Câu 1:
a: \(\sqrt{64}-3\cdot\sqrt{25}-\sqrt{100}\)
\(=8-3\cdot5-10\)
=8-25
=-17
b: \(\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}+4}{x-4}\)
\(=\dfrac{3\left(\sqrt{x}+2\right)-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{x-4}{\sqrt{x}+4}\)
\(=\dfrac{3\sqrt{x}+6-\sqrt{x}+2}{\sqrt{x}+4}=\dfrac{2\sqrt{x}+8}{\sqrt{x}+4}=2\)
c: Thay x=0 và y=-3 vào y=ax+b, ta được:
\(a\cdot0+b=-3\)
=>b=-3
Vậy: y=ax-3
Thay x=1 và y=2 vào y=ax-3, ta được:
\(a\cdot1-3=2\)
=>a-3=2
=>a=5
Câu 2:
a: \(4x^2-7x+1=0\)
\(\text{Δ}=\left(-7\right)^2-4\cdot4\cdot1=49-16=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{33}}{2\cdot4}=\dfrac{7-\sqrt{33}}{8}\\x_2=\dfrac{7+\sqrt{33}}{8}\end{matrix}\right.\)
b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=3\\x_1x_2=\dfrac{c}{a}=-5\end{matrix}\right.\)
\(T=\dfrac{x_1-2}{x_2}+\dfrac{x_2-2}{x_1}\)
\(=\dfrac{x_1^2-2x_1+x_2^2-2x_2}{x_1x_2}\)
\(=\dfrac{\left(x_1^2+x_2^2\right)-2\left(x_1+x_2\right)}{x_1x_2}\)
\(=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-2\left(x_1+x_2\right)}{x_1x_2}\)
\(=\dfrac{3^2-2\cdot\left(-5\right)-2\cdot3}{-5}=\dfrac{9+10-6}{-5}=\dfrac{13}{-5}\)
