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ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
Sửa đề: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
=\(\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{2-3\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11+\left(2-3\sqrt{x}\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11+2\sqrt{x}+6-3x-9\sqrt{x}-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\left(x\ge0;x\ne1\right)\)
\(P=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2x-2\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{15\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{-\left(\sqrt{x}-1\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
a: Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: Ta có: \(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}-1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}+1+1\)
\(=a-\sqrt{a}+2\)
a,ĐKXĐ: tự tìm :v
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(x+2\sqrt{x}+1\right)-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+1\right)^2-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6+2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{9\sqrt{x}-x-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(9\sqrt{x}-9\right)-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{9\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(10-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\dfrac{10-\sqrt{x}}{\sqrt{x}+3}\)
1:
\(A=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
3: A nguyên
=>-5căn x-15+17 chia hết cho căn x+3
=>căn x+3 thuộc Ư(17)
=>căn x+3=17
=>x=196
a: \(=6+2\sqrt{11}-4+\sqrt{11}=2+3\sqrt{11}\)
b: \(=\dfrac{3x+9\sqrt{x}-2x+4\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-2\sqrt{x}\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}=\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) Thay x=0 vào A, ta được:
\(A=\dfrac{15\cdot\sqrt{0}-11}{0+2\sqrt{0}-3}-\dfrac{3\sqrt{0}-2}{\sqrt{0}-1}-\dfrac{2\sqrt{0}+3}{\sqrt{0}+3}\)
\(=\dfrac{-11}{-3}-\dfrac{-2}{-1}-\dfrac{3}{3}\)
\(=\dfrac{11}{3}-2-1\)
\(=\dfrac{11}{3}-\dfrac{9}{3}=\dfrac{2}{3}\)
Ta có: M=A+B
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)
\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}-2}{\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\) (ĐK: \(x\ne1;x\ge0\))
\(A=\dfrac{15\sqrt{x}-11}{x+3\sqrt{x}-\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(A=\dfrac{15\sqrt{x}-11}{\sqrt{x}\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{\left(15\sqrt{x}-11\right)-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{-\left(5\sqrt{x}-2\right)}{\sqrt{x}+3}\)
\(A=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)