Ôn tập phương trình bậc hai một ẩn

\(x^2-3x-1=0\)

Theo vi-ét: \(\left\{{}\begin{matrix}x_1x_2=-1\\x_1+x_2=3\end{matrix}\right.\)

\(A=\dfrac{x_1}{x_1-2x_2}+\dfrac{x_2}{x_2-2x_1}\)

\(=\dfrac{x_1\left(x_2-2x_1\right)}{\left(x_1-2x_2\right)\left(x_2-2x_1\right)}+\dfrac{x_2\left(x_1-2x_2\right)}{\left(x_1-2x_2\right)\left(x_2-2x_1\right)}\)

\(=\dfrac{x_1\left(x_2-2x_1\right)+x_2\left(x_1-2x_2\right)}{\left(x_1-2x_2\right)\left(x_2-2x_1\right)}\)

\(=\dfrac{x_1x_2-2x_1^2+x_1x_2-2x_2^2}{x_1x_2-2x_1^2-2x_2^2+4x_1x_2}\)

\(=\dfrac{2x_1x_2-2\left(x_1^2+x_2^2\right)}{5x_1x_2-2\left(x_1^2+x_2^2\right)}\)

\(=\dfrac{2x_1x_2-2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}{5x_1x_2-2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}\)

\(=\dfrac{2x_1x_2-2\left(x_1+x_2\right)^2+4x_1x_2}{5x_1x_2-2\left(x_1+x_2\right)^2+4x_1x_2}\)

\(=\dfrac{6x_1x_2-2\left(x_1+x_2\right)^2}{9x_1x_2-2\left(x_1+x_2\right)^2}\)

\(=\dfrac{6\cdot\left(-1\right)-2\cdot3^2}{9\cdot\left(-1\right)-2\cdot3^2}\) 

\(=\dfrac{-6-18}{-9-18}\)

\(=\dfrac{8}{9}\)

\(\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=xy-1\\\left(x-3\right)\left(y-3\right)=xy-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy-x+y-1=xy-1\\xy-3x-3y+9=xy-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x+y=0\\-3x-3y=-12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-y=0\\x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\x=y\end{matrix}\right.\)

=>\(x=y=\dfrac{4}{2}=2\)

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

=>\(\left\{{}\begin{matrix}2\left(x+y\right)=x-y\\5x=3y+15\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+2y-x+y=0\\5x-3y=15\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+3y=0\\5x-3y=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y+5x-3y=15\\x+3y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}6x=15\\x+3y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2,5\\3y=-x=-2,5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2,5\\y=-\dfrac{2.5}{3}=-\dfrac{5}{6}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x-4y=3\\2x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-8y=6\\2x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-8y-2x+y=6-4\\2x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-7y=2\\2x=y+4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{2}{7}\\2x=-\dfrac{2}{7}+4=\dfrac{26}{7}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{2}{7}\\x=\dfrac{13}{7}\end{matrix}\right.\)

Pt: \(3x^2-6x-2=0\)

Theo vi-ét: \(\left\{{}\begin{matrix}x_1x_2=\dfrac{-2}{3}\\x_1+x_2=\dfrac{-\left(-6\right)}{3}=2\end{matrix}\right.\) 

\(A=\dfrac{2x_1+1}{x_2}+\dfrac{2x_2+1}{x_1}\)

\(A=\dfrac{2x^2_1+x_1}{x_1x_2}+\dfrac{2x^2_2+x_2}{x_1x_2}\)

\(A=\dfrac{2\left(x^2_1+x_2^2\right)+\left(x_1+x_2\right)}{x_1x_2}\)

\(A=\dfrac{2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+\left(x_1+x_2\right)}{x_1x_2}\)

\(A=\dfrac{2\cdot\left(2^2-2\cdot-\dfrac{2}{3}\right)+2}{-\dfrac{2}{3}}\)

\(A=\dfrac{2\cdot\left(4+\dfrac{4}{3}\right)+2}{-\dfrac{2}{3}}\)

\(A=-19\)

Phương trình đã cho có a = 3; c = -2 nên luôn có 2 nghiệm phân biệt

Theo định lý Vi-ét, ta có:

x₁ + x₂ = 2

x₁x₂ = -2/3

Ta có:

A = (2x₁ + 1)/x₂ + (2x₂ + 1)/x₁ 

= (2x₁² + x₁ + 2x₂² + x₂)/x₁x₂

= 2(x₁² + x₂²)/x₁x₂ + (x₁ + x₂)/x₁x₂

= [2(x₁ + x₂)² - 4x₁x₂]/x₁x₂ + (x₁ + x₂)/x₁x₂

= (2.2² + 4.2/3)/(-2/3) + 2/(-2/3)

= -16 - 3

= -19

Đặt x/y + y/x = a -> x2/y2 + y2/x2 + 2 = a2.

Dễ dàng chứng minh x2/y2 + y2/x2 >= 2 nên a2 >= 4, do đó |a| >= 2 (1).

Bất đẳng thức phải chứng minh tương đương với: a2 - 2 + 4 >= 3a -> a2 - 3a + 2 >= 0 tương đương (a - 1)(a - 2) >= 0 (2)

Từ (1) suy ra a >= 2 ,hoặc a <= -2. Nếu a >= 2 thì (2) đúng. Nếu a <= - 2 thì (2) cũng đúng.

Bài 7:

Gọi AB là bóng của cây cột đèn trên mặt đất, AC là chiều cao của cây cột đèn

Theo đề, ta có: AB=4m; AC=7m; ΔABC vuông tại A

Xét ΔABC vuông tại A có \(tanB=\dfrac{AC}{AB}=\dfrac{7}{4}\)

nên \(\widehat{B}\simeq60^015'\)

Bài 11:

Xét ΔABH vuông tại H có \(tanB=\dfrac{AH}{HB}\)

=>\(HB=\dfrac{AH}{tanB}=\dfrac{1.2}{tan18}\left(m\right)\)

Xét ΔAHC vuông tại H có \(tanC=\dfrac{AH}{HC}\)

=>\(HC=\dfrac{AH}{tanC}=\dfrac{1.2}{tan15}\)(m)

\(BC=BH+CH\)

\(=1.2\left(\dfrac{1}{tan18}+\dfrac{1}{tan15}\right)\)

\(\simeq8,17\left(m\right)=817\left(cm\right)\)

Bài 12:

Kẻ CH\(\perp\)AB tại H

=>CH là độ cao của máy bay so với mặt đất

Xét ΔCAB có \(\widehat{ACB}+\widehat{A}+\widehat{B}=180^0\)

=>\(\widehat{ACB}=180^0-55^0-40^0=85^0\)

Xét ΔCAB có \(\dfrac{AB}{sinACB}=\dfrac{CB}{sinA}=\dfrac{AC}{sinB}\)

=>\(\dfrac{100}{sin85}=\dfrac{CB}{sin55}=\dfrac{AC}{sin40}\)

=>\(CB=100\cdot\dfrac{sin55}{sin85}\simeq82,23\left(m\right);AC=100\cdot\dfrac{sin40}{sin85}\simeq64,52\left(m\right)\)

Diện tích tam giác CAB là:

\(S_{CAB}=\dfrac{1}{2}\cdot CA\cdot CB\cdot sinACB\)

=>\(S_{ACB}=\dfrac{1}{2}\cdot82,23\cdot64,52\cdot sin85\simeq2642,65\left(m^2\right)\)

Ta có:CH là đường cao của ΔCAB

=>\(S_{CAB}=\dfrac{1}{2}\cdot CH\cdot AB\)

=>\(\dfrac{1}{2}\cdot CH\cdot100=2642,65\)

=>\(CH\simeq52,9\left(m\right)\)