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Giải:
a) \(\sqrt{a^4.\left(3-a\right)^2}\)
\(=\sqrt{\left(a^2\left(3-a\right)\right)^2}\)
\(=\left|a^2\left(3-a\right)\right|\)
b) \(\sqrt{27.48.\left(a-3\right)^2}\)
\(=\sqrt{3.9.16.3.\left(a-3\right)^2}\)
\(=\sqrt{3.3.9.16\left(a-3\right)^2}\)
\(=\sqrt{\left(9.4\left(a-3\right)\right)^2}\)
\(=\left|9.4\left(a-3\right)\right|\)
\(=\left|36\left(a-3\right)\right|\)
c) \(\sqrt{48.75a^2}\)
\(=\sqrt{16.3.25.3a^2}\)
\(=\sqrt{\left(4.3.5a\right)^2}\)
\(=\left|4.3.5a\right|\)
\(=\left|60a\right|\)
d) \(\sqrt{2^4.\left(-9\right)^2}\)
\(=\sqrt{2^4.9^2}\)
\(=\sqrt{\left(2^2.9\right)^2}\)
\(=\left|2^2.9\right|\)
\(=\left|36\right|=36\)
Vậy ...
a) \(\sqrt{25\cdot96}=\sqrt{5^2\cdot2^5\cdot3}=\sqrt{5^2\cdot\left(2^2\right)^2\cdot2\cdot3}\)
\(=20\sqrt{6}\)
b) \(\sqrt{21\cdot75\cdot14}=\sqrt{2\cdot3^2\cdot5^2\cdot7^2}=105\sqrt{2}\)
c) \(y^2\sqrt{x^6\cdot y^8}=\sqrt{x^6\cdot y^4\cdot y^8}=\sqrt{\left(x^3\right)^2\cdot\left(y^6\right)^2}=x^3\cdot y^6\)
hì,giúp bn đc phần a thôi nha!!!
\(a,\sqrt{25.96}=\sqrt{25.16.6}=\sqrt{25}.\sqrt{16}.\sqrt{6}=5.4.\sqrt{6}=20\sqrt{6}\)
=.= hok tốt!!!
a, \(\left(\sqrt{2006}-\sqrt{2005}\right).\left(\sqrt{2006}+\sqrt{2005}\right)=\left(2006-2005\right)=1\)
b.
=\(\frac{7+4\sqrt{3}+14-8\sqrt{3}}{49-48}\left(21+4\sqrt{3}\right)\)
=\(\left(21-4\sqrt{3}\right)\left(21+4\sqrt{3}\right)\)
=441-48
393
vậy.......
hc tốt
a)Ta có: \(2\sqrt{5}< 5\sqrt{2}\)\(2\sqrt{5}=\sqrt{2^2.5}=\sqrt{20}\)
\(5\sqrt{2}=\sqrt{5^2.2}=\sqrt{50}\)
Vì \(\sqrt{20}< \sqrt{50}\)
Nên \(2\sqrt{5}< 5\sqrt{2}\)
b)Ta có: \(3\sqrt{13}=\sqrt{3^2.13}=\sqrt{117}\)
\(4\sqrt{11}=\sqrt{4^2.11}=\sqrt{176}\)
Vì \(\sqrt{117}< \sqrt{176}\)
Nên \(3\sqrt{13}< 4\sqrt{11}\)
c) Ta có: \(\frac{3}{4}.\sqrt{7}=\sqrt{\left(\frac{3}{4}\right)^2.7}=\sqrt{\frac{63}{16}}\)
\(\frac{2}{5}.\sqrt{5}=\sqrt{\left(\frac{2}{5}\right)^2.5}=\sqrt{\frac{4}{5}}\)
Vì \(\sqrt{\frac{63}{16}}>1\)
\(\sqrt{\frac{4}{5}}< 1\)
Nên \(\sqrt{\frac{63}{16}}>\sqrt{\frac{4}{5}}\)
Vậy \(\frac{3}{4}.\sqrt{7}>\frac{2}{5}.\sqrt{5}\)
a/
ĐK \(x^2-6x+6\ge0\)
\(\text{pt }\Leftrightarrow\left(x^2-6x+6\right)-4\sqrt{x^2-6x+6}+3=0\)
Đặt \(t=\sqrt{x^2-6x+6};t\ge0\)
pt thành \(t^2-4t+3=0\Leftrightarrow t=3\text{ hoặc }t=1\)
\(+t=1\Rightarrow x^2-6x+6=1^2\Leftrightarrow x^2-6x+7=0\Leftrightarrow t=3+\sqrt{2}\text{ hoặc }t=3-\sqrt{2}\)
\(+t=3\Rightarrow x^2-6x+6=3^2\Leftrightarrow x^2-6x-3=0\Leftrightarrow x=3+2\sqrt{3}\text{ hoặc }x=3-2\sqrt{3}\)
Vậy ....
b/
ĐK: \(x^2+3x\ge0\)
\(\left(x+5\right)\left(2-x\right)=3\sqrt{x^2+3x}\Leftrightarrow-\left(x^2+3x\right)-3\sqrt{x^2+3x}+10=0\)
\(\Leftrightarrow\left(\sqrt{x^2+3x}-2\right)\left(\sqrt{x^2+3x}+5\right)=0\)
\(\Leftrightarrow\sqrt{x^2+3x}=2\text{ hoặc }\sqrt{x^2+3x}=-5\text{ (loại)}\)
\(\Leftrightarrow x^2+3x-2^2=0\Leftrightarrow x=1\text{ hoặc }x=-4\)
Vậy ....
4 x 4 : 4 = 16 : 4 = 4 ≠ 6