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a: \(=6\sqrt{a}+\dfrac{1}{3}\sqrt{a}-3\sqrt{a}+\sqrt{7}=\dfrac{10}{3}\sqrt{a}+\sqrt{7}\)
b: \(=5a\cdot5b\sqrt{ab}+\sqrt{3}\cdot2\sqrt{3}\cdot ab\sqrt{ab}+9ab\cdot3\sqrt{ab}-5b\cdot9a\sqrt{ab}\)
\(=25ab\sqrt{ab}+12ab\sqrt{ab}+27ab\sqrt{ab}-45ab\sqrt{ab}\)
\(=19ab\sqrt{ab}\)
c: \(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}-\dfrac{a}{b}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}\)
\(=\sqrt{ab}\left(\dfrac{1}{b}+1\right)-\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=\sqrt{ab}\)
d: \(=11\sqrt{5a}-5\sqrt{5a}+2\sqrt{5a}-12\sqrt{5a}+9\sqrt{a}\)
\(=-4\sqrt{5a}+9\sqrt{a}\)
2:
\(VT=\dfrac{a^2b}{a-b}\cdot\dfrac{2\sqrt{2}\left(a-b\right)}{5\sqrt{3}\cdot a^2\sqrt{b}}=\dfrac{2}{15}\cdot\sqrt{6b}=VP\)
1: \(=9\sqrt{ab}+\dfrac{7\sqrt{ab}}{b}-\dfrac{5\sqrt{ab}}{a}-3\sqrt{ab}=\)6căn ab+căn ab(7/b-5/a)
=căn ab(6+7/b-5/a)
a. \(\sqrt{\dfrac{63y^3}{7y}}\)=\(\sqrt{9y^2}\)=3y
b.\(\sqrt{\dfrac{48x^3}{3x^5}}\)=\(\sqrt{16\cdot\dfrac{1}{X^2}}\)= \(\sqrt{16}\cdot\sqrt{\dfrac{1}{X^2}}\)=\(4\cdot\dfrac{1}{X}=\dfrac{4}{X}\)
c.\(\sqrt{\dfrac{45mn^2}{20m}}=\sqrt{\dfrac{9n^2}{4}}=\dfrac{\sqrt{9n^2}}{\sqrt{4}}=\dfrac{3n}{2}\)
d. \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\sqrt{2}a}\)
a) \(\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3y\)
b) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\dfrac{4}{x}\)
c) \(\dfrac{\sqrt{45mn^2}}{\sqrt{20m}}=\sqrt{\dfrac{45mn^2}{20m}}=\sqrt{\dfrac{9n^2}{4}}=\dfrac{3n}{2}\)
d) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\left|a\right|\sqrt{2}}=\dfrac{-1}{2a\sqrt{2}}\)
a) ab2.√3a2b4=ab2.√3√a2b4ab2.3a2b4=ab2.3a2b4
=ab2.√3√a2.√b4=ab2.√3|a|.|b2|=ab2.3a2.b4=ab2.3|a|.|b2|
=ab2.√3(−a).b2=ab2.3(−a).b2 (Do a<0a<0 nên |a|=−a|a|=−a và b≠0b≠0 nên b2>0b2>0 ⇒⇒ ∣∣b2∣∣=b2|b2|=b2)
=−√3=−3.
b) √27(a−3)248=√9(a−3)21627(a−3)248=9(a−3)216
=√9.√(a−3)2√16=3.|a−3|4=9.(a−3)216=3.|a−3|4
=3(a−3)4=3(a−3)4.
(Do a>3a>3 nên |a−3|=a−3|a−3|=a−3)
c) √9+12a+4a2b2=√32+2.3.2a+(2a)2√b29+12a+4a2b2=32+2.3.2a+(2a)2b2
=√(3+2a)2√b2=|3+2a||b|=(3+2a)2b2=|3+2a||b|
=3+2a−b=−2a+3b=3+2a−b=−2a+3b.
(Do a≥−1,5a≥−1,5 ⇒⇒ 3+2a≥03+2a≥0 nên |3+2a|=3+2a|3+2a|=3+2a và b<0b<0 nên |b|=−b|b|=−b)
d) (a−b).√ab(a−b)2=(a−b).√ab√(a−b)2(a−b).ab(a−b)2=(a−b).ab(a−b)2
=(a−b).√ab|a−b|=(a−b).√ab−(a−b)=(a−b).ab|a−b|=(a−b).ab−(a−b)
=−√ab=−ab.
(Do a<b<0a<b<0 nên |a−b|=−(a−b)|a−b|=−(a−b) và ab>0ab>0)
a) ab2.√3a2b4=ab2.√3√a2b4ab2.3a2b4=ab2.3a2b4
=ab2.√3√a2.√b4=ab2.√3|a|.|b2|=ab2.3a2.b4=ab2.3|a|.|b2|
=ab2.√3(−a).b2=ab2.3(−a).b2 (Do a<0a<0 nên |a|=−a|a|=−a và b≠0b≠0 nên b2>0b2>0 ⇒⇒ ∣∣b2∣∣=b2|b2|=b2)
=−√3=−3.
b) √27(a−3)248=√9(a−3)21627(a−3)248=9(a−3)216
=√9.√(a−3)2√16=3.|a−3|4=9.(a−3)216=3.|a−3|4
=3(a−3)4=3(a−3)4.
(Do a>3a>3 nên |a−3|=a−3|a−3|=a−3)
c) √9+12a+4a2b2=√32+2.3.2a+(2a)2√b29+12a+4a2b2=32+2.3.2a+(2a)2b2
=√(3+2a)2√b2=|3+2a||b|=(3+2a)2b2=|3+2a||b|
=3+2a−b=−2a+3b=3+2a−b=−2a+3b.
(Do a≥−1,5a≥−1,5 ⇒⇒ 3+2a≥03+2a≥0 nên |3+2a|=3+2a|3+2a|=3+2a và b<0b<0 nên |b|=−b|b|=−b)
d) (a−b).√ab(a−b)2=(a−b).√ab√(a−b)2(a−b).ab(a−b)2=(a−b).ab(a−b)2
=(a−b).√ab|a−b|=(a−b).√ab−(a−b)=(a−b).ab|a−b|=(a−b).ab−(a−b)
=−√ab=−ab.
(Do a<b<0a<b<0 nên |a−b|=−(a−b)|a−b|=−(a−b) và ab>0ab>0)
a, \(A=\left(\sqrt{12}-2\sqrt{5}\right)\sqrt{3}+\sqrt{60}\)
\(=\left(2\sqrt{3}-2\sqrt{5}\right)\sqrt{3}+2\sqrt{15}\)
\(=2\sqrt{9}-2\sqrt{15}+2\sqrt{15}=2\sqrt{9}\)
b, \(B=\frac{\sqrt{4x}}{x-3}\sqrt{\frac{x^2-6x+9}{x}}=\frac{2\sqrt{x}}{x-3}.\sqrt{\frac{\left(x-3\right)^2}{x}}\)
\(=\frac{2\sqrt{x}}{x-3}.\frac{x-3}{\sqrt{x}}=2\)
Với x > 0 ; x \(\ne\)9
a, \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}+\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{x-9}\right):\left(\frac{3\sqrt{x}+1+2\left(\sqrt{x}-3\right)}{x-3\sqrt{x}}\right)\)
\(=\left(\frac{-3\sqrt{x}-9}{x-9}\right):\left(\frac{5\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)=\frac{-3}{\sqrt{x}-3}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{5\left(\sqrt{x}-1\right)}=\frac{-3\sqrt{x}}{5\left(\sqrt{x}-1\right)}\)
b, Ta có : \(B< 0\Rightarrow\frac{-3\sqrt{x}}{5\left(\sqrt{x}-1\right)}< 0\Rightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)
Kết hợp vói đk vậy x > 1 ; x \(\ne\)9
a,\(ab^2\sqrt{\dfrac{3}{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{ab^2}=\sqrt{3}\)
b,\(\sqrt{\dfrac{27\left(a-3\right)^2}{48}}=\dfrac{3\sqrt{3}\left(a-3\right)}{4\sqrt{3}}=\dfrac{3}{4}\left(a-3\right)\)
c,\(\sqrt{\dfrac{9+12a+4a^2}{b^2}}=\dfrac{\sqrt{\left(3+2a\right)^2}}{\sqrt{b^2}}=\dfrac{3+2a}{b}\)
d, \(\left(a-b\right).\sqrt{\dfrac{ab}{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\sqrt{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\left(a-b\right)}=\sqrt{ab}\)
đkxđ a>=0 a khác 1
\(C=\left(\frac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(C=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+3}{a-1}\)
\(C=\frac{\left(a-1\right).\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
b)
\(a=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
\(\sqrt{a}=\sqrt{3}-1\)
thay vào nha
c) \(C=\frac{\left(a-1\right).\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
để c<0 thì \(\frac{\left(a-1\right).\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}< 0\)
mà \(\sqrt{a}\left(\sqrt{a}+3\right)>0\)
\(\left(a-1\right)\left(\sqrt{a}+1\right)< 0\)
mà \(\sqrt{a}+1>0\)
nên a-1<0
\(0\le a< 1\)
\(=9\sqrt{ab}-6b\cdot\dfrac{\sqrt{a}}{\sqrt{b}}-\dfrac{1}{b}\cdot3b\sqrt{ab}\)
\(=9\sqrt{ab}-6\sqrt{ab}-3\sqrt{ab}=0\)