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a: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
c: Để A=1/2 thì \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{1}{2}\)
=>\(-10\sqrt{x}+4=\sqrt{x}+3\)
=>x=1/121
d: \(A-\dfrac{2}{3}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{2}{3}\)
\(=\dfrac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}< =0\)
=>A<=2/3
1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
Bài 1:
a. ta có \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
=\(\sqrt{xy}\)
b.ĐK: x ≠ 1
Ta có: A= \(\sqrt{\dfrac{x+2\sqrt{x}+1}{x-2\sqrt{x}+1}}\)=\(\sqrt{\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}}\)=\(\dfrac{\sqrt{x}+1}{\left|\sqrt{x}-1\right|}\)
*Nếu \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge1\)
⇒ A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
*Nếu \(\sqrt{x}-1< 0\Rightarrow\sqrt{x}< 1\)
⇒ A=\(\dfrac{\sqrt{x}+1}{-\sqrt{x}+1}\)
c.Ta có:
Q=\(\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)+\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
Q=\(\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6-x+4\sqrt{x}-4+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
Q=\(\dfrac{x\sqrt{x}-2x+2-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)=\(\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
Q=\(\dfrac{x-1}{\sqrt{x}-1}=\sqrt{x}+1\)
\(Q=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{x-3\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{2-\sqrt{x}}=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{x-3\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{x\sqrt{x}-2x-4\sqrt{x}+6-x+4\sqrt{x}-4+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{x\sqrt{x}-2x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{x\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\sqrt{x}+1\)
a: \(P=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
b: ĐểP<15/4 thì P-15/4<0
\(\Leftrightarrow4\left(3\sqrt{x}+8\right)-15\left(\sqrt{x}+2\right)< 0\)
=>12 căn +32-15 căn x+30<0
=>-3 căn x<-62
=>căn x>62/3
=>x>3844/9
Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)
a: \(A=\dfrac{3\sqrt{x}-3-\sqrt{x}-1-\sqrt{x}+3}{x-1}\)
\(=\dfrac{\sqrt{x}-1}{x-1}=\dfrac{1}{\sqrt{x}+1}\)
b: Thay \(x=3-2\sqrt{2}\) vào A,ta được:
\(A=\dfrac{1}{\sqrt{2}-1+1}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
a) \(P=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2x+6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x\sqrt{x}+24\sqrt{x}-3x-22}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(x+25\sqrt{x}+22\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x+25\sqrt{x}+22}{\sqrt{x}+3}\)
b) chả biết nữa, nghĩ ra thì gửi cho
96neko, vì tớ ko biết làm nên chờ ở đây nhá: Câu hỏi của Cold Wind - Toán lớp 9 | Học trực tuyến
Xin cảm ơn!