ngoctruong236

$cho a,b,c la 3 số thực ko am thoa man a^2+4b^2+9c^2=14.CMR:3b+8c+abc\leq 12$$cho a,b,c la 3 số thực ko am thoa man a^2+4b^2+9c^2=14.CMR:3b+8c+abc\leq 12$

$ap dụng bdt cauchy-schward ta coVT\geq \frac{a^2+b^2+c^2}^2{\sum a^3+\sum 2a^2b^2}.Cần Cm Bdt\geq 1\rightarrow phai cm \sum a^4\geq \sum a^3.vi a+b+c=3\rightarrow dpcm$

$ta co :4^{a}-2008=4^a-1-2007.Ta co4^a-1luon chia het cho 3,2007 chia het cho 3\rightarrow 4^a-2008 luon chia het cho 3.Mat khac 4^a-2008 chia het cho 2\rightarrow 4^a-2008 chia het cho 6.ta co 4^a-2008=4^a+a+b-(a+1+b+2007).Tu day suy ra4^a+a+b chia het cho 6(dpcm)$

$\sqrt{\frac{b+c}{a}} +\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\geq \sqrt{\frac{16(a+b+c)^{3}}{3(a+b)(b+c)(a+c)}}$

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1. x^3-x^2-x=1/3tuong duong voi 3x^3-3x^2-3x=1  tuong duong voi 4x^3=(x+1)^3ta tim ra dc x ok nhe ban
2.