cho n thuốc N. chứng minh nếu 4n^3+27 chia hết cho 3 thì n không chia hết cho 3 ( chứng minh bằng phản chứng ạ )
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a,b dể tự làm nha
c)ta có: \(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2-2ab\ge0\Leftrightarrow a^2+2ab+b^2-2ab-2ab\ge0\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\) mà a+b=1
\(\Rightarrow1\ge4ab\Leftrightarrow ab\le\frac{1}{4}\)
lại có: \(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2\ge2ab\) mà \(ab\le\frac{1}{4}\)
tahy vào có \(a^2+b^2\ge2\times\frac{1}{4}\Leftrightarrow a^2+b^2\ge\frac{1}{2}\left(dpcm\right)\)
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\(a,\sqrt{68^2-32^2}\)
\(=\sqrt{\left(68-32\right)\left(68+32\right)}\)
\(=\sqrt{3600}=60\)
\(b,\sqrt{37^2-12^2}\)
\(=\sqrt{\left(37-12\right)\left(37+12\right)}\)
\(=\sqrt{25.49}=5.7=35\)
\(c,\sqrt{21,8^2-18,2^2}\)
\(=\sqrt{\left(21,8+18,2\right)\left(21,8-18,2\right)}\)
\(=\sqrt{40.3,6}=12\)
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a) AM ứng với cạnh huyền BC nên AM = \(\frac{1}{2}\) x BC = \(\frac{4}{2}\) = 2 cm
AH = tan\(\widehat{ACH}\)x HM = tan 150 x 2 = \(4-2\sqrt{3}\)cm
Sin \(\widehat{AMH}\)= \(\frac{AH}{AM}\)= \(\frac{4-2\sqrt{3}}{2}\) = \(2-\sqrt{3}\) cm
Định lí Pitago : AM2 = AH2 + HM2
HC = tan \(\widehat{ACH}\)x AH
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\(\frac{x+5}{\sqrt{x}+1}=\frac{\left(x-1\right)+6}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+6}{\sqrt{x}+1}=\sqrt{x}-1+\frac{6}{\sqrt{x}+1}\)
\(=\sqrt{x}+1+\frac{6}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right)\frac{6}{\sqrt{x}+1}}-2=2\sqrt{6}-2\)(Cauchy)
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\(=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\right]:\left(\frac{x+1}{x-1}-\frac{x-1}{x-1}\right)\)
\(=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\right]:\frac{2}{x-1}\)
\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{x-1}{2}\)
\(=\left[\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{x-1}{2}\)
\(=\frac{x-\sqrt{x}-2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)
\(=\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)
\(=\frac{1}{\sqrt{x}+1}\le1\)
Dấu "=" xảy ra khi x=0
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