\(\dfrac{a}{2}\)= \(\dfrac{b}{3}\), \(\dfrac{b}{4}\)=\(\dfrac{c}{5}\) và a+b-2c
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Đặt a/2=b/3=c/4=k
=>a=2k; b=3k; c=4k
Ta có: \(a^2+3b^2-2c^2=-16\)
\(\Leftrightarrow4k^2+27k^2-32k^2=-16\)
\(\Leftrightarrow k^2=16\)
Trường hợp 1: k=4
=>a=8; b=12; c=16
Trường hợp 2: k=-4
=>a=-8; b=-12; c=-16
tham khảo!!
https://lazi.vn/edu/exercise/tim-cac-so-a-b-c-biet-rang-a-2-b-3-c-4-va-a-2-b-2-2c-2-108
Ta có:\(\dfrac{x^2}{4}=\dfrac{x}{2};\dfrac{y^2}{9}=\dfrac{y}{3};\dfrac{z^2}{25}=\dfrac{z}{5}\)
Aps dụng tính chất dãy tỉ số bằn nhau:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)
=>\(\dfrac{x}{2}=1=>x=2\)
\(\dfrac{y}{3}=1=>y=3\)
\(\dfrac{z}{5}=1=>z=5\)
Vậy x=2, y=3, z=5
Ta có : \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)
\(\Leftrightarrow x=2;y=3;z=5\)
\(\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow\dfrac{a}{8}=\dfrac{b}{12}\) (1)
\(\dfrac{b}{4}=\dfrac{c}{5}\Rightarrow\dfrac{b}{12}=\dfrac{c}{15}\) (2)
Từ (1) và (2) ta có: \(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\Rightarrow\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{2c}{30}=\dfrac{a+b-2c}{8+12-30}=\dfrac{10}{-10}=-1\)
Vậy \(\left\{{}\begin{matrix}a=-8\\b=-12\\c=-15\end{matrix}\right.\)
\(\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow\dfrac{a}{8}=\dfrac{b}{12}\left(1\right)\)
\(\dfrac{b}{4}=\dfrac{c}{5}\Rightarrow\dfrac{b}{12}=\dfrac{c}{15}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{2c}{15}=\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{30}\)
Theo bài ra ta có:
\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{30}\) và \(a+b-2c=10\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\Rightarrow\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{2c}{30}=\dfrac{a+b-2c}{8+12-30}=\dfrac{10}{-10}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{8}=-1\Rightarrow a=-8\\\dfrac{b}{12}=-1\Rightarrow b=-12\\\dfrac{c}{15}=-1\Rightarrow c=-15\end{matrix}\right.\)
Chúc bạn học tốt!
\(2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\right)\ge1+\dfrac{b}{b+1a}+\dfrac{c}{c+2b}+\dfrac{a}{a+2c}\)
\(\Leftrightarrow2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}+\dfrac{c}{a+2c}\right)\ge1+\dfrac{b+2a}{b+2a}+\dfrac{c+2b}{c+2b}+\dfrac{a+2c}{a+2c}=1+1+1+1=4\)Thật vậy:
\(\dfrac{a}{b+2c}+\dfrac{a}{b+2a}+\dfrac{b}{c+2a}+\dfrac{b}{c+2b}+\dfrac{c}{a+2b}+\dfrac{c}{a+2c}=a\left(\dfrac{1}{b+2c}+\dfrac{1}{b+2a}\right)+b\left(\dfrac{1}{c+2a}+\dfrac{1}{c+2b}\right)+c\left(\dfrac{1}{a+2b}+\dfrac{1}{a+2c}\right)\)
\(\ge\dfrac{4a}{2\left(a+b+c\right)}+\dfrac{4b}{2\left(a+b+c\right)}+\dfrac{4c}{2\left(a+b+c\right)}=2\)
\(\Rightarrow VT\ge2.2=4\)
\(\RightarrowĐPCM\)
Áp dụng BĐt cô-si, ta có \(\frac{2\left(a+b\right)^2}{2a+3b}\ge\frac{8ab}{2a+3b}=\frac{8}{\frac{2}{b}+\frac{3}{a}}\)
\(\frac{\left(b+2c\right)^2}{2b+c}\ge\frac{8bc}{2b+c}=\frac{8}{\frac{2}{c}+\frac{1}{b}}\)
\(\frac{\left(2c+a\right)^2}{c+2a}\ge\frac{8ac}{c+2a}\ge\frac{8}{\frac{1}{a}+\frac{2}{c}}\)
Cộng 3 cái vào, ta có
A\(\ge8\left(\frac{1}{\frac{2}{b}+\frac{3}{a}}+\frac{1}{\frac{1}{b}+\frac{2}{c}}+\frac{1}{\frac{1}{a}+\frac{2}{c}}\right)\ge8\left(\frac{9}{\frac{3}{b}+\frac{4}{c}+\frac{4}{a}}\right)=8.\frac{9}{3}=24\)
Vậy A min = 24
Neetkun ^^
Ta có :
(1) \(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)=>\(\dfrac{a}{2}\).\(\dfrac{1}{4}\)=\(\dfrac{b}{3}\).\(\dfrac{1}{4}\)=>\(\dfrac{a}{8}\)=\(\dfrac{b}{12}\)
(2)\(\dfrac{b}{4}\)=\(\dfrac{c}{5}\)=>\(\dfrac{b}{4}\).\(\dfrac{1}{3}\)=\(\dfrac{c}{5}\).\(\dfrac{1}{3}\)=>\(\dfrac{b}{12}\)=\(\dfrac{c}{15}\)
Từ (1),(2) =>\(\dfrac{a}{8}\)=\(\dfrac{b}{12}\)=\(\dfrac{c}{15}\)
Ta có:\(\dfrac{a}{8}\)giữ nguyên , \(\dfrac{b}{12}\)giữ nguyên, \(\dfrac{c}{15}\)=\(\dfrac{2c}{30}\)
Áp dụng T/C dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{8}\)=\(\dfrac{b}{12}\)=\(\dfrac{2c}{30}\)=>\(\dfrac{a}{8}+\dfrac{b}{12}-\dfrac{2c}{30}\)= ?
bạn ơi cái chỗ mà bạn ghi :"Và a+b-2c" cái câu này phải có đáp án sẵn r
vd:"a+b-2c=30" chứ bạn ghi "a+b-2c" chứ ko có đáp án ren tính đc bạn