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1. \(VT=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2+\sqrt{3}-2+\sqrt{3}=VP\)
Bài 1.
Ta có : \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3+4\sqrt{3}+4}-\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|\sqrt{3}-2\right|\)
\(=\sqrt{3}+2-\left(2-\sqrt{3}\right)\)
\(=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\left(đpcm\right)\)
Sửa đề: A>-4
\(A=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x-1+1-\sqrt{x}}{x+\sqrt{x}}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(A+4=\dfrac{x-2\sqrt{x}+1+4\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}>0\)
=>A>-4
@Mai.T.Loan câu a pha cuối hơi tắt đó nhìn khó hiểu lắm
còn câu b kl sai r nha
a: \(T=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=1:\dfrac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: \(T-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>T>3
\(=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+\left(1-\sqrt{x}\right)}{\sqrt{x}+x}\right)\)
\(=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\left(\sqrt{x}-1\right)\sqrt{x}}{\left(\sqrt{x}+1\right)\sqrt{x}}\right)\)
\(=\dfrac{x-1}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{\left(1+\sqrt{x}\right)^2}{\sqrt{x}}\)
ma \(\left(1+\sqrt{x}\right)^2>4\) voi moi x
\(\Rightarrow A>4\)
a,\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(P=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
Vậy \(P=\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có \(x+\sqrt{x}+1=\left(x+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)Suy ra \(\dfrac{2}{x+\sqrt{x}+1}>0\forall x>0,x\ne1\)
hay \(P>0\forall x>0,x\ne1\)(đpcm)
Với x ≥ 0 , x ≠ 1 , x ≠ 4 ta có:
Q = x + 27 . P x + 3 x − 2 = x + 27 x + 3 = x − 9 + 36 x + 3 = x − 3 + 36 x + 3 = − 6 + x + 3 + 36 x + 3 ≥ − 6 + 12 = 6