Câu 1 :
\(\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right)\sqrt{6}=\dfrac{\sqrt{18}}{\sqrt{2}}-\dfrac{\sqrt{12}}{\sqrt{3}}=3-2=1\)
b, y = ax + b đi qua A(2;3)
<=> 3 = 2a + b (1)
y = ax + b đi qua b(-2;1)
<=> -2a + b = 1 (2)
Từ (1) ; (2) ta được \(\left\{{}\begin{matrix}2a+b=3\\-2a+b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=2\\a=\dfrac{1}{2}\end{matrix}\right.\)
Câu 2 :
a, \(x^2-3x+1=0\Leftrightarrow x^2-\dfrac{2.3}{2}x+\dfrac{9}{4}-\dfrac{5}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}=0\Leftrightarrow\left(x-\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)\left(x-\dfrac{3}{2}+\dfrac{\sqrt{5}}{2}\right)=0\)
\(\Leftrightarrow x=\dfrac{3+\sqrt{5}}{2};x=\dfrac{3-\sqrt{5}}{2}\)
b, đk x khác 1 ; -1
\(\dfrac{x}{x-1}-\dfrac{2}{x+1}=\dfrac{4}{x^2-1}\Rightarrow x^2+x-2x+2=4\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow x=-1;x=2\)
