\(\sqrt{ }\)a4×(3-a)2
\(\sqrt{ }\)27×48×(a-3)2
\(\sqrt{ }\)48×75a2
\(\sqrt{ }\)24×(-9)2
đề bài : đưa thừa số ra ngoài dấu căn
đưa thừa số ra ngoài dấu căn :
a) a2\(\sqrt{\dfrac{2}{3a}}\)( a > 0 )
b) \(\dfrac{x-3}{x}\)\(\sqrt{\dfrac{x^3}{9-x^2}}\)(0<x<3)
a: \(a^2\cdot\sqrt{\dfrac{2}{3a}}=a^2\cdot\dfrac{\sqrt{2}}{\sqrt{3}\cdot\sqrt{a}}=\dfrac{a\sqrt{2}}{\sqrt{3}}=\dfrac{a\sqrt{6}}{3}\)
b: \(\dfrac{x-3}{x}\cdot\sqrt{\dfrac{x^3}{9-x^2}}\)
\(=\dfrac{x-3}{x}\cdot\dfrac{x\sqrt{x}}{\sqrt{x-3}\cdot\sqrt{x+3}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{x-3}}{\sqrt{x+3}}\)
Đưa thừa số ra ngoài dấu căn
a.
\(\left(\sqrt{28}-5\sqrt{35}+7\sqrt{112}\right)2\sqrt{7}\)
b. \(\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}\)
a) \(\left(\sqrt{28}-5\sqrt{35}+7\sqrt{112}\right)2\sqrt{7}=2\sqrt{196}-10\sqrt{245}+14\sqrt{784}\)
\(=28-10\sqrt{49.5}+392=420-70\sqrt{5}\)
b) \(\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}=\sqrt{144}-3\sqrt{48}+5\sqrt{16}+4\sqrt{9.3}\)
\(=12-3\sqrt{16.3}+20+12\sqrt{3}=32-12\sqrt{3}+12\sqrt{3}=32\)
\(\sqrt{48.45}\) Đưa thừa số ra ngoài dấu căn:
\(\sqrt{225.17}\)
\(\sqrt{a^3b^7}với\) \(a\ge0;b\ge0\)
\(\sqrt{x^5\left(x-3\right)^2}\) với \(x>0\)
\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)
\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)
\(\sqrt{225\cdot17}=15\sqrt{17}\)
Bài 43 (trang 27 SGK Toán 9 Tập 1)
Viết các số hoặc biểu thức dưới dấu căn thành dạng tích rồi đưa thừa số ra ngoài dấu căn
a) $\sqrt{54}$ ; b) $\sqrt{108}$ ; c) $0,1 \sqrt{20000}$ ;
d) $-0,05 \sqrt{28800}$ ; e) $\sqrt{7.63 . a^2}$.
a, \(\sqrt{54}=\sqrt{9.6}=3\sqrt{6}\)
b, \(\sqrt{108}=\sqrt{36.3}=6\sqrt{3}\)
c, \(0,1\sqrt{20000}=0,1\sqrt{2.10000}=10\sqrt{2}\)
d, \(-0,05\sqrt{28800}=-0,05\sqrt{288.100}=-0,05.10.\sqrt{144.2}\)
\(=-0,5.12\sqrt{2}=-6\sqrt{2}\)
e, \(\sqrt{7.63.a^2}=\sqrt{7.7.9.a^2}=21\left|a\right|\)
a) .
b) .
c)
.
d)
.
e)
a) \(\sqrt{54}=\sqrt{9}.\sqrt{6}=3\sqrt{6}\)
b) \(\sqrt{108}=\sqrt{9}.\sqrt{4}.\sqrt{3}=6\sqrt{3}\)
c) \(0,1\sqrt{20000}=0,1.\sqrt{2}.\sqrt{10000}=0,1.100.\sqrt{2}=10\sqrt{2}\)
d) \(-0,05\sqrt{28800}=-0,05\sqrt{400}.\sqrt{9}.\sqrt[]{4}.\sqrt{2}=-0,05.20.3.2.\sqrt{2}=-6\sqrt{2}\)
e) \(\sqrt{7.63.a^2}=21\sqrt{a^2}\left\{{}\begin{matrix}=-21a\\=21a\end{matrix}\right.\)
đưa thừa số ra ngoài dấu căn :
\(\sqrt{18b^3\left(1-2a\right)^2}\)( a≥\(\dfrac{1}{2}\); b ≥0)
\(\sqrt{18b^3\cdot\left(1-2a\right)^2}\)
\(=3\sqrt{2}\cdot b\sqrt{b}\cdot\left|1-2a\right|\)
\(=3\sqrt{2}\left(2a-1\right)\cdot b\sqrt{b}\)
Đưa thừa số ra ngoài dấu căn:
\(\sqrt{34+24\sqrt{2}}\)
\(\sqrt{34+24\sqrt{2}}=\sqrt{18+2.3\sqrt{2}.4+16}\)
\(=\sqrt{\left(3\sqrt{2}+4\right)^2}=3\sqrt{2}+4\)
Đưa thừa số vào trong dấu căn:\(\dfrac{2+2\sqrt{5}}{3-\sqrt{5}}\).\(\sqrt{\dfrac{24-8\sqrt{5}}{3+3\sqrt{5}}}\)
Đưa thừ số ra ngoài dấu căn
\(a.\sqrt{27x^2}
b.\sqrt{8xy^2
}
c.\sqrt{25x^3}d.\sqrt{48xy^4}\)
Help
a) \(\sqrt{27x^2}\)
\(=\sqrt{3^2\cdot3x^2}\)
\(=\left|3x\right|\sqrt{3}\)
\(=3\left|x\right|\sqrt{3}\)
b) \(\sqrt{8xy^2}\)
\(=\sqrt{2^2\cdot2\cdot x\cdot y^2}\)
\(=\left|2y\right|\sqrt{2x}\)
\(=2\left|y\right|\sqrt{2x}\)
c) \(\sqrt{25x^3}\)
\(=\sqrt{5^2\cdot x^2\cdot x}\)
\(=\left|5x\right|\sqrt{x}\)
\(=5\left|x\right|\sqrt{x}\)
d) \(\sqrt{48xy^4}\)
\(=\sqrt{4^2\cdot3x\cdot\left(y^2\right)^2}\)
\(=\left|4y^2\right|\sqrt{3x}\)
\(=4y^2\sqrt{3x}\)
`a, sqrt(27x^2b) = sqrt(3^2. 3.x^2b) = 3|x|sqrt(3b)`.
`b, sqrt(8xy^2) =sqrt(2^2.2xy^2)= 2|y|sqrt(2x)`
`c, sqrt(25x^3d) = sqrt(5^2.x^2.x.d) = 5|x|sqrt(xd)`.
`d, sqrt(48xy^4) = sqrt(4^2.3 . xy^4) = 4y^2sqrt(3x)`.
Đưa thừa số ra ngoài dấu căn
a) \(6\left(\sqrt{15}-4\right)\sqrt{\frac{31+8\sqrt{15}}{12}}\)
b) \(\frac{2+2\sqrt{5}}{3-\sqrt{5}}\sqrt{\frac{24-8\sqrt{5}}{3+3\sqrt{5}}}\)
_Trục căn thức ở mẫu
c) \(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
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