Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-1\right)\left(x+2\right)}=\dfrac{-4x^2+11x-2}{\left(x+2\right)\left(x-1\right)}\)
\(\Leftrightarrow x^2+4x+4+4x^2-11x+2=0\)
\(\Leftrightarrow5x^2-7x+6=0\)
hay \(x\in\varnothing\)
c: \(\Leftrightarrow\left(3x^2+2\right)^2-5x\left(3x^2+2\right)=0\)
=>3x^2-5x+2=0
=>3x^2-3x-2x+2=0
=>(x-1)(3x-2)=0
=>x=2/3 hoặc x=1
a: \(=\sqrt{\dfrac{1}{10}}+\sqrt{\dfrac{1}{60}}-\dfrac{2\sqrt{15}}{15}\)
\(=\dfrac{\sqrt{10}}{10}-\dfrac{2\sqrt{15}}{15}+\dfrac{\sqrt{15}}{30}\)
\(=\dfrac{3\sqrt{10}-3\sqrt{15}}{30}=\dfrac{\sqrt{10}-\sqrt{15}}{10}\)
b: \(=\dfrac{\left(\sqrt{5}+\dfrac{1}{2}\cdot2\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)}{2\sqrt{5}}\)
\(=\dfrac{\left(\sqrt{5}+\sqrt{5}-\dfrac{1}{2}\sqrt{5}+\sqrt{5}\right)}{2\sqrt{5}}\)
\(=\dfrac{5}{2}:2=\dfrac{5}{4}\)
Bài 1 :
Câu a : \(\sqrt{\dfrac{1,44}{3,61}}=\sqrt{\dfrac{144}{361}}=\dfrac{\sqrt{144}}{\sqrt{361}}=\dfrac{12}{19}\)
Câu b : \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{25}{900}}=\dfrac{\sqrt{25}}{\sqrt{900}}=\dfrac{5}{30}=\dfrac{1}{6}\)
Câu c : \(\sqrt{1\dfrac{13}{36}}.\sqrt{3\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}.\sqrt{\dfrac{121}{46}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{121}}{36}=\dfrac{7}{6}.\dfrac{11}{6}=\dfrac{77}{36}\)
Câu d : \(\sqrt{\dfrac{1}{121}.3\dfrac{6}{25}}=\sqrt{\dfrac{1}{121}.\dfrac{81}{25}}=\dfrac{1}{\sqrt{121}}.\dfrac{\sqrt{81}}{\sqrt{25}}=\dfrac{1}{11}.\dfrac{9}{5}=\dfrac{9}{55}\)
Câu e : \(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}=\sqrt{\dfrac{49}{36}.\dfrac{100}{49}.\dfrac{25}{9}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{100}}{\sqrt{49}}.\dfrac{\sqrt{25}}{\sqrt{9}}=\dfrac{7}{6}.\dfrac{10}{7}.\dfrac{5}{3}=\dfrac{25}{9}\)
Bài 2 :
Câu a : \(\dfrac{\sqrt{245}}{\sqrt{5}}=\sqrt{\dfrac{245}{5}}=\sqrt{49}=7\)
Câu b : \(\dfrac{\sqrt{3}}{\sqrt{75}}=\sqrt{\dfrac{3}{75}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
Câu c : \(\dfrac{\sqrt{10,8}}{\sqrt{0,3}}=\sqrt{\dfrac{10,8}{0,3}}=\sqrt{\dfrac{108}{3}}=\sqrt{36}=6\)
Câu d : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\dfrac{18\left(x+7-x-4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)
\(18.3=\left(x+4\right)\left(x+7\right)\)
\(x^2+11x+28-54=0\)
\(x^2+11x-26=0\)
\(\left(x-2\right)\left(x+13\right)=0\)
\(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
Theo đề x < 0 nên x = -13
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
a: \(P=\sqrt{x}\left(\dfrac{\sqrt{x}}{x^2-1}+\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}\right)-\dfrac{5x}{x^2-1}\)
\(=\sqrt{x}\left(\dfrac{\sqrt{x}}{x^2-1}+\dfrac{4\sqrt{x}}{x-1}\right)-\dfrac{5x}{x^2-1}\)
\(=\sqrt{x}\left(\dfrac{\sqrt{x}+4\sqrt{x}\left(x+1\right)}{\left(x^2-1\right)}\right)-\dfrac{5x}{x^2-1}\)
\(=\dfrac{x+4x\left(x+1\right)}{x^2-1}-\dfrac{5x}{x^2-1}\)
\(=\dfrac{x+4x^2+4x-5x}{x^2-1}\)
\(=\dfrac{4x^2}{x^2-1}\)
Khi x=4 thì \(P=\dfrac{4\cdot16}{16-1}=\dfrac{64}{15}\)
b: Để P/Q=0 thì P=0
=>x=0
Đề bài sai: Khi \(x=4\) thì \(A=\dfrac{1}{2};B=\dfrac{28}{9};\dfrac{A}{B}=\dfrac{9}{56};\dfrac{x-2}{4\sqrt{x}}=\dfrac{1}{4}\Rightarrow\dfrac{A}{B}\ne\dfrac{x-2}{4\sqrt{x}}\)
b) \(B=\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-\dfrac{x\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x-\sqrt{x}\right)-x\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\) = \(\dfrac{x\sqrt{x}-x-x\sqrt{x}+x^2}{\sqrt{x}-x}=\dfrac{x^2-x}{\sqrt{x}-x}\)
c) \(C=\dfrac{x+2\sqrt{x}}{\sqrt{x}-x}-\dfrac{x\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)-x\sqrt{x}\left(\sqrt{x}-x\right)}{\left(\sqrt{x}-x\right)\left(\sqrt{x}+1\right)}=x+2\sqrt{x}-x\sqrt{x}\)
\(d,D=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\) \(\dfrac{\left(x+2\sqrt{x}\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+7\sqrt{x}-2}{\sqrt{x}+2}\)
e) \(E=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{2\sqrt{x}-24}{\sqrt{x}+3}\)
F) F = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{23-2\sqrt{x}}{\sqrt{x}+5}\)
\(\dfrac{60}{x}-\dfrac{60}{x+2}=\dfrac{1}{20}\left(đk:x\ne0,x\ne-2\right)\)
\(\Leftrightarrow\dfrac{60x+120-60x}{x\left(x+2\right)}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{120}{x^2+2x}=\dfrac{1}{20}\Leftrightarrow x^2+2x=2400\)
\(\Leftrightarrow\left(x+1\right)^2=2401\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=49\\x+1=-49\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=48\\x=-50\end{matrix}\right.\)(thỏa đk)
Ta có: \(\dfrac{60}{x}-\dfrac{60}{x+2}=\dfrac{1}{20}\)
\(\Leftrightarrow x\left(x+2\right)=1200x+2400-1200x\)
\(\Leftrightarrow x^2+2x-2400=0\)
\(\Delta=2^2-4\cdot1\cdot\left(-2400\right)=9604\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-2-98}{2}=-50\left(nhận\right)\\x_2=\dfrac{-2+98}{2}=48\left(nhận\right)\end{matrix}\right.\)