asqrt{ }a-8+2a-4sqrt{ }a trên a-4
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a\(\sqrt{ }\)a-8+2a-4\(\sqrt{ }\)a trên a-4
Những câu hỏi liên quan
a) \(\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\)
b) \(\dfrac{12\sqrt{6}}{\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}}\)
\(a,\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\left(dk:a\ne4\right)\)
\(=\dfrac{a\sqrt{a}-4\sqrt{a}-8+2a}{a-4}\)
\(=\dfrac{\sqrt{a}\left(a-4\right)+2\left(a-4\right)}{a-4}\)
\(=\dfrac{\left(a-4\right)\left(\sqrt{a}+2\right)}{a-4}\)
\(=\sqrt{a}+2\)
\(b,\dfrac{12\sqrt{6}}{\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}}\\ =\dfrac{12\sqrt{6}}{\sqrt{\left(\sqrt{6}+1\right)^2}-\sqrt{\left(\sqrt{6}-1\right)^2}}\\ =\dfrac{12\sqrt{6}}{\left|\sqrt{6}+1\right|-\left|\sqrt{6}-1\right|}\\ =\dfrac{12\sqrt{6}}{\sqrt{6}+1-\sqrt{6}+1}\\ =\dfrac{12\sqrt{6}}{2}\\ =6\sqrt{6}\)
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Tính: a) A= \(\frac{1+2a}{1+\sqrt{1+2a}}+\frac{1-2a}{1-\sqrt{1-2a}}\)với a= \(\frac{\sqrt{3}}{4}\)
b) B= \(\frac{x\sqrt{x}-2x+28}{x-3\sqrt{x}-4}-\frac{\sqrt{x}-4}{\sqrt{x}+1}+\frac{\sqrt{x}+8}{4-\sqrt{x}}\)với 0 < x khác 16)
a \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
b \(\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}\) với a>0
c \(\sqrt{5a.45a}-3a\) với a<0
a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
b: \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{6a^2}{24}}=\sqrt{\dfrac{a^2}{4}}=\dfrac{a}{2}\)
c: \(\sqrt{5a\cdot45a}-3a=-15a-3a=-18a\)
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Rút gọn: \(\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\)
\(\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}=\dfrac{\sqrt{a}\left(a-4\right)+2\left(a-4\right)}{a-4}=\dfrac{\left(a-4\right)\left(\sqrt{a}+2\right)}{a-4}=\sqrt{a}+2\)
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Rút gọn biểu thức:
a) A frac{sqrt{5-2sqrt{6}}+sqrt{8-2sqrt{15}}}{sqrt{7+2sqrt{10}}}
b) B left(1+frac{a+sqrt{a}}{sqrt{a}+1}right).left(1-frac{a-sqrt{a}}{sqrt{a}-1}right) a0 va a # 1
c) C frac{asqrt{a}-8+2a-4sqrt{a}}{a-4}
d) D frac{1}{2a-1}.sqrt{5a^4.left(-4a+4a^2right)}
e) E frac{2}{x^2-y^2}.sqrt{frac{3x^2+6xy+3y^2}{4}}
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Rút gọn biểu thức:
a) A = \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
b) B = \(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\) a>0 va a # 1
c) C = \(\frac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\)
d) D = \(\frac{1}{2a-1}.\sqrt{5a^4.\left(-4a+4a^2\right)}\)
e) E = \(\frac{2}{x^2-y^2}.\sqrt{\frac{3x^2+6xy+3y^2}{4}}\)
a : \(\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}\) với a ≥ 0
b : \(\sqrt{3a}.\sqrt{\dfrac{52}{a}}\)với a ≥ 0
c : \(2y^2.\sqrt{\dfrac{x^4}{4y^2}}\)với y ≤ 0
a) \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}\)
\(=\sqrt{\dfrac{2a\cdot3a}{3\cdot8}}\)
\(=\sqrt{\dfrac{6a^2}{24}}\)
\(=\sqrt{\dfrac{a^2}{4}}\)
\(=\dfrac{\sqrt{a^2}}{\sqrt{4}}\)
\(=\dfrac{a}{2}\)
b) \(\sqrt{3a}\cdot\sqrt{\dfrac{52}{a}}\)
\(=\sqrt{3a\cdot\dfrac{52}{a}}\)
\(=\sqrt{3\cdot52}\)
\(=\sqrt{13\cdot3\cdot4}\)
\(=2\sqrt{39}\)
c) \(2y^2\cdot\sqrt{\dfrac{x^4}{4y^2}}\)
\(=2y^2\cdot\dfrac{\sqrt{\left(x^2\right)^2}}{\sqrt{\left(2y\right)^2}}\)
\(=2y^2\cdot\dfrac{x^2}{-2y}\)
\(=\dfrac{2y^2\cdot x^2}{-2y}\)
\(=-x^2y\)
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RÚT GỌN:
1)\(\sqrt{3+2\sqrt{2}}\) +\(\sqrt{6-4\sqrt{2}}\)
2)\(\frac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\)
Kết quả của phép tính \(\sqrt{\left(a-b\right)^2}+\sqrt{\left(a+b\right)^2}\)
A. 2a B. 2b C. -2a D. -2b E. cả 4 câu trên đều sai
rút gọn
a, \(\sqrt{12-2\sqrt{11}}-\sqrt{11}\)
b, \(x-4+\sqrt{16-8x+x^2}\left(x>4\right)\)
c, \(\sqrt{3-2\sqrt{2}+}\sqrt{3+2\sqrt{2}}\)
d, \(A=\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\)
a ) \(\sqrt{12-2\sqrt{11}}-\sqrt{11}=\sqrt{11}-1-\sqrt{11}=-1\)
b ) \(x-4+\sqrt{16-8x+x^2}\left(x>4\right)\)
\(=x-4+x-4=-8\)
c ) \(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}=\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}\)
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a,
\(\sqrt{12-2\sqrt{11}}-\sqrt{11}\\ =\sqrt{\left(\sqrt{11}-1\right)^2}-\sqrt{11}\\ =\left|\sqrt{11}-1\right|-\sqrt{11}\\ =\sqrt{11}-1-\sqrt{11}\\ =-1\)
b,
\(x-4+\sqrt{16-8x+x^2}\\ =x-4+\sqrt{\left(4-x\right)^2}\\ =x-4+\left|4-x\right|\\ =x-4+x-4\\ =2x-8\\ =2\left(x-4\right)\)
c,
\(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\\ =\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{2}+1\right)^2}\\ =\left|\sqrt{2}-1\right|+\left|\sqrt{2}+1\right|\\ =\sqrt{2}-1+\sqrt{2}+1\\ =2\sqrt{2}\)
d,
\(A=\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\\ =\dfrac{\left(a\sqrt{a}-4\sqrt{a}\right)+\left(2a-8\right)}{a-4}\\ =\dfrac{\left(a-4\right)\sqrt{a}+2\left(a-4\right)}{a-4}\\ =\dfrac{\left(a-4\right)\left(\sqrt{a}+2\right)}{\left(a-4\right)}\\ =\sqrt{a}+2\)
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