Bài 1:
1: \(2\sqrt{10}=\sqrt{10\cdot4}=\sqrt{40}\)
\(5\sqrt{2}=\sqrt{25\cdot2}=\sqrt{50}\)
mà 40<50
nên \(2\sqrt{10}< 5\sqrt{2}\)
2: \(\dfrac{\sqrt{a}-a}{1-\sqrt{a}}\)
\(=\dfrac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=\sqrt{a}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}=\dfrac{a+\sqrt{a}}{1+\sqrt{a}}\)
=>ĐPCM
3: Phương trình hoành độ giao điểm là:
\(-2x^2=4x+b\)
=>\(2x^2+4x+b=0\)
\(\text{Δ}=4^2-4\cdot2\cdot b=16-8b\)
Để (P) tiếp xúc với (d) thì Δ=0
=>16-8b=0
=>b=2
Khi b=2 thì (d): y=4x+2
Phương trình hoành độ giao điểm là:
\(-2x^2=4x+2\)
=>\(2x^2+4x+2=0\)
=>\(\left(x+1\right)^2=0\)
=>x+1=0
=>x=-1
Khi x=-1 thì \(y=4\cdot\left(-1\right)+2=-4+2=-2\)
Vậy: Tọa độ tiếp điểm là A(-1;-2)
Bài 2:
1:
a: \(\left\{{}\begin{matrix}2x+y=6\\x-3y=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x+3y=18\\x-3y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=28\\x-3y=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=4\\3y=x-10=4-10=-6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=4\\y=-2\end{matrix}\right.\)
b: \(x^4-11x^2-12=0\)
=>\(x^4-12x^2+x^2-12=0\)
=>\(\left(x^2-12\right)\left(x^2+1\right)=0\)
=>\(x^2-12=0\)
=>\(x^2=12\)
=>\(x=\pm2\sqrt{3}\)
2: \(\text{Δ}=\left(2m-5\right)^2-4\cdot2\cdot\left(m-3\right)\)
\(=4m^2-20m+25-8m+24\)
\(=4m^2-28m+49=\left(2m-7\right)^2>=0\forall m\)
=>Phương trình luôn có hai nghiệm
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-2m+5}{2}\\x_1x_2=\dfrac{c}{a}=\dfrac{m-3}{2}\end{matrix}\right.\)
\(2x_1-2x_2=5\)
=>\(x_1-x_2=\dfrac{5}{2}\)
=>\(\left(x_1-x_2\right)^2=\dfrac{25}{4}\)
=>\(\left(x_1+x_2\right)^2-4x_1x_2=\dfrac{25}{4}\)
=>\(\left(\dfrac{-2m+5}{2}\right)^2-4\cdot\dfrac{m-3}{2}=\dfrac{25}{4}\)
=>\(\dfrac{1}{4}\left(4m^2-20m+25\right)-2\left(m-3\right)-\dfrac{25}{4}=0\)
=>\(m^2-5m+\dfrac{25}{4}-2m+6-\dfrac{25}{4}=0\)
=>\(m^2-7m+6=0\)
=>(m-1)(m-6)=0
=>\(\left[{}\begin{matrix}m=1\\m=6\end{matrix}\right.\)
