Cho a,b,c >1 Chứng minh rằng \(\frac{a}{\sqrt{b}-1}\)+ \(\frac{b}{\sqrt{c}-1}\)+ \(\frac{c}{\sqrt{a}-1}\)> 12
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\(x^2-6x+10\)
\(=x^2-6x+9+1\)
\(=\left(x-3\right)^2+1\ge1\forall x\)
Mà 1>0
\(\Rightarrow x^2-6x+10\) luôn dương \(\forall x\left(đpcm\right)\)
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\(2^3.19-2^3.14+1^{2018}\)
= \(8.19-8.14+1\)
= \(8.\left(19-14\right)+1\)
= \(8.5+1\)
= \(40+1\)
= \(41\)
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\(ab\left(a-b\right)-ac\left(a+c\right)+bc\left(2a-b+c\right)\)
\(=a^2b-ab^2-a^2c-ac^2+2abc-b^2c+bc^2\)
\(=a^2b-ab^2-a^2c-ac^2+abc+abc-b^2c+bc^2\)
\(=\left(bc^2-ac^2+abc-a^2c\right)-\left(b^2c-abc-ab^2+a^2b\right)\)
\(=c\left(bc-ac+ab-a^2\right)-b\left(bc-ac-ab+a^2\right)\)
\(=\left(c-b\right)\left(bc-ac+ab-a^2\right)\)
\(=\left(c-b\right)\left[c\left(b-a\right)+a\left(b-a\right)\right]\)
\(=\left(c-b\right)\left(c+a\right)\left(b-a\right)\)
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\(x^2-3\left|x\right|-4=0\)
\(\Leftrightarrow3\left|x\right|=x^2-4\)
\(\Leftrightarrow3x=\pm\left(x^2-4\right)\)
\(\Leftrightarrow x^2-3x-4=0\) hoặc \(x^2+3x-4=0\)
Ta giải 2 phương trình này được \(s=\left\{-4;4\right\}\)
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\(x^2-3\left|x\right|-4=0\)
\(\Leftrightarrow x^2-3\left|x\right|=4\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3x=4\\x^2-3\left(-x\right)=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3x-4=0\\x^2+3x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-4\right)\left(x+1\right)=0\\\left(x+4\right)\left(x-1\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\pm4\left(tm\right)\\x=\pm1\left(ktm\right)\end{cases}}\)
\(\Rightarrow x=\pm4\)
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\(B=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=5^{32}-1< 5^{32}\)
Vậy \(B< A\)