bài1:

Từ \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{x}{90}=\dfrac{y}{80}=\dfrac{z}{75}\)

Áp dụng t/c của dãy tỉ số bằng nhau,ta có:

\(\dfrac{x}{90}=\dfrac{y}{80}=\dfrac{z}{75}=\dfrac{x+y+z}{90+80+75}=\dfrac{49}{245}=\dfrac{1}{5}\)

=>x=18;b=16;c=15

Vậy...

\(\left(-\dfrac{2}{5}\right)^2\cdot\left|\dfrac{1}{3}-\dfrac{3}{5}\right|-\dfrac{2}{5}\cdot\sqrt{\dfrac{1}{25}}+\dfrac{4}{3}\)

\(=\dfrac{4}{25}\cdot\dfrac{4}{15}-\dfrac{2}{5}\cdot\dfrac{1}{5}+\dfrac{4}{3}\)

\(=\dfrac{16}{375}-\dfrac{2}{25}+\dfrac{4}{3}\)

\(=\dfrac{16}{375}-\dfrac{30}{375}+\dfrac{500}{375}\)

\(=\dfrac{486}{375}=\dfrac{162}{125}\)

\(\dfrac{1}{2}A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{2023}\)

\(A-\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^{2023}-1\)

\(\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^{2023}-1\)

\(A=\dfrac{1}{2^{2022}}-2\)

A−12A=(12)2023−1A−12A=(12)2023−1

A=122022−2

`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`

`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`

`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`

`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`

`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`

b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)

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

\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)

a, \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)

\(=\sqrt{2}\left(2\sqrt{2}+4\sqrt{2}-7\sqrt{2}\right)\)

\(=\sqrt{2}.\left(-\sqrt{2}\right)=-2\)

c, \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)

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

\(=\left(2+\sqrt{3}\right)\left(3-\sqrt{2}\right)\)

\(=6-2\sqrt{2}+3\sqrt{3}-\sqrt{6}\)

\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)

\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)

Ta có: \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\cdot...\cdot\left(1-\dfrac{1}{10^2}\right)\)

\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-99}{100}\)

\(=-\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)

\(=-\dfrac{10+1}{2\cdot10}=\dfrac{-11}{20}\)

A=4/63

\(=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}-\dfrac{132}{132}-\dfrac{84}{132}\right)\)

\(=\dfrac{115}{-161}=-\dfrac{115}{161}\)