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8 tháng 12 2022
a: \(G=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b: Khi x=0,16 thì \(G=-0,4\left(0,4-1\right)=-0,4\cdot\left(-0,6\right)=0,24\)
9 tháng 7 2022
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
1 tháng 9 2018
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 : \(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}=\sqrt{\dfrac{49}{36}.\dfrac{100}{49}.\dfrac{25}{9}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{100}}{\sqrt{49}}.\dfrac{\sqrt{25}}{\sqrt{9}}=\dfrac{7}{6}.\dfrac{10}{7}.\dfrac{5}{3}=\dfrac{25}{9}\)
1 tháng 9 2018
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 : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
8 tháng 12 2022
a: ĐKXĐ: x>0; x<>1
\(M=\dfrac{x-\sqrt{x}-x-\sqrt{x}-1}{x-1}\cdot\dfrac{x}{2\sqrt{x}+1}\)
\(=\dfrac{-\left(2\sqrt{x}+1\right)}{x-1}\cdot\dfrac{x}{2\sqrt{x}+1}=\dfrac{-x}{x-1}\)
b: Khi \(x=\dfrac{\sqrt{3}-1}{2}\) thì \(M=\dfrac{-\sqrt{3}+1}{2}:\dfrac{-\sqrt{3}+1-2}{2}\)
\(=\dfrac{-\sqrt{3}+1}{-1-\sqrt{3}}=2-\sqrt{3}\)
20 tháng 8 2022
Câu 3:
\(C=\dfrac{3\sqrt{x}-x+x+9}{9-x}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Để C<-1 thì C+1<0
=>-3 căn x+2 căn x+4<0
=>-căn x<-4
=>x>16
23 tháng 6 2019
a,
\(\sqrt{0,0004}=0.02\)
\(\sqrt{\frac{16}{81}}=\frac{\sqrt{16}}{\sqrt{81}}=\frac{4}{9}\)
\(\sqrt{25}=5\)
\(\sqrt{0,16}=0,4\)
b,\(\sqrt{\frac{9}{16}}+\sqrt{\frac{25}{9}}\)
= \(\frac{\sqrt{9}}{\sqrt{16}}+\frac{\sqrt{25}}{\sqrt{9}}\)
= \(\frac{3}{4}+\frac{5}{3}\)
=\(\frac{29}{12}\)
31 tháng 8 2022
a: ĐKXĐ: x>=0; x<>1
b: \(G=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
c: KHi x=0,16 thì \(G=-\dfrac{2}{5}\cdot\left(\dfrac{2}{5}-1\right)=\dfrac{6}{25}\)
23 tháng 5 2022
a; ĐKXĐ: x>=0; x<>1
\(P=\dfrac{x+\sqrt{x}-6\sqrt{x}+4+3\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b: Thay \(x=3-\sqrt{8}\) vào P, ta được:
\(P=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1+1}=\dfrac{\sqrt{2}-2}{\sqrt{2}}=1-\sqrt{2}\)
16 tháng 8 2018
a , Thu gọn :
\(Q=\left(\dfrac{1}{x-\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{x-\sqrt{2}+1}{\sqrt{x}-1}\)
\(Q=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right].\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\)
\(Q=\dfrac{-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)\)
\(Q=\dfrac{-1.\left(\sqrt{x}-1\right)}{\sqrt{x}}\)
\(Q=\dfrac{1-\sqrt{x}}{\sqrt{x}}\)
b , Với x = 9 ta có :
\(Q=\dfrac{1-\sqrt{9}}{\sqrt{9}}=\dfrac{1-3}{3}=-\dfrac{2}{3}\)
a