Rút gọn các biểu thức sau:
a, \(\sqrt{\left(120-11\right)^2}+\sqrt{\left(10-\sqrt{120}\right)^2}\)
b, \(\sqrt{x+2+2\sqrt{x+1}-\sqrt{x+2+2\sqrt{x+1}}}\) ( với đk x \(\ge\) -1 )
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Rút gọn các biểu thức sau:
a) \(\left(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)\)
b) \(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\) với x>0
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=3-1=2\)
b: \(=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)
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a, \(=\left(\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
b, với x > 0
\(=\left(\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\)
\(=-\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+2\right)\sqrt{x+1}}=\dfrac{4}{\left(\sqrt{x}+2\right)\sqrt{x^2+x}}\)
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Rút gọn các biểu thức sau:a) A left(dfrac{sqrt{x}}{x-4}+dfrac{2}{2-sqrt{x}}+dfrac{1}{sqrt{x}+2}right):left(sqrt{x}-2+dfrac{10-x}{sqrt{x}+2}right)b) B left(dfrac{xsqrt{x}+ysqrt{y}}{sqrt{x}+sqrt{y}}-sqrt{xy}right):left(x-yright)+dfrac{2sqrt{y}}{sqrt{x}+sqrt{y}}c) C left(1-dfrac{sqrt{x}}{1+sqrt{x}}right):left(dfrac{sqrt{x}+3}{sqrt{x}-2}+dfrac{2+sqrt{x}}{3-sqrt{x}}+dfrac{sqrt{x}+2}{x-5sqrt{x}+6}right)d) D sqrt{dfrac{a+x^2}{x}-2sqrt{a}}-sqrt{dfrac{a+x^2}{x}+2sqrt{a}} với a 0, x 0.
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Rút gọn các biểu thức sau:
a) A = \(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
b) B = \(\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
c) C = \(\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2+\sqrt{x}}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
d) D = \(\sqrt{\dfrac{a+x^2}{x}-2\sqrt{a}}-\sqrt{\dfrac{a+x^2}{x}+2\sqrt{a}}\) với a > 0, x > 0.
Câu 1: Rút gọn các biểu thức sau:
a. \(\sqrt{36\left(x-5\right)^2}\) với x ≥ 5
b. \(\sqrt{\dfrac{1}{4}\left(1-x\right)^2}\) với x > 1
c. \(\sqrt{x^2\left(2x-4\right)^2}\) với a ≥ 2
d. \(\dfrac{1}{x}\sqrt{x^2\left(1+x\right)^2}\) với x < -1
a) \(\sqrt{36\left(x-5\right)^2}\left(x\ge5\right)=6\left|x-5\right|=6\left(x-5\right)=6x-30\)
b) \(\sqrt{\dfrac{1}{4}\left(1-x\right)^2}\left(x>1\right)=\dfrac{1}{2}\left|1-x\right|=\dfrac{1}{2}\left(x-1\right)=\dfrac{1}{2}x-\dfrac{1}{2}\)
c) \(\sqrt{x^2\left(2x-4\right)^2}\left(x\ge2\right)=\left|x\left(2x-4\right)\right|=x\left(2x-4\right)=2x^2-4x\)
d) \(\dfrac{1}{x}\sqrt{x^2\left(1+x\right)^2}\left(x< -1\right)=\dfrac{1}{x}\left|x\left(1+x\right)\right|=\dfrac{1}{x}x\left(1+x\right)=1+x\)
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Rút gọn các biểu thức sau:
a, \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
b, \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\) với \(x\ge1\)
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\dfrac{x\sqrt{x}+y\sqrt{y}-\left(x-y\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}+x\sqrt{y}+y\sqrt{x}-y\sqrt{y}}{\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)( do \(x\ge1\))
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a: Ta có: \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
\(=\sqrt{xy}\)
b: Ta có: \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
\(=\dfrac{ \left|\sqrt{x}-1\right|}{\left|\sqrt{x}+1\right|}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
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3 tháng 3 2022 lúc 21:02
Rút gọn biểu thức :
\(B=\dfrac{\sqrt{1+\sqrt{1-x^2}}\left[\left(1+x\right)\sqrt{1+x}-\left(1-x\right)\sqrt{1-x}\right]}{x\left(2+\sqrt{1-x^2}\right)}\)
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29 tháng 5 2022 lúc 18:49
Rút gọn biểu thức sau:
A = \(\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}\right)\) . \(\left(\sqrt{x}-1\right)\)
\(A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\left(\sqrt{x}-1\right)\);\(ĐK:x\ge0;x\ne1\)
\(A=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(A=\left(\dfrac{x-\sqrt{x}+2\sqrt{x}-2-2\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(A=\left(\dfrac{-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
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\(A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}\right).\left(\sqrt{x-1}\right)\left(đk:x\ne1\right)\\ A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\left(\sqrt{x}-1\right)\\ A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-2\right).\left(\sqrt{x}-1\right)\)
\(A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\\ A=\left(\dfrac{\sqrt{x}+2-2\sqrt{x}-2}{\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\\ A=\dfrac{-\sqrt{x}}{\sqrt{x}+1}.\left(\sqrt{x}-1\right)\\ A=\dfrac{-x+\sqrt{x}}{\sqrt{x}+1}\)
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rút gọn các biểu thức sau:
a) \(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)-\sqrt{x^3}\) với x lớn hơn hoặc = 0
b) \(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\) với a lớn hơn hoặc = 0
a: \(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)-\sqrt{x^3}\)
\(=1-x\sqrt{x}-x\sqrt{x}\)
\(=1-2x\sqrt{x}\)
b: \(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\cdot\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\left(\dfrac{\left(1-\sqrt{a}\right)\cdot\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right)\)
\(=\left(\dfrac{1}{\sqrt{a}+1}\right)^2\cdot\left(a+\sqrt{a}+1+\sqrt{a}\right)\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)
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Rút gọn các biểu thức sau:
a. A = \(\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
b. B = \(\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\) (x > 0 ; x ≠ 1)
\(a,A=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
\(=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{2^2-\sqrt{3}^2}\)
\(=\dfrac{4}{1}=4\)
Vậy \(A=4\)
\(b,B=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\left(\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Vậy \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\) với \(x>0,x\ne1\)
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Rút gọn các biểu thức sau:
a. \(A=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
b. \(B=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\) (x > 0 ; x ≠ 1)
a: \(=2+\sqrt{3}+2-\sqrt{3}=4\)
b: \(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
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