2(x+1)+(x+1)^2
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\(\text{x3 - 19x - 30}=x^3+2x^2-4x-15x-30\)
\(=x^2\left(x+2\right)-2x\left(x+2\right)-15\left(x+2\right)\)
\(=\left(x^2-2x-15\right)\left(x+2\right)\)
\(=\left[x^2-5x+3x-15\right]\left(x+2\right)\)
\(=\left[x\left(x-5\right)+3\left(x-5\right)\right]\left(x+2\right)\)
\(=\left(x+3\right)\left(x-5\right)\left(x+2\right)\)
\(x^3-19x+30\)
\(=x^3-9x-10x+30\)
\(=x\left(x^2-9\right)-10\left(x-3\right)\)
\(=x\left(x-3\right)\left(x+3\right)-10\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x-10\right)\)
\(=\left(x-3\right)\left(x^2-2x+5x-10\right)\)
\(=\left(x-3\right)\left[x\left(x-2\right)+5\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(x-3\right)\left(x+5\right)\)
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Số gà bác An nuôi là:
28 + 72 = 100 (con)
Trung bình số gà mỗi bác nuôi là:
(100 + 12) : 1 = 112 (con)
Số gà bác Minh nuôi là:
112 + 12 = 124 (con)
Số gà trống bác Minh nuôi là:
(124 - 8) : 2 = 58 (con)
Đáp số: 58 con gà trống
k mình nhé!!!!
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\(3=a+b+c\ge3\sqrt[3]{abc}\)\(\Leftrightarrow\)\(abc\le1\)
\(VT=\frac{a^3\left(a+1\right)+b^3\left(b+1\right)+c^3\left(c+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}=\frac{a^4+b^4+c^4+a^3+b^3+c^3}{a+b+c+ab+bc+ca+abc+1}\)
\(\ge\frac{\frac{\left(a^2+b^2+c^2\right)^2}{3}+\frac{\left(a^2+b^2+c^2\right)^2}{a+b+c}}{\frac{\left(a+b+c\right)^2}{3}+5}=\frac{\frac{\frac{\left(a+b+c\right)^4}{9}}{3}+\frac{\frac{\left(a+b+c\right)^4}{9}}{3}}{8}\)
\(=\frac{\frac{\frac{3^4}{9}}{3}}{4}=\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
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Mk giải hộ
5^255 = 5^(3*75) = (5^3)^75 = 125^75
2^572 > 2^525
2^525 = 2^(7*75) = (2^7)^75 = 128^75
=> 2^572 > 128^75
128^75 > 125^75
=> 2^572 > 2^525 > 125^75
Đáp số:
2^572 > 5^255
Ta có : 54 > 29 ( vì 625 > 512 )
=> ( 54 )64 > ( 29 )64
=> 5256 > 2576
=> 5256 : 5 > 2576 : 5 > 2576 : 8
=> 5255 > 2576 : 23
=> 5255 > 2573 > 2572
=> 5255 > 2572
Vậy 5255 > 2572
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Ta có \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\ge\frac{\left(ab+bc+ac\right)^2}{ab+bc+ac}=ab+bc+ac\left(1\right)\)
Áp dụng bất đẳng thức buniacoxki ta có :
\(\left(\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\right)\left(ab+bc+ac\right)\ge\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)^2\)
Kết hợp với (1)
=> \(\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c
![](https://rs.olm.vn/images/avt/0.png?1311)
#)Giải :
\(2\left(x+1\right)+\left(x+1\right)^2=\left(2x+2-1\right)\left(2x+2+1\right)=\left(\left(2x+2\right)^2-1\right)=4\left(x+1\right)^4\)