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\(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(\Leftrightarrow P=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2-8\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-x-3-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(\Leftrightarrow P=\left(\dfrac{\left(\sqrt{x}+1+\sqrt{x}-1\right)\left(\sqrt{x}+1-\sqrt{x}+1\right)-8\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{-x-4}\right)\)
\(\Leftrightarrow P=\dfrac{-4\sqrt{x}}{-x-4}=\dfrac{4\sqrt{x}}{x+4}\)
Thay x = \(3+2\sqrt{2}\) ta được :
\(P=\dfrac{4\sqrt{3+2\sqrt{2}}}{3+2\sqrt{2}+4}=\dfrac{4\left(\sqrt{2}+1\right)}{7+2\sqrt{2}}=\dfrac{4\sqrt{2}+4}{7+2\sqrt{2}}\)
\(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-8\sqrt{x}}{x-1}:\dfrac{\sqrt{x}-x-3-\sqrt{x}-1}{x-1}=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{x-1}{-x-4}=\dfrac{4\sqrt{x}}{x+4}\left(x\ne4;x\ge0;x\ne1\right)\)
Ta có : \(x=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\left(TMĐKXĐ\right)\)
\(P=\dfrac{4\left(\sqrt{2}+1\right)}{3+2\sqrt{2}+4}=\dfrac{4+4\sqrt{2}}{7+2\sqrt{2}}\)
Đặt \(\sqrt{x-1}=a\), khi đó ta có:
\(P=\left(\dfrac{\sqrt{x-1}}{3+\sqrt{x-1}}+\dfrac{x+8}{10-x}\right):\left(\dfrac{3\sqrt{x-1}+1}{x-3\sqrt{x-1}-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
\(=\left[\dfrac{\sqrt{x-1}}{\sqrt{x-1}+3}+\dfrac{\left(x-1\right)+9}{9-\left(x-1\right)}\right]:\left[\dfrac{3\sqrt{x-1}+1}{\left(x-1\right)-3\sqrt{x-1}}-\dfrac{1}{\sqrt{x-1}}\right]\)
\(=\left(\dfrac{a}{a+3}+\dfrac{a^2+9}{9-a^2}\right):\left(\dfrac{3a+1}{a^2-3a}-\dfrac{1}{a}\right)\)
\(=\dfrac{a\left(3-a\right)+\left(a^2+9\right)}{\left(3+a\right)\left(3-a\right)}:\dfrac{\left(3a-1\right)-\left(a-3\right)}{a\left(a-3\right)}\)
\(=\dfrac{3a-a^2+a^2+9}{\left(3+a\right)\left(3-a\right)}:\dfrac{3a-1-a+3}{a\left(a-3\right)}\)
\(=\dfrac{3a+9}{\left(3+a\right)\left(3-a\right)}:\dfrac{2a+4}{a\left(a-3\right)}\)
\(=\dfrac{3\left(a+3\right)}{\left(a+3\right)\left(a-3\right)}.\dfrac{a\left(a-3\right)}{2\left(a+2\right)}\)
\(=\dfrac{-3a}{2\left(a+2\right)}\).
Suy ra: P \(=\dfrac{-3\sqrt{x-1}}{2\left(\sqrt{x-1}+2\right)}\).
Ta lại có: \(x=\sqrt[4]{\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}}-\sqrt[4]{\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}}\)
\(=\sqrt[4]{\dfrac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)^2}}-\sqrt[4]{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}-1}}-\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}-1}}\)
\(=\sqrt{\dfrac{\left(\sqrt{2}+1\right)^2}{2-1}}-\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{2-1}}\)
\(=\left(\sqrt{2}+1\right)-\left(\sqrt{2}-1\right)\)
\(=2\).
Suy ra: \(P=\dfrac{-3\sqrt{2-1}}{2\left(\sqrt{2-1}+2\right)}=\dfrac{-3}{2.3}=-\dfrac{1}{2}\).
a: \(A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}\cdot\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+2\sqrt{x}+2\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\)
\(=\left(x-\sqrt{x}\right)\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\)
\(=2x\sqrt{x}+x-2x-\sqrt{x}+2\sqrt{x}+2\)
\(=2x\sqrt{x}-x+\sqrt{x}+2\)
b: \(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{-4}=-\sqrt{x}+1\)
c: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-3x+8\sqrt{x}+5-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}+8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(a.\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)=\dfrac{x+1+\sqrt{x}}{x\sqrt{x}-1}.\dfrac{x\sqrt{x}+1-\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}=\dfrac{1}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)^2=\sqrt{x}-1\)
\(b.ĐK:x>2\) ( thường là những bài rút gọn sẽ kèm theo ĐK nhé , mình thêm như vậy , nếu không bạn chia TH ra )
\(\dfrac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{\dfrac{1}{x^2}-\dfrac{2}{x}+1}}=\dfrac{\sqrt{x-1}-1+\sqrt{x-1}+1}{1-\dfrac{1}{x}}=\dfrac{2\sqrt{x-1}}{1-\dfrac{1}{x}}\)
\(c.\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{x}-\sqrt{y}+2\sqrt{y}}{\sqrt{x}+\sqrt{y}}=1\)
\(d.Tuong-tự\)
bạnn giải giúp mik lun câu d lun nha?!:)))))cảm ơn nhiw!:))))))
a) Ta có:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x-3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{x-9}\right):\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\left(\frac{2x-6}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3x+3}{x-9}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2x-6+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\)
b) \(P< \frac{-1}{2}\Rightarrow\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}< \frac{-1}{2}\)
.....Chưa nghĩ ra....
c) Ta có: \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-3=0\Rightarrow x=9\)
Vậy Min P = 0 khi x =9.
k - kb với tớ nhia mn!
a: \(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{x-1}=\dfrac{-2\left(\sqrt{x}-1\right)}{x-1}=\dfrac{-2}{\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}+\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}}{1-xy}:\left(\dfrac{x+y+2xy+1-xy}{1-xy}\right)\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{x+y+xy+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
c: \(=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{x-1}:\dfrac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}=\dfrac{x+2\sqrt{x}}{x-1}.\dfrac{x\left(\sqrt{x}+1\right)}{x+2\sqrt{x}}=\dfrac{x}{\sqrt{x}-1}\) ( x > 0 ; x # 1 )
Ta có : \(x=\dfrac{2}{2-\sqrt{3}}-2\sqrt{3}=\dfrac{2-4\sqrt{3}+6}{2-\sqrt{3}}=\dfrac{8-4\sqrt{3}}{2-\sqrt{3}}=\dfrac{4\left(2-\sqrt{3}\right)}{2-\sqrt{3}}=4\left(TMĐKXĐ\right)\)Khi đó : \(P=\dfrac{x}{\sqrt{x}-1}=\dfrac{4}{2-1}=4\)