Gọi chiều dài đoạn dây điện kéo từ A→B là: x(km) (0<x<5)

Khi đó chiều dài dây điện kéo từ B→C là 

\(BC=\sqrt{1+\left(5-x\right)^2}=\sqrt{x^2-10x+26}\left(km\right)\)

Tổng số tiền công là \(3\sqrt{x^2-10x+26}+2x=13\)

Theo bài ra ta có:

\(3\sqrt{x^2-10x+26}+2x=13\)

\(\Leftrightarrow9x^2-90x+234=169-52x+4x^2\)

\(\Leftrightarrow5x^2-38x+65=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(l\right)\\x=\dfrac{13}{5}\left(tm\right)\end{matrix}\right.\)

⇒AB=x=\(\dfrac{13}{5}\)

Vậy tổng chiều dài dây điện đã kéo từ A đến C là:

\(AB+BC=\dfrac{26}{5}=5,2\left(km\right)\)

a)  VTPT cua △ la \(\overrightarrow{n1}=\left(3,-4\right)\)

VTPT Cua \(\Delta_1\) la \(\overrightarrow{n2}=\left(12,-5\right)\)

Cosα= \(\dfrac{\left|3.12+\left(-4\right).\left(-5\right)\right|}{\sqrt{3^2+\left(-4\right)^2}.\sqrt{12^2+\left(-5\right)^2}}=\dfrac{56}{65}\)

Vay \(Cos\alpha=\dfrac{56}{65}\)

b) Do duong thang a//△ nen pt a co dang 3x-4y+c=0

Tam (C) la I(-3,2) Ban kinh R=6

Khoang cach tu I den a la d(I,a)=\(\dfrac{\left|3.\left(-3\right)-4.2+c\right|}{\sqrt{3^2+\left(-4\right)^2}}=\dfrac{\left|-17+c\right|}{5}\)

Để a tiếp xúc (C) thì khoảng cách từ I đến a=R

⇒\(\dfrac{\left|-17+c\right|}{5}5=6\)

⇔\(\left|-17+c\right|=30\)

⇔\(\left[{}\begin{matrix}-17+c=30\\-17+c=-30\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}c=47\\c=-13\end{matrix}\right.\)

⇒a: 3x-4y+47=0

⇒a:3x-4y-13=0

a)\(^{-2x^2}\)\(^{+18x+20}\)≥0

⇔\(^{x^2}\)\(^{-9x-10\le0}\)

⇔(x-10)(x+1)

⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10\le0\\x+1\ge0\end{matrix}\right.\\\\\\\left\{{}\begin{matrix}x-10\ge0\\x+1\le0\end{matrix}\right.\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le10\\x\ge-1\end{matrix}\right.\\\\\\\left\{{}\begin{matrix}x\ge10\\x\le-1\end{matrix}\right.\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}-1\le x\le10\\\\\\Loai\end{matrix}\right.\)

 Vay bpt co nghiem \(-1\le x\le10\)

b)\(\sqrt{2x^2+18x+20}=x-2\left(x\ge2\right)\)

⇔\(2x^2-8x+4=x^2-4x+4\)

⇔\(x^2-4x=0\)

⇔\(x\left(x-4\right)=0\)

⇔\(\left[{}\begin{matrix}x=0\left(Loai\right)\\x=4\left(TM\right)\end{matrix}\right.\)

Vay pt co nghiem x=4

a) ta co delta=m2-6m-19

De tam thuc f(x) lon duong voi moi gia tri cua x khi va chi khi a>0=>a=1>0

va delta =x2-6m-19<0<=>0,4<m<5,6

b)binh phuong 2 ve ta co 

2x^2-8x+4=x^2-4x+4<=>x^2-4x=0

<=>x=0 hoac x=4

a)\(\overrightarrow{n_{\Delta}}\)=(3,4)

\(\overrightarrow{n_{\Delta1}}\)=(5,-12)

→\(_{n_{\Delta}}\)=\(\sqrt{3^2+4^2}\)=5

\(_{n_{\Delta1}}\)=\(\sqrt{5^2+12^2}\)=13

→cosα= \(\dfrac{\overrightarrow{n_{\Delta}}\cdot\overrightarrow{n_{\Delta2}}}{|n_{\Delta}\cdot n_{\Delta1}|}=-\dfrac{33}{65}\)

Do rong vien xung quanh la x(cm) (x>0)

Chieu dai hoan chinh la (25+2x) (cm)

Chieu rong hoan chinh la (17+2x) (cm)

Dien tich hoan chinh la (25+2x)*(17+2x)
Ma dien tich khung anh lon nhat la 513 (\(cm^2\)) nen ta co:

(25+2X)*(17+2x)=513

⇔\(4x^2\)+84x-88=0

⇔\(x^2\)+21x-22=0

⇔\(\left[{}\begin{matrix}x=-22\left(vli\right)\\x=1\left(tm\right)\end{matrix}\right.\)