Thu gọn
\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{11\sqrt{x}-3}{9-x}\)
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3 tháng 8 2021 lúc 10:26
Cho A= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)và B = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)
Thu gọn B và tính M=A+B
\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\left(x\ne\sqrt{3},x\ge0\right)\)
\(B=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\dfrac{x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\dfrac{x+14\sqrt{x}}{x-9}\)
\(M=A+B=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+14\sqrt{x}}{x-9}\)
\(M=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+x+14\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2x-6\sqrt{x}+x+14\sqrt{x}}{x-9}\)
\(M=\dfrac{3x+8\sqrt{x}}{x-9}\)
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7 tháng 6 2021 lúc 10:03
thu gọn giúp mk với ạ
\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)
\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)\(\left(ĐK:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
Chúc bạn học tốt ^.^
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\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11}{9-x}\left(x\ge0,x\ne9\right)\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
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rút gọn biểu thức: \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)và B=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)vs x≥0;x≠9
rút gọn biểu thức M=A+B
Ta có: M=A+B
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)
\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
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7 tháng 6 2021 lúc 9:48
Thu gọn và cho bt tập xác định của biểu thứcA dfrac{2sqrt{x}-1}{sqrt{x}+1}+dfrac{3sqrt{x}-2}{x-sqrt{x}+1}-dfrac{2xsqrt{x}+2sqrt{x-5}}{xsqrt{x}+1}B dfrac{15sqrt{x}-11}{x+2sqrt{x}-3}-dfrac{3sqrt{x}-2}{sqrt{x}-1}-dfrac{2sqrt{x}+3}{sqrt{x}+3}C dfrac{sqrt{x}-1}{sqrt{x}+1}-dfrac{sqrt{x}+3}{sqrt{x}-2}-dfrac{x+5}{x-sqrt{x}-2}D dfrac{sqrt{x}+2}{sqrt{x}-3}-dfrac{sqrt{x}+1}{sqrt{x}-2}-dfrac{3sqrt{x}-1}{x-5sqrt{x}+2}E dfrac{sqrt{x}-3}{sqrt{x}-2}-dfrac{2sqrt{x}-1}{sqrt{x}-1}+dfrac{x-2}{x-3sqrt{x}+2}
Đọc tiếp
Thu gọn và cho bt tập xác định của biểu thức
A= \(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{3\sqrt{x}-2}{x-\sqrt{x}+1}-\dfrac{2x\sqrt{x}+2\sqrt{x-5}}{x\sqrt{x}+1}\)
B= \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
C= \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\)
D= \(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{3\sqrt{x}-1}{x-5\sqrt{x}+2}\)
E= \(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)
a, ĐKXĐ: \(x\ge0,\)
b, ĐKXĐ: \(x\ge0,x\ne1\)
c, ĐKXĐ: \(x\ge0,x\ne4\)
d,ĐKXĐ:\(x\ge0,x\ne9,x\ne4\)
e,ĐKXĐ:\(x\ge0,x\ne1,x\ne4\)
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26 tháng 7 2021 lúc 16:48
Thu gọn
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
Đk:\(x\ge0;x\ne1\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
Vậy...
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Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
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A=\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)và B=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\) với x\(\ge\)0 ,x\(\ne\)9
rút gọn biểu thức M=A+B
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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21 tháng 11 2021 lúc 20:20
Thu gọn P
\(P=\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)
a) Tính P biết \(x=\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
b) Tính P biết \(x=\dfrac{1}{\sqrt{2}-1}-\dfrac{1}{\sqrt{2}+1}\)
24 tháng 7 2021 lúc 21:38
Thu gọn A
A= \(\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}-1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)
Ta có: \(A=\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}-1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\dfrac{x+2\sqrt{x}-7-\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+2\sqrt{x}-7-x-2\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{-4}\)
\(=\dfrac{-4\cdot\left(\sqrt{x}-1\right)}{-4\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
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