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Ta có \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)
=> \(\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\)
=> \(\left(\frac{a}{b}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\)
=> \(\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\left(\frac{a+b+c}{b+c+d}\right)^3\)
=> \(\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\left(\frac{a+b+c}{b+c+d}\right)^3\) (Vì \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\))
=> \(\frac{a}{d}=\left(\frac{a+b+c}{b+c+d}\right)^3\)(đpcm)
Theo đầu bài ta có:
\(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
Do \(a+b+c=259\Rightarrow\hept{\begin{cases}a=259-\left(b+c\right)\\b=259-\left(a+c\right)\\c=259-\left(a+b\right)\end{cases}}\)
Từ đó suy ra:
\(\Leftrightarrow Q=\frac{259-\left(b+c\right)}{b+c}+\frac{259-\left(a+c\right)}{a+c}+\frac{259-\left(a+b\right)}{a+b}\)
\(\Leftrightarrow Q=\left(\frac{259}{b+c}-\frac{b+c}{b+c}\right)+\left(\frac{259}{a+c}-\frac{a+c}{a+c}\right)+\left(\frac{259}{a+b}-\frac{a+b}{a+b}\right)\)
\(\Leftrightarrow Q=\left(259\cdot\frac{1}{b+c}+259\cdot\frac{1}{a+c}+259\cdot\frac{1}{a+b}\right)-\left(\frac{b+c}{b+c}+\frac{a+c}{a+c}+\frac{a+b}{a+b}\right)\)
\(\Leftrightarrow Q=259\cdot\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-\left(1+1+1\right)\)
Do \(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}=15\) nên:
\(\Leftrightarrow Q=259\cdot15-3\)
\(\Leftrightarrow Q=3882\)
a=259-(b+c)
b=259-(c+a)
c=259-(a+b)
Thay vào Q
Q=259-(a+b)/a+b+259-(b+c)/b+c+259-(c+a)/c+a
Q=259/a+b+259/b+c+259/c+a-3
Q=259.(1/a+b+1/c+a+1/b)+c-3
Q=259x15-3
Q=3882
\(A=\left|x+\frac{2}{3}\right|\)
Ta có: \(\left|x+\frac{2}{3}\right|\ge0\forall x\)
\(A=0\Leftrightarrow\left|x+\frac{2}{3}\right|=0\Leftrightarrow x=-\frac{2}{3}\)
Vậy \(A_{min}=0\Leftrightarrow x=-\frac{2}{3}\)
\(B=\left|x\right|+\frac{1}{2}\)
Ta có: \(\left|x\right|\ge0\forall x\)
\(\Rightarrow\left|x\right|+\frac{1}{2}\ge\frac{1}{2}\forall x\)
\(B=\frac{1}{2}\Leftrightarrow\left|x\right|=0\Leftrightarrow x=0\)
Vậy \(B_{min}=\frac{1}{2}\Leftrightarrow x=0\)
Câu c,d tương tự
P/S tất cả những bài trên chỉ tìm được min, ko tìm được max.
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Leftrightarrow6a-3b=2a+2b\)
\(\Leftrightarrow6a-2a=2b+3b\)
\(\Leftrightarrow4a=5b\)
\(\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Leftrightarrow4a-2b=3b-3c+3a\)
\(\Leftrightarrow4a-3a=3b-3c+2b\)
\(\Leftrightarrow a=5b-3c\)
\(\Leftrightarrow a=4a-3c\)
\(\Leftrightarrow3a=3c\)
\(\Rightarrow a=c\)
\(\Rightarrow P=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^5}{\left(8a\right)^2\left(4a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=\frac{8^3}{4^3}=2^3=8\)
tớ cũng đang học cái đấy nè
225^10
a = _________ = 3^10
75^10
2^15 * 3^8 2^15 * 3^8
c= _____________= -------------------- =2^3 * 3^5
2^3 * 3^3 * 2^9 2^12 * 3^3
Đặt \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}=k\left(k\in R\right)\)
\(\Rightarrow\hept{\begin{cases}a=2k\\b=5k\\c=7k\end{cases}}\)
Thay vào A ta được \(A=\frac{2k-5k+7k}{2k+2\cdot5k-7k}=\frac{4k}{5k}=\frac{4}{5}\)
Vậy A=\(\frac{4}{5}\)
Đặt a/2 = b/5 = c/7 = k => a = 2k
b = 5k
c = 7k
=> a - b + c / a + 2b - c = 2k - 5k + 7k / 2k + 2 * 5k - 7k = 4k / 5k = 4/5
Vậy A = 4/5