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2/
a) Ta có:
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)
Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) Ta có:
\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)
\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)
Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)
3/
a)ĐKXĐ: \(x\ne1;x\ge0\)
b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)
\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)
\(A=1^2-\left(\sqrt{x}\right)^2\)
\(A=1-x\)
Mình ghi nhầm đề bài 1 tí đề bài là :
So sánh 2 số A và B biết :
A = (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) và B = 3^32 - 1
\(3,\\ a,=a^2+2a+1-a^2+2a-1-3a^2+3=-3a^2+4a+3\\ b,=\left(m^3-m+1-m^2+3\right)^2=\left(m^3-m^2-m+4\right)^2\\ 4,\\ a,\Leftrightarrow25x^2+10x+1-25x^2+9=3\\ \Leftrightarrow10x=-7\Leftrightarrow x=-\dfrac{7}{10}\\ b,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\\ c,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)
A= 80.(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A = (34 - 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A = (38 - 1)(38 + 1)(316 + 1)(332 + 1)
A = (316 - 1)(316 + 1)(332 + 1)
A = (332 - 1)(332 + 1)
A = 364 - 1 < 364 = B
=> A < B
\(A=4.\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=\frac{3^{32}-1}{2}< 3^{32}-1=B\)
Vậy \(A< B\)