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1)
a) \(\left(-48\right)^3:16^3\)
\(=\left(-48:16\right)^3\)
\(=\left(-3\right)^3\)
\(=-27.\)
b) \(\left(\frac{9}{10}\right)^6:\left(\frac{17}{-20}\right)^6\)
\(=\left(\frac{9}{10}:\frac{17}{-20}\right)^6\)
\(=\left(-\frac{18}{17}\right)^6\)
Chúc em học tốt!
\(\frac{-13^3}{\left(2^3\right)^3}:\frac{\left(-2^5\right)^4}{13^4}\)1.
a, (-48)3:163
= \(\left(\frac{-48}{16}\right)^3\)
= (-3)3
b,\(\left(\frac{9}{10}\right)^6\):\(\left(\frac{17}{-20}\right)^6\)
= \(\left(\frac{9}{10}:\frac{17}{-20}\right)^6\)
=\(\left(\frac{-18}{17}\right)^6\)
c, \(\left(\frac{-13}{8}\right)^3:\left(\frac{-32}{13}\right)^4\)
= \(\frac{-13^3}{\left(2^3\right)^3}:\frac{\left(-2^5\right)^4}{13^4}\)
= \(\frac{-13^3}{2^9}.\frac{-13^4}{2^{20}}\)
=\(\frac{13^7}{2^{29}}\)
Đặt \(A=\left(\frac{a}{b}+1\right)\left(\frac{b}{c}+1\right)\left(\frac{c}{d}+1\right)\left(\frac{d}{a}+1\right)\)
\(\frac{-a+b+c+d}{a}=\frac{a-b+c+d}{b}=\frac{a+b-c+d}{c}=\frac{a+b+c-d}{d}=\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\)( tc dãy tỉ số bằng nhau )
\(\Rightarrow\hept{\begin{cases}-a+b+c+d=2a\\a-b+c+d=2b\\a+b-c+d=2c\end{cases}}\)và \(a+b+c-d=2d\)
\(\Rightarrow\hept{\begin{cases}a+b+c+d=4a\\a+b+c+d=4b\\a+b+c+d=4c\end{cases}}\)và \(a+b+c+d=4d\)
\(\Rightarrow4a=4b=4c=4d\)
\(\Rightarrow a=b=c=d\)thay vào bt A ta được:
\(A=2.2.2.2=16\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3=\left(a+b\right)^3+3\left(a+b\right)c\left(a+b+c\right)-a^3-b^3.\)\(=3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)=3\left(a+b\right)\left(ab+ac+bc+c^2\right)=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
#)Giải :
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b+c\right)^3-a^3\right]-\left(b^3+c^3\right)\)
\(=\left(a+b+c-a\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b-c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ca+a^2+ab+ac+a^2\right)-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+2bc+b^2+c^2\right)-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+2ab+b^2+c^2-b^2+bc-c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+3bc\right)\)
\(=3\left(b+c\right)\left(a^2+ab+ac+bc\right)\)
\(=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)
\(=3\left(b+c\right)\left(a+b\right)\left(a+c\right)\Rightarrowđpcm\)
\(\text{a)Để C đạt GTNN}\)
\(\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\\\left(y-\frac{1}{5}\right)^2\end{cases}\ge0}\)
\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)
\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge0-10\)
\(\Rightarrow C\ge-10\)
\(\text{Vậy minC=-10 khi x=-2;y= }\frac{1}{5}\)
b)\(\text{Để D đạt GTLN}\)
=>(2x-3)2+5 đạt GTNN
Mà (2x-3)2\(\ge\)5
\(\Rightarrow GTLN\)của \(A=\frac{4}{5}\)khi \(x=\frac{3}{2}\)
`VT = (b-c)/((a-b)(a-c)) + (c-a)/((b-c)(b-a)) +(a-b)/((c-a)(c-b)) = 2/(a-b) + 2/(b-c) + 2/(c-a)`
`=-((a-b-a+c)/((a-b)(a-c))+(b-c-b+a)/((b-c)(b-a))+(c-a-c+b)/((c-a)(c-b)))`
`=-((a-b)/((a-b)(a-c))-(a-c)/((a-b)(a-c))+(b-c)/((b-c)(b-a))-(b-a)/((b-c)(b-a))+(c-a)/((c-a)(c-b))-(c-b)/((c-a)(c-b)))`
`= 1/(c-a)+1/(a-b)+1/(a-b)+1/(b-c)+1/(b-c)+1/(c-a)`
`=2/(a-b)+2/(b-c)+2/(c-a)=VP(đpcm)`
a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)
\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)
Mà \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\le0\)
Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)
- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)
Vậy ...
b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;
hàiiii chán quá