\(\left(\dfrac{2x}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
rút gọn biểu thức trên
Những câu hỏi liên quan
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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P=\(\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
Rút gọn biểu thức trên
Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\dfrac{3x+3\sqrt{x}}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)
\(=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
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\(\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{9x-1}:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}=\dfrac{3x+3\sqrt{x}-1}{9x-1}.\dfrac{3\sqrt{x}+1}{3}=\dfrac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)
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cho biểu thức P=\(\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a,rút gọn biểu thức
b,tính giá trị của biểu thức với x=3 - \(2\sqrt{2}\)
a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
\(=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-\left(x-1\right)}-\dfrac{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}\right)\cdot\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\left(\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)
b) Ta có: \(x=3-2\sqrt{2}\)
\(=2-2\cdot\sqrt{2}\cdot1+1\)
\(=\left(\sqrt{2}-1\right)^2\)
Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\), ta được:
\(P=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)
\(=\dfrac{\sqrt{2}-\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{2}-\sqrt{2}+1}{\sqrt{2}-1}\)
\(=\dfrac{1}{\sqrt{2}-1}\)
\(=\sqrt{2}+1\)
Vậy: Khi \(x=3-2\sqrt{2}\) thì \(P=\sqrt{2}+1\)
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7 tháng 6 2023 lúc 18:17
cho biểu thức P=\(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right)\):\(\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
1. Rút gọn biểu thức P
2. Tính giá trị của P biết x=\(\sqrt{7+4\sqrt{3}}\)+\(\sqrt{7-4\sqrt{3}}\)
\(1,P=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(dkxd:x\ge0,x\ne9\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}-2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-1-2\sqrt{x}+6}\)
\(=\dfrac{-x-3\sqrt{x}}{\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{-\sqrt{x}+5}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{5-\sqrt{x}}\)
\(=-\dfrac{x}{5-\sqrt{x}}\)
\(2,x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\)
\(=2+\sqrt{3}+2-\sqrt{3}=4\)
\(x=4\Rightarrow P=-\dfrac{4}{5-\sqrt{4}}=\dfrac{-4}{5-2}=-\dfrac{4}{3}\)
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Rút gọn biểu thức
\(\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right).\dfrac{x-6\sqrt{x}+9}{\sqrt{x}}\)
\(\dfrac{x}{\sqrt{x}+1}+\dfrac{2x+\sqrt{x}}{x+\sqrt{x}}\)
\(a,=\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}}\left(x>0;x\ne9\right)\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\\ b,=\dfrac{x}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\left(x>0\right)\\ =\dfrac{x}{\sqrt{x}+1}+\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}=\sqrt{x}+1\)
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28 tháng 9 2023 lúc 19:42
Rút gọn biểu thức:
1, Bleft(dfrac{x.sqrt{x}+x+sqrt{x}}{x.sqrt{x}-1}-dfrac{sqrt{x}+3}{1-sqrt{x}}right).dfrac{x-1}{2x+sqrt{x}-1}với x-0, x khác 1, x khác dfrac{1}{4}
2, Adfrac{left(sqrt{x}-1right)^2.left(sqrt{x}+1right)^2}{left(sqrt{x}-1right)left(sqrt{x}+1right)}-dfrac{3sqrt{x}+1}{x-1} với xge0:xne0
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Rút gọn biểu thức:
1, \(B=\left(\dfrac{x.\sqrt{x}+x+\sqrt{x}}{x.\sqrt{x}-1}-\dfrac{\sqrt{x}+3}{1-\sqrt{x}}\right).\dfrac{x-1}{2x+\sqrt{x}-1}\)với x>-0, x khác 1, x khác \(\dfrac{1}{4}\)
2, \(A=\dfrac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{x-1}\) với x\(\ge\)0:x\(\ne\)0
Rút gọn các biểu thức sau:Aleft(dfrac{1}{sqrt{x}-3}+dfrac{1}{sqrt{x}+3}right)left(1-dfrac{3}{sqrt{x}}right)Bleft(dfrac{sqrt{x}}{sqrt{x}-2}+dfrac{1}{sqrt{x}+2}+dfrac{6-7sqrt{x}}{x-4}right)left(sqrt{x}+2right)Cleft(dfrac{sqrt{a}}{sqrt{a}-1}-dfrac{sqrt{a}}{a-sqrt{1}}right):dfrac{sqrt{a}+1}{a-1}Dleft(dfrac{x-2}{x+2sqrt{x}}+dfrac{1}{sqrt{x}+2}right).dfrac{sqrt{x}+1}{sqrt{x}-1}Eleft(1+dfrac{x+sqrt{x}}{sqrt{x}+1}right)left(1+dfrac{x-sqrt{x}}{1-sqrt{x}}right)giúp mình với ạ!mình đang cần gấp
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Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}+\dfrac{6-7\sqrt{x}}{x-4}\right)\left(\sqrt{x}+2\right)\)
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{1}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(D=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(E=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)
giúp mình với ạ!mình đang cần gấp
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
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2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
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3. ĐKXĐ: $a\geq 0; a\neq 1$
\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)
\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)
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Xem thêm câu trả lời
A=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
Rút gọn biểu thức trên
\(x\ge0,x\ne9\)
\(A=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right]:\)
\(\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(A=\left[\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right].\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(A=\dfrac{-3\left(\sqrt{x}+1\right).\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(\sqrt{x}+1\right)}=\dfrac{-3}{\sqrt{x}+3}\)
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1) Rút gọn biểu thức
P=\(\left(\dfrac{3x-6\sqrt{x}}{x\sqrt{x}-2x}-\dfrac{1}{2-\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}\right).\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
Ta có: \(P=\left(\dfrac{3x-6\sqrt{x}}{x\sqrt{x}-2x}-\dfrac{1}{2-\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}\right)\cdot\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{x\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}-3}{\sqrt{x}}\right)\cdot\left(\dfrac{\sqrt{x}-2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{3\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\)\(=\dfrac{3\sqrt{x}-6+\sqrt{x}+x-5\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)^2}\)
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