Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a.x=3-2\sqrt{2}\\ \Rightarrow\sqrt{x}=\sqrt{3-2\sqrt{2}}\\ =\sqrt{2-2\sqrt{2}+1}\\ =\sqrt{\left(\sqrt{2}-1\right)^2}\\ =\left|\sqrt{2}-1\right|\\ =\sqrt{2}-1\left(vì\sqrt{2}>1\right)\)
Thay \(\sqrt{x}=\sqrt{2}-1\) vào A ta được
\(A=\dfrac{\sqrt{2}-1}{1+\sqrt{2}-1}=\dfrac{\sqrt{2}-1}{\sqrt{2}}=\dfrac{\sqrt{2}-2}{2}\)
\(b.B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}-\dfrac{10-5\sqrt{x}}{x-5\sqrt{x}+6}\\ B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{10-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{10-5\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ B=\dfrac{x-3\sqrt{x}-\sqrt{x}+3-x+4-10+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{1}{\sqrt{x}-2}\)
\(c,P=A:B\\ P=\dfrac{\sqrt{x}}{1+\sqrt{x}}:\dfrac{1}{\sqrt{x}-2}\\ P=\dfrac{x-2\sqrt{x}}{1+\sqrt{x}}\)
\(P=\dfrac{-\sqrt{x}\left(-\sqrt{x}+2\right)}{\sqrt{x}+1}\)
Có: \(\sqrt{x}\ge0\)
\(\Rightarrow\sqrt{x}+1\ge1\left(I\right)\)
Lại có: \(\sqrt{x}\ge0\)
\(\Rightarrow-\sqrt{x}\le0\\ \Rightarrow-\sqrt{x}+2\le2\)
mà \(-\sqrt{x}\le0\)
\(\Rightarrow-\sqrt{x}\left(-\sqrt{x}+2\right)\ge2\)
Kết hợp với \(\left(I\right)\) \(\Rightarrow\) \(P=\dfrac{-\sqrt{x}\left(-\sqrt{x}+2\right)}{\sqrt{x}+1}\ge2\)
Vậy gtnn của P = \(2\) khi \(x=10+4\sqrt{6}\)
a: Khi \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\) thì
\(A=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{1+\sqrt{\left(\sqrt{2}-1\right)^2}}=\dfrac{\sqrt{2}-1}{1+\sqrt{2}-1}=\dfrac{\sqrt{2}-1}{\sqrt{2}}=\dfrac{2-\sqrt{2}}{2}\)
b: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}-\dfrac{10-5\sqrt{x}}{x-5\sqrt{x}+6}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}-2}\)
b: Ta có: \(B=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\left(x+\sqrt{x}+1+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)
\(=4-2\sqrt{3}+2\sqrt{3}\)
=4
Thay x=4 vào B, ta được:
\(B=\dfrac{2-4}{2}=-1\)
\(P=A.B=\dfrac{\sqrt{x}}{\sqrt{x}+1}.\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
Ta có : \(\left|P\right|-P=0\) \(\Leftrightarrow\left|P\right|=P\Leftrightarrow\left|\dfrac{\sqrt{x}}{\sqrt{x}-2}\right|=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(+TH_1:x\ge0\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\) (luôn đúng)
\(+TH_2:x< 0\Leftrightarrow-\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(\Leftrightarrow-\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}=0\)
\(\Leftrightarrow-2.\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)=0\)
\(\Leftrightarrow x=0\)
a, Ta có : \(x=9\Rightarrow\sqrt{x}=3\)
Thay vào biểu thức A ta được : \(A=\frac{2}{3-2}=2\)
b, Với \(x\ge0;x\ne4\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{4\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}-2\right)+4\sqrt{x}}{x-4}\)
\(=\frac{x+2\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}=\frac{\sqrt{x}}{\sqrt{x}-2}\)( đpcm )
c, Ta có : \(A+B=\frac{3x}{\sqrt{x}-2}\)hay
\(\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{2+\sqrt{x}}{\sqrt{x}-2}=\frac{3x}{\sqrt{x}-2}\)
\(\Rightarrow2+\sqrt{x}=3x\Leftrightarrow3x-2-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}=3x-2\Leftrightarrow x=9x^2-12x+4\)
\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\Leftrightarrow x=\frac{4}{9}\left(ktm\right);x=1\)( đk : \(x\ge\frac{2}{3}\))
Bạn cần làm gì với 2 biểu thức này?