giups minh voi ; gap aa
câu 1: a) \(x^2+9x+8=0\)
\(\Delta=b^2-4ac=9^2-4\cdot1\cdot8=49>0\\ x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-9-\sqrt{49}}{2\cdot1}=-8\\ x_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-9+\sqrt{49}}{2\cdot1}=-1\)
vậy phương trình có 2 nghiệm là: \(x_1=-8;x_2=-1\)
b) \(\left\{{}\begin{matrix}2x+y=3\left(1\right)\\2\cdot\left(x+1\right)=x-3y+1\left(2\right)\end{matrix}\right.\)
từ (1) ta có: y = 3 - 2x
thay vào (2) ta được: \(2\left(x+1\right)=x-3\left(3-2x\right)+1\)
\(\Rightarrow2x+2=x-9+6x+1\\ \Rightarrow2x-x-6x=-9+1-2\\ \Rightarrow-5x=-10\\ \Rightarrow x=2\\ \Rightarrow y=3-2\cdot2=-1\)
vậy x = 2; y = -1
câu 2: \(A=\dfrac{3+\sqrt{3}}{\sqrt{3}}-\dfrac{1}{2+\sqrt{3}}-\sqrt{12}\)
\(=\left(\sqrt{3}+1\right)-\left(2-\sqrt{3}\right)-2\sqrt{3}\\ =\sqrt{3}+1-2+\sqrt{3}-2\sqrt{3}\\ =1-2+2\sqrt{3}-2\sqrt{3}=-1\)
\(B=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ =\left[\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ =\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ =\dfrac{2}{\sqrt{x}-2}\)
để B đạt giá trị nguyên thì \(\sqrt{x}-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
mà 2 > 0 thì \(\sqrt{x}-2< 0\)
\(\sqrt{x}-2=-1\Rightarrow x=1\\ \sqrt{x}-2=-2\Rightarrow x=0\)
vậy x = 1; 0 thì B có giá trị âm
giups mk voi 
