Cho \(A=\frac{x-1}{x+3}\&B=\frac{1}{x+3}+\frac{x}{x-1}-\frac{4x}{x^2+2x-3}\left(x\ge0;x\ne1\right)\)
a, Rút gọn B
Tìm x để \(\frac{A-1}{B}\le\frac{-1}{2}\)
K
Khách
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1 tháng 11 2020
a, \(\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\frac{-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):(\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}+3\right)})\)
\(=\frac{-3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{2\sqrt{x}+4}\)
\(=\frac{-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
câu b mình quên cách tính rồi sorry
7 tháng 7 2018
mk làm luôn
a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right).\)
=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)
=\(\frac{\left(3x+3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right).3}\)
=\(\frac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)
=
HH
6 tháng 7 2018
a/ \(A=\frac{\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)-\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}+1-3}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{3x-4\sqrt{x}+1-3\sqrt{x}-1}{\left(3\sqrt{x}\right)^2-1}-\frac{8\sqrt{x}}{9x-1}}{1-1-\frac{3}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{3x-7\sqrt{x}}{9x-1}-\frac{8\sqrt{x}}{9x-1}}{-\frac{3}{3\sqrt{x}+1}}\)
\(A=\frac{3x-7\sqrt{x}-8\sqrt{x}}{9x-1}\left(\frac{-\left(3\sqrt{x}+1\right)}{3}\right)\)
\(A=\frac{3x-15\sqrt{x}}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)
\(A=\frac{3\left(x-3\sqrt{x}\right)}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)
\(A=\frac{\left(x-3\sqrt{x}\right)\left(-3\sqrt{x}-1\right)}{9x-1}\)
\(A=\frac{3x\sqrt{x}+8x+3\sqrt{x}}{9x-1}\)
\(A=\frac{3x\sqrt{x}}{9x-1}+\frac{8x}{9x-1}+\frac{3\sqrt{x}}{9x-1}\)
\(A=\frac{x\sqrt{x}}{x-\frac{1}{3}}+\frac{8x}{9x-1}+\frac{\sqrt{x}}{x-\frac{1}{3}}\)
\(A=\frac{\sqrt{x}\left(x-1\right)}{x-\frac{1}{3}}+\frac{\frac{8}{3}x}{x-\frac{1}{3}}\)
\(A=\frac{\sqrt{x}\left(x-1\right)+\frac{8}{3}x}{x-\frac{1}{3}}\)
a) \(B=\frac{1}{x+3}+\frac{x}{x-1}-\frac{4x}{x^2+2x-3}=\frac{x-1}{x^2+2x-3}+\frac{x^2+3x}{x^2+2x-3}-\frac{4x}{x^2+2x-3}\)
\(\Leftrightarrow B=\frac{x-1+x^2+3x-4x}{x^2+2x-3}=\frac{x^2-1}{x^2+2x+1-4}=\frac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)^2-2^2}\)
\(\Leftrightarrow B=\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{x+1}{x+3}\)
b) \(\frac{A-1}{B}=\frac{\frac{x-1}{x+3}-1}{\frac{x+1}{x+3}}=\frac{\frac{-4}{x+3}}{\frac{x+1}{x+3}}=\frac{-4}{x+1}\le\frac{1}{2}\Leftrightarrow-8\le x+1\Leftrightarrow x\ge-9\)