P=\(\dfrac{3\sqrt{x}+6}{x-4}\) -\(\dfrac{\sqrt{x}}{2-\sqrt{x}}\)
Rút gọn
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( \(\dfrac{3\sqrt{x}+6}{x-4}\) + \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\) ) : \(\dfrac{x-9}{\sqrt{x}-3}\)
rút gọn biểu thức
\(\left(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{x-9}{\sqrt{x}-3}\left(dkxd:x\ne9,x\ne4,x\ge0\right)\)
\(=\left(\dfrac{3\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}-3}\)
\(=\left(\dfrac{3\sqrt{x}+6+\sqrt{x}\left(\sqrt{x}+2\right)}{(\sqrt{x}-2)\left(\sqrt{x}+2\right)}\right).\dfrac{1}{\sqrt{x}+3}\)
\(=\dfrac{3\sqrt{x}+6+x+2\sqrt{x}}{x-4}.\dfrac{1}{\sqrt{x}+3}\)
\(=\dfrac{x+5\sqrt{x}+6}{x-4}.\dfrac{1}{\sqrt{x}+3}\)
\(=\dfrac{x+2\sqrt{x}+3\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{1}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{1}{\sqrt{x}+3}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{1}{\sqrt{x}+3}\)
\(=\dfrac{1}{\sqrt{x}-2}\)
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Câu 1: Rút gọn biểu thức: Bleft(dfrac{x}{x+3sqrt{x}}+dfrac{1}{sqrt{x}+3}right):left(1-dfrac{2}{sqrt{2}}+dfrac{6}{x+3sqrt{x}}right) với x 0Câu 2: Rút gọn biểu thức:Pdfrac{xsqrt{2}}{2sqrt{x}+xsqrt{2}}+dfrac{sqrt{2x}-2}{x-2} với x 0; x ne 2Câu 3: Rút gọn biểu thức:Qleft(dfrac{a}{a-2sqrt{a}}+dfrac{a}{sqrt{a}-2}right):dfrac{sqrt{a}+1}{a-4sqrt{a}+4} với a 0; a ne 4
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Câu 1: Rút gọn biểu thức: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{2}}+\dfrac{6}{x+3\sqrt{x}}\right)\) với x > 0
Câu 2: Rút gọn biểu thức:
\(P=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}\) với x > 0; x \(\ne\) 2
Câu 3: Rút gọn biểu thức:
\(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\) với a > 0; a \(\ne\) 4
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: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
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Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
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1) Rút gọn biểu thức
P=\(\left(\dfrac{x}{x\sqrt{x}-4\sqrt{x}}-\dfrac{6}{3\sqrt{x}-6}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
\(P=\left(\dfrac{x}{x\sqrt{x}-4\sqrt{x}}-\dfrac{6}{3\sqrt{x}-6}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
\(P=\left(\dfrac{\sqrt{x}}{x-4}-\dfrac{2\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}-2}{x-4}\right):\left(\dfrac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(P=\left(\dfrac{-6}{x-4}\right):\left(\dfrac{6}{\sqrt{x}+2}\right)=\dfrac{-1}{\sqrt{x}-2}\)
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Rút gọn các biểu thức sau:
a) \(\dfrac{4}{\sqrt{11}-3}-\dfrac{5}{4+\sqrt{11}}\)
b) \(\left(\dfrac{3\sqrt{x}}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{x+6\sqrt{x}+9}\) với x>0;x\(\ne\)4
a: \(=6+2\sqrt{11}-4+\sqrt{11}=2+3\sqrt{11}\)
b: \(=\dfrac{3x+9\sqrt{x}-2x+4\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-2\sqrt{x}\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}=\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\)
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1.cho biểu thức A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}-\dfrac{1}{\sqrt{x}-2}\)với(x≥0;x≠4)
a)rút gọn A
b)tính A khi x=6+4\(\sqrt{2}\)
2.cho biểu thức P=\(\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}+3\right)\)với x≥0;x≠1;x≠4
a)rút gọn P
b)tìm x để P=-4
Bài 3: Cho biểu thức A(dfrac{sqrt{x}}{2-sqrt{x}} + dfrac{sqrt{x}}{2+sqrt{x}}) - dfrac{6+sqrt{x}}{4-x} 1. Rút gọn A 2. Tìm giá trị nhỏ nhất của A Mình cần gấp
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Bài 3: Cho biểu thức A=(\(\dfrac{\sqrt{x}}{2-\sqrt{x}}\) + \(\dfrac{\sqrt{x}}{2+\sqrt{x}}\)) - \(\dfrac{6+\sqrt{x}}{4-x}\) 1. Rút gọn A 2. Tìm giá trị nhỏ nhất của A Mình cần gấp
1: Ta có: \(A=\left(\dfrac{\sqrt{x}}{2-\sqrt{x}}+\dfrac{\sqrt{x}}{2+\sqrt{x}}\right)-\dfrac{\sqrt{x}+6}{4-x}\)
\(=\dfrac{2\sqrt{x}+x+2\sqrt{x}-x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\dfrac{\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-3}{\sqrt{x}+2}\)
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13 tháng 8 2021 lúc 0:28
Rút gọn:
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\) với x ≠4, x ≠16, x >0
Lời giải:
\(A=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}+3)(\sqrt{x}-2)}-\frac{5}{(\sqrt{x}+3)(\sqrt{x}-2)}-\frac{\sqrt{x}+3}{(\sqrt{x}-2)(\sqrt{x}+3)}\)
\(=\frac{x-4-5-\sqrt{x}-3}{(\sqrt{x}-2)(\sqrt{x}+3)}=\frac{x-\sqrt{x}-12}{(\sqrt{x}-2)(\sqrt{x}+3)}=\frac{(\sqrt{x}+3)(\sqrt{x}-4)}{(\sqrt{x}-2)(\sqrt{x}+3)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
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A=(\(\dfrac{2\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}-1}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\))(\(3\sqrt{x}-\dfrac{\sqrt{x}+4}{\sqrt{x}-1}\))
a,rút gọn A b,tìm x để A<2
a: \(A=\dfrac{2x-6\sqrt{x}+\sqrt{x}-3-2x+4\sqrt{x}+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{3x-3\sqrt{x}-\sqrt{x}-4}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1}{3x-4\sqrt{x}-4}\)
\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{3x-6\sqrt{x}+2\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+2}{\sqrt{x}-1}\)
b: Để A<2 thì \(\dfrac{3\sqrt{x}+2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)}< 0\)
=>x<1
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Rút gọn A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{2\sqrt{x}+6}{x-9}\right):\dfrac{x-2\sqrt{x}}{\sqrt{x}-3}\) với x>0,x khác 4 và 9
\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{2\sqrt{x}+6}{x-9}\right):\dfrac{x-2\sqrt{x}}{\sqrt{x}-3}\left(x>3;x\ne9\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{2\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\dfrac{\sqrt{x}-3}{x-2\sqrt{x}}\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left( \sqrt{x}+3\right)}-\dfrac{2\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\cdot\dfrac{x-2\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{x+3\sqrt{x}-2\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{x-2\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)-2\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\cdot\dfrac{1}{x-2\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{1}{\sqrt{x}}\)
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