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\(\sqrt{20}\cdot\sqrt{72}\cdot\sqrt{4,9}=\sqrt{20\cdot72\cdot4,9}=\sqrt{2\cdot10\cdot72\cdot4,9}\\ =\sqrt{144\cdot49}=\sqrt{144}\cdot\sqrt{49}=12\cdot7=84\)
Bài 2:
a) \(\sqrt{3a^3}\cdot\sqrt{12a}=\sqrt{3a^3\cdot12a}=\sqrt{36a^4}=6a^2\)
b) \(\sqrt{2a\cdot32ab^2}=\sqrt{64a^2b^2}=8ab\)
Ta có :
\(\sqrt{2a.32ab^2}\)
\(=\)\(\sqrt{64a^2b^2}\)
\(=\)\(\sqrt{8^2a^2b^2}\)
\(=\)\(\sqrt{\left(8ab\right)^2}\)
\(=\)\(\left|8ab\right|\)
Chúc bạn học tốt ~
a: \(=4x-4x\sqrt{2}-2x\sqrt{2}+2x=6x-6x\sqrt{2}\)
b: \(=6x-4\sqrt{xy}+3\sqrt{xy}-2y=6x-\sqrt{xy}-2y\)
\(a,\frac{a-4\sqrt{a}+4-1}{\sqrt{a}-3}=\frac{\left(\sqrt{a}-2\right)^2-1}{\sqrt{a}-3}.\)
\(=\frac{\left(\sqrt{a}-3\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-3}\)
\(=\sqrt{a}-1\)
\(b,\frac{a+\sqrt{a^2-6a+9}}{2a-3}=\frac{a+\sqrt{\left(a-3\right)^2}}{2a-3}\)
\(=\frac{a+a-3}{2a-3}=\frac{2a-3}{2a-3}\)
\(=1\)
Câu 1:
\(x^2-19=x^2-\left(\sqrt{19}^2\right)\left(x+\sqrt{19}\right)\)
Câu 2:
\(\sqrt{8t}.\sqrt{32t^3}=\sqrt{8t.32t^3}=\sqrt{\left(16.t^2\right)^2}=16.t^2\)
Câu 3 :
\(\sqrt{a^8\left(4-a\right)^2}=\sqrt{a.8}.\sqrt{\left(4-a\right)^2}=a^4\left|4-a\right|\)
( do \(a\le4\))
câu 1
\(x^2-19=\left(x-\sqrt{19}\right)\left(x+\sqrt{19}\right)\)
câu 2
\(\sqrt{8t}.\sqrt{32t^3}=\sqrt{8t.32t^3}=\sqrt{256t^4}=\sqrt{\left(16t^2\right)^2}=16t^2\)
câu 3
\(\sqrt{a^8\left(4-a\right)^2}=\sqrt{\left[a^4\left(4-a\right)\right]^2}=a^4\left(4-a\right)=4a^4-a^5\)
nếu mk sai thì bỏ qua nha <3
a: \(=1-\left(\sqrt{x}\right)^3=1-x\sqrt{x}\)
b: \(=\left(\sqrt{x}\right)^3+2^3=x\sqrt{x}+8\)
c: \(=\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3=x\sqrt{x}-y\sqrt{y}\)
d: \(=x^3+\left(\sqrt{y}\right)^3=x^3+y\sqrt{y}\)
a)sin a-sin a.cos^2 a=sin a(1-cos^2 a)=sin a(sin^2 a)=sin^3 a
b)sin^4a+cos^4a+2sin^2acos^2a=(sin^2a+cos^2a)^2=1^2=1
√(2a . 32ab2) = √(64a2b2 )
= √((8ab)2) = 8ab (do a ≥ 0; b ≥ 0)