Rút gọn biểu thức:
a) A = \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
b) B = \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\), với a > 0, a ≠ 4
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rút gọn biểu thức A=\(\dfrac{\left(2-\sqrt{a}\right)-\left(\sqrt{a+3}\right)}{1+2\sqrt{a}}\) (với a>0) ; B=\(\dfrac{1}{1-\sqrt{2}+\sqrt{3}}-\dfrac{1}{1-\sqrt{2-\sqrt{3}}}\); C=\(\dfrac{1}{\sqrt{5-2}}+\dfrac{1}{\sqrt{5+\sqrt{2}}}\)
\(A=\dfrac{2-\sqrt{a}-\sqrt{a}-3}{2\sqrt{a}+1}=-1\)
\(B=\dfrac{1}{1-\sqrt{2+\sqrt{3}}}-\dfrac{1}{1-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}-1}-\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}+1}\)
\(=\dfrac{2-\sqrt{6}+\sqrt{2}-2+\sqrt{6}+\sqrt{2}}{5-2\sqrt{6}-1}\)
\(=\dfrac{2\sqrt{2}}{4-2\sqrt{6}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}=-\sqrt{2}-\sqrt{3}\)
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thực hiện phép tính ( rút gọn biểu thức )
a) \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{3\sqrt{6}}{\sqrt{2}}+\dfrac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
b) \(\left(\dfrac{2-2\sqrt{5}}{\sqrt{5}-2}-\dfrac{\sqrt{6}-3}{\sqrt{3}-\sqrt{2}}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
a: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-3\sqrt{3}+\dfrac{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\)
\(=\sqrt{3}-3\sqrt{3}+\sqrt{3}=-\sqrt{3}\)
b: \(=\left(\left(2-2\sqrt{5}\right)\left(\sqrt{5}+2\right)+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(2\sqrt{5}+4-10-4\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(-2\sqrt{5}+\sqrt{3}-6\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=-20+2\sqrt{15}+\sqrt{15}-3-6\sqrt{5}+6\sqrt{3}\)
\(=-23+3\sqrt{15}-6\sqrt{5}+6\sqrt{3}\)
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Rút gọn các biểu thức:
\(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{x-4}{3\sqrt{x}}\)
\(B=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}+\dfrac{6-7\sqrt{a}}{a-4}\right).\left(\sqrt{a}+2\right)\)
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{2}{3}\)
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rút gọn biểu thứcGdfrac{sqrt[3]{a}.a^{dfrac{2}{3}}}{left(a^{4-2sqrt{3}}right)^{4+2sqrt{3}}}Gdfrac{a^{sqrt{7}+1}.a^{2-sqrt{7}}}{left(a^{sqrt{2}-2}right)^{sqrt{2}+2}}Hdfrac{a^2.left(a^{-2}.b^3right).b^{-1}}{left(a^{-1}.bright)^3.a^{-5}.b^{-2}}Hdfrac{b^3.a^{-4}.left(ab^2right)^3}{left(a^2right)^{-2}.left(ab^3right)^2.b^2} Hdfrac{b^3.a^{-4}.left(ab^2right)^3}{left(a^2right)^{-2}.left(ab^3right)^2.b^2} Hdfrac{b^3.a^{-4}.left(ab^2right)^3}{left(a^2right)^{-2}.left(ab^3right)^2.b^2}
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rút gọn biểu thức
\(G=\dfrac{\sqrt[3]{a}.a^{\dfrac{2}{3}}}{\left(a^{4-2\sqrt{3}}\right)^{4+2\sqrt{3}}}\)
\(G=\dfrac{a^{\sqrt{7}+1}.a^{2-\sqrt{7}}}{\left(a^{\sqrt{2}-2}\right)^{\sqrt{2}+2}}\)
\(H=\dfrac{a^2.\left(a^{-2}.b^3\right).b^{-1}}{\left(a^{-1}.b\right)^3.a^{-5}.b^{-2}}\)
\(H=\dfrac{b^3.a^{-4}.\left(ab^2\right)^3}{\left(a^2\right)^{-2}.\left(ab^3\right)^2.b^2}\)
\(H=\dfrac{b^3.a^{-4}.\left(ab^2\right)^3}{\left(a^2\right)^{-2}.\left(ab^3\right)^2.b^2}\)
\(H=\dfrac{b^3.a^{-4}.\left(ab^2\right)^3}{\left(a^2\right)^{-2}.\left(ab^3\right)^2.b^2}\)
Bài 1.Rút gọn A sqrt{x^2+dfrac{2x^2}{3}} với x0Bài 2.Rút gọn biểu thức left(dfrac{10+2sqrt{10}}{sqrt{5}+sqrt{2}}+dfrac{sqrt{30}-sqrt{6}}{sqrt{5}-1}right):dfrac{2}{2sqrt{5}-sqrt{6}}Bài 3.Cho ba biểu thức A asqrt{b} + bsqrt{a};B asqrt{a}-bsqrt{b} ;C a-b.Trong ba biểu thức trên biểu thức bằng biểu thức left(sqrt{a}-sqrt{b}right)left(sqrt{a}+sqrt{b}right) với a,b0Bài 7.Cho B dfrac{1}{sqrt{1}+sqrt{2}}+dfrac{1}{sqrt{2}+sqrt{3}}+dfrac{1}{sqrt{3}+sqrt{4}}+...+dfrac{1}{sqrt{98}+sqrt{99}}+dfrac{1}{sq...
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Bài 1.Rút gọn A = \(\sqrt{x^2+\dfrac{2x^2}{3}}\) với x<0
Bài 2.Rút gọn biểu thức \(\left(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{30}-\sqrt{6}}{\sqrt{5}-1}\right)\):\(\dfrac{2}{2\sqrt{5}-\sqrt{6}}\)
Bài 3.Cho ba biểu thức A = a\(\sqrt{b}\) + b\(\sqrt{a}\);B = \(a\sqrt{a}-b\sqrt{b}\) ;C = a-b.Trong ba biểu thức trên biểu thức bằng biểu thức \(\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\) với a,b>0
Bài 7.Cho B = \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{98}+\sqrt{99}}+\dfrac{1}{\sqrt{99}+\sqrt{100}}\).Giá trị của biểu thức B là
Bài 8.Gọi M là giá trị nhỏ nhất của \(\dfrac{\sqrt{x}+1}{\sqrt{x}+4}\) và N là giá trị lớn nhất của \(\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\).Tìm M và N
Giúp mình với!Mình đang cần gấp
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
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Rút gọn các biểu thức sau:Aleft(dfrac{1}{sqrt{a}+2}+dfrac{1}{sqrt{a}-2}right):dfrac{sqrt{a}}{a-4}Bleft(dfrac{4x}{sqrt{x}-1}-dfrac{sqrt{x}-2}{x-3sqrt{x}+2}right).dfrac{sqrt{x}-1}{x^2}Cleft(dfrac{1}{sqrt{x}+3}+dfrac{2sqrt{x}}{x-9}right).dfrac{2sqrt{x}+6}{sqrt{x}-1}Dleft(dfrac{5sqrt{x}-6}{x-9}-dfrac{2}{sqrt{x}+3}right):left(1+dfrac{6}{x-9}right)Eleft(dfrac{sqrt{x}}{3+sqrt{x}}+dfrac{9+x}{9-x}right).left(3sqrt{x}-xright) help
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Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right).\dfrac{\sqrt{x}-1}{x^2}\)
\(C=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right).\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)
\(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(E=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{9+x}{9-x}\right).\left(3\sqrt{x}-x\right)\)
help
Rút gọn các biểu thức sau:Aleft(dfrac{1}{sqrt{a}+2}+dfrac{1}{sqrt{a}-2}right):dfrac{sqrt{a}}{a-4}Bleft(dfrac{4x}{sqrt{x}-1}-dfrac{sqrt{x}-2}{x-3sqrt{x}+2}right).dfrac{sqrt{x}-1}{x^2}Cleft(dfrac{1}{sqrt{x}+3}+dfrac{2sqrt{x}}{x-9}right).dfrac{2sqrt{x}+6}{sqrt{x}-1}Dleft(dfrac{5sqrt{x}-6}{x-9}-dfrac{2}{sqrt{x}+3}right):left(1+dfrac{6}{x-9}right)Eleft(dfrac{sqrt{x}}{3+sqrt{x}}+dfrac{9+x}{9-x}right).left(3sqrt{x}-xright) help
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Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right).\dfrac{\sqrt{x}-1}{x^2}\)
\(C=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right).\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)
\(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(E=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{9+x}{9-x}\right).\left(3\sqrt{x}-x\right)\)
help
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)
=2
b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{x^2}\)
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rút gọn biểu thức sau:
a.\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
b.\(A=\dfrac{\sqrt{a}}{\sqrt{a}-5}-\dfrac{10\sqrt{a}}{a-25}-\dfrac{5}{\sqrt{a}+5}\) với a\(\ge\)0; a\(\ne25\)
\(a,\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)
\(b,A=\dfrac{\sqrt{a}}{\sqrt{a}-5}-\dfrac{10\sqrt{a}}{a-25}-\dfrac{5}{\sqrt{a}+5}\)
\(\Rightarrow A=\dfrac{\sqrt{a}\left(\sqrt{a}+5\right)}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{10\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{5\left(\sqrt{a}-5\right)}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)
\(\Rightarrow A=\dfrac{a+5\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{10\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{5\sqrt{a}-25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)
\(\Rightarrow A=\dfrac{a+5\sqrt{a}-10\sqrt{a}-5\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)
\(\Rightarrow A=\dfrac{a-10\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)
\(\Rightarrow A=\dfrac{\left(\sqrt{a}-5\right)^2}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{a}-5}{\sqrt{a}+5}\)
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a: \(=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)
b: \(A=\dfrac{a+5\sqrt{a}-10\sqrt{a}-5\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}=\dfrac{\left(\sqrt{a}-5\right)^2}{a-25}=\dfrac{\sqrt{a}-5}{\sqrt{a}+5}\)
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Rút Gọn Biểu Thứcdfrac{2}{sqrt{3}-1}-dfrac{2}{sqrt{3}+1}dfrac{sqrt{12}-sqrt{6}}{sqrt{30}-sqrt{15}}sqrt{9a}+sqrt{81a}+3sqrt{25a}-16sqrt{49a} (a ≥ 0)dfrac{ab-bc}{sqrt{ab}-sqrt{bc}}left(asqrt{dfrac{a}{b}+2sqrt{ab}+bsqrt{dfrac{a}{b}}}right)sqrt{ab}left(dfrac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right)left(dfrac{1+asqrt{a}}{1+sqrt{a}}right)
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Rút Gọn Biểu Thức
\(\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\)
\(\dfrac{\sqrt{12}-\sqrt{6}}{\sqrt{30}-\sqrt{15}}\)
\(\sqrt{9a}+\sqrt{81a}+3\sqrt{25a}-16\sqrt{49a}\) (a ≥ 0)
\(\dfrac{ab-bc}{\sqrt{ab}-\sqrt{bc}}\)
\(\left(a\sqrt{\dfrac{a}{b}+2\sqrt{ab}+b\sqrt{\dfrac{a}{b}}}\right)\sqrt{ab}\)
\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}\right)\)
a: \(\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}\)
\(=\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{2}=\dfrac{4}{2}=2\)
b: \(\dfrac{\sqrt{12}-\sqrt{6}}{\sqrt{30}-\sqrt{15}}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{\sqrt{15}\left(\sqrt{2}-1\right)}\)
\(=\dfrac{\sqrt{6}}{\sqrt{15}}=\sqrt{\dfrac{6}{15}}=\sqrt{\dfrac{2}{5}}=\dfrac{\sqrt{10}}{5}\)
c: \(\sqrt{9a}+\sqrt{81a}+3\sqrt{25a}-16\sqrt{49a}\)
\(=3\sqrt{a}+9\sqrt{a}+3\cdot5\sqrt{a}-16\cdot7\sqrt{a}\)
\(=27\sqrt{a}-112\sqrt{a}=-85\sqrt{a}\)
d: \(\dfrac{ab-bc}{\sqrt{ab}-\sqrt{bc}}=\dfrac{\left(\sqrt{ab}-\sqrt{bc}\right)\left(\sqrt{ab}+\sqrt{bc}\right)}{\sqrt{ab}-\sqrt{bc}}\)
\(=\sqrt{ab}+\sqrt{bc}\)
e: \(a\left(\sqrt{\dfrac{a}{b}+2\sqrt{ab}+b\cdot\sqrt{\dfrac{a}{b}}}\right)\cdot\sqrt{ab}\)
\(=a\cdot\sqrt{\dfrac{a}{b}\cdot ab+2\sqrt{ab}\cdot ab+b\cdot\sqrt{\dfrac{a}{b}}\cdot ab}\)
\(=a\cdot\sqrt{a^2+2\cdot ab\cdot\sqrt{ab}+a\sqrt{a}\cdot b\sqrt{b}}\)
\(=a\cdot\sqrt{a^2+3\cdot a\cdot\sqrt{a}\cdot b\cdot\sqrt{b}}\)
e: ĐKXĐ: a>=0 và a<>1
\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}+1}\)
\(=\left(1+\sqrt{a}+\sqrt{a}+a\right)\cdot\left(a-\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)^2\cdot\left(a-\sqrt{a}+1\right)\)
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Câu 3: Rút gọn biểu thức sau:
a. \(\dfrac{1}{\sqrt{5}-1}+\dfrac{1}{1+\sqrt{5}}\)
b. \(\sqrt{14-6\sqrt{5}}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
c. \(\dfrac{2}{\sqrt{5}+\sqrt{3}}-\dfrac{3-\sqrt{15}}{\sqrt{5}-\sqrt{3}}\)
\(a,=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)^2}+\left|2-\sqrt{5}\right|=3-\sqrt{5}+\sqrt{5}-2=1\\ c,=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}-\dfrac{-\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=\sqrt{5}-\sqrt{3}+\sqrt{3}=\sqrt{5}\)
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