3/4(x2y)2:1/8xy2
Giúp e vs. E cảm ơn ạ
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\(\left(3\sqrt{7}\right)^2=63>28=\left(\sqrt{28}\right)^2\) hoặc \(3\sqrt{7}>2\sqrt{7}=\sqrt{28}\)
\(\lim\dfrac{\left(3n^2+1\right)\left(1-4n\right)}{n^3-2n+5}=\lim\dfrac{\left(3+\dfrac{1}{n^2}\right)\left(\dfrac{1}{n}-4\right)}{1-\dfrac{2}{n^2}+\dfrac{5}{n^3}}=\dfrac{3.\left(-4\right)}{1}=-12\)
\(\lim\dfrac{\sqrt[]{4n^2-1}+\sqrt[]{n^2-5}}{n+\sqrt[3]{n^3-2n^2}}=\lim\dfrac{\sqrt[]{4-\dfrac{1}{n^2}}+\sqrt[]{1-\dfrac{5}{n^2}}}{1+\sqrt[3]{1-\dfrac{2}{n}}}=\dfrac{\sqrt[]{4}+\sqrt[]{1}}{1+\sqrt[3]{1}}=\dfrac{5}{2}\)
\(\lim\dfrac{\left(3-n\right)^7\left(2+n\right)^3}{\left(n^2+1\right)\left(n^8+3\right)}=\lim\dfrac{\left(\dfrac{3}{n}-1\right)^7\left(\dfrac{2}{n}+1\right)^3}{\left(1+\dfrac{1}{n^2}\right)\left(1+\dfrac{3}{n^8}\right)}=\dfrac{\left(-1\right)^7.1^3}{1.1}=-1\)
Bài 2: Chọn C
Bài 4:
a: \(\widehat{C}=180^0-80^0-50^0=50^0\)
Xét ΔABC có \(\widehat{A}=\widehat{C}< \widehat{B}\)
nên BC=AB<AC
b: Xét ΔABC có AB<BC<AC
nên \(\widehat{C}< \widehat{A}< \widehat{B}\)
\(\dfrac{3}{4}\left(x^2y\right)^2:\dfrac{1}{8}xy^2\\ =\dfrac{3}{4}x^4y^2:\dfrac{1}{8}xy^2\\ =\left(\dfrac{3}{4}:\dfrac{1}{8}\right)\left(x^4:x\right)\left(y^2:y^2\right)\\ =6x^3\)
\(\dfrac{3}{4}\left(x^2y\right)^2\div\dfrac{1}{8}xy^2\)
\(=\dfrac{3}{4}x^4y^2\div\dfrac{1}{8}xy^2\)
\(=6x^3\)