\(\frac{180}{v}+\frac{180}{v-5}=\frac{8}{5}\)

\(900\left(v-5\right)+900v=8v\left(v-5\right)\)

\(1800v-4500=8v^2-40v\)

\(1800v-4500-8v^2+40v=0\)

\(1840v-4500-8v^2=0\)

\(\left[{}\begin{matrix}v=-\frac{-1840+\sqrt{1840^2-144000}}{16}\\v=-\frac{-1840-\sqrt{1840^2-144000}}{16}\end{matrix}\right.\)

\(5x=3y=15z\Leftrightarrow\dfrac{5x}{15}=\dfrac{3y}{15}=\dfrac{15z}{15}\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{1}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{1}=\dfrac{x+y+z}{3+5+1}=\dfrac{180}{9}=20\)

\(\Rightarrow\left\{{}\begin{matrix}x=20.3=60\\y=20.5=100\\z=20.1=20\end{matrix}\right.\)

1) Ta có: \(\left(x^2+2x-5\right)^2=\left(x^2-x+5\right)^2.\)

<=> \(\left(x^2+2x-5\right)^2-\left(x^2-x+5\right)^2=0\)

<=> \(\left(3x-10\right)\left(2x^2+x\right)=0\)

<=> \(\left(3x-10\right)\cdot x\cdot\left(2x+1\right)=0\)

TH1: 3x-10=0 <=> x=10/3

TH2: x=0

TH3: 2x+1=0 <=>  x=-1/2

2) Ta có: \(\left(x-5\right)\left(x-6\right)\left(x+2\right)\left(x+3\right)=180\)

<=> \(\left(x-5\right)\left(x+2\right)\cdot\left(x-6\right)\left(x+3\right)=180\)

<=> \(\left(x^2-3x-10\right)\left(x^2-3x-18\right)=180\)

Đặt t = \(x^2-3x-14\)

Ta được pt <=> \(\left(t-4\right)\left(t+4\right)=180\)

<=> \(t^2-16=180\)

<=> \(t^2=196\)<=> \(\orbr{\begin{cases}t=14\\t=-14\end{cases}}\)

TH1: t=14 <=> \(x^2-3x-14=14\)

             <=> \(x^2-3x-28=0\)

            <=>  \(\orbr{\begin{cases}x=-4\\x=7\end{cases}}\)

TH2: t=-14 <=> \(x^2-3x-14=-14\)

<=>  \(x\left(x-3\right)=0\)

<=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

0 < α < 90 => cosα > 0

Ta có: sin²α + cos²α = 1 => cosα = \(\frac{3}{5}\)

90 < β < 180 => cosβ < 0

Ta có: sin²β + cos²β = 1 => cosβ = \(\frac{-15}{17}\)

a = cos(α + β) = cosαcosβ - sinαsinβ = \(\frac{-77}{85}\)

\(x\in\left\{0;40;80;120;160\right\}\)

4=2mũ2

5=5

8=2mu3

TSNT chung và riêng là 2 5 

BCNN (4 5 8 ) =2mu3.5=40 

BC (4 5 8 ) =B (40) ={ 0 40 80 120 160 200 ,... }

Vì-20<x<180 nên x = { 0 40 80 120 160 }

\(x\in\left\{0;40;80;120;160\right\}\)

b: \(\dfrac{3x}{5}=\dfrac{y}{4}\)

nên \(\dfrac{x}{\dfrac{5}{3}}=\dfrac{y}{4}\)

=>x/5=y/12

Đặt x/5=y/12=k

=>x=5k; y=12k

Ta có: xy=180

=>60k^2=180

=>k^2=3

TH1: \(k=\sqrt{3}\)

=>\(x=5\sqrt{3};y=12\sqrt{3}\)

TH2: \(k=-\sqrt{3}\)

=>\(x=-5\sqrt{3};y=-12\sqrt{3}\)