Tìm số a,b,c biết
1, a/4=b/3=b/9 và a-3b+4c=62
2, a/b=7/20, b/c=5/8 và 2a+5b-2c=1000
3, a/12=b/9=c/5 và abc=20
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Bài 5 :
a) \(\dfrac{y}{4}=\dfrac{9}{y}\)
\(\Rightarrow y^2=36\left(y\ne0\right)\)
\(\Rightarrow y=\pm6\)
b) \(\dfrac{y+7}{20}=\dfrac{5}{y+7}\left(y\ne-7\right)\)
\(\Rightarrow\left(y+7\right)^2=100=10^2\)
\(\Rightarrow\left[{}\begin{matrix}y+7=10\\y+7=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=3\\y=-17\end{matrix}\right.\)
c) \(\dfrac{4-5y}{3}=\dfrac{y+2}{5}\)
\(\Rightarrow5\left(4-5y\right)=3\left(y+2\right)\)
\(\Rightarrow20-25y=3y+6\)
\(\Rightarrow28y=14\)
\(\Rightarrow y=\dfrac{14}{28}=\dfrac{1}{2}\)
Bài 4 :
\(\dfrac{a}{5}=\dfrac{b}{7}=\dfrac{c}{10}\)
\(\Rightarrow\dfrac{2a}{10}=\dfrac{3b}{21}=\dfrac{4c}{40}=\dfrac{2a+3b-4c}{10+21-40}=\dfrac{81}{-9}=-9\)
\(\Rightarrow\left\{{}\begin{matrix}a=-9.5=-45\\b=-9.7=-63\\c=-9.10=-90\end{matrix}\right.\)
Tìm các số a, b, c biết rằng :
1 . Ta có: \(\frac{a}{20}=\frac{b}{9}=\frac{c}{6}=\frac{a}{20}=\frac{2b}{9.2}=\frac{4c}{6.4}=\frac{a}{20}=\frac{2b}{18}=\frac{4c}{24}\)
Ap dụng tính chất dãy tỉ số bắng nhau ta dược :
\(\frac{a}{20}=\frac{2b}{18}=\frac{4c}{24}\)=\(\frac{a-2b+4c}{20-18+24}=\frac{13}{26}=\frac{1}{3}\)( do x+2b+4c=13)
Nên : a/20=1/3\(\Leftrightarrow\) a=1/3.20 \(\Leftrightarrow\)a=20/3
b/9=1/3 \(\Leftrightarrow\) b=1/3.9 \(\Leftrightarrow\) b=3
c/6=1/3 \(\Leftrightarrow\) c=1/3.6 \(\Leftrightarrow\) c= 2
Sẵn tiện mk chỉ cho bn luôn dạng này nhé.
Phân tích:
Với \(\alpha,\beta,\gamma>0\) thỏa \(\alpha< 2,\beta< 3,\gamma< 4\) ta có:
\(A=2a+3b+4c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
\(=\left[\left(2-\alpha\right)a+\dfrac{3}{a}\right]+\left[\left(3-\beta\right)b+\dfrac{9}{2b}\right]+\left[\left(4-\gamma\right)c+\dfrac{4}{c}\right]+\left(\alpha a+\beta b+\gamma c\right)\)
\(\ge2\sqrt{3.\left(2-\alpha\right)}+2\sqrt{\dfrac{9}{2}.\left(3-\beta\right)}+2\sqrt{4.\left(4-\gamma\right)}+\left(\alpha a+\beta b+\gamma c\right)\)
Chọn \(\alpha,\beta,\gamma\) (thỏa đk trên) sao cho:
\(\left\{{}\begin{matrix}\left(2-\alpha\right)a=\dfrac{3}{a}\\\left(3-\beta\right)b=\dfrac{9}{2b}\\\left(4-\gamma\right)c=\dfrac{4}{c}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\sqrt{\dfrac{3}{2-\alpha}}\\b=\sqrt{\dfrac{9}{2\left(3-\beta\right)}}\\c=\sqrt{\dfrac{4}{\left(4-\gamma\right)}}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\sqrt{\dfrac{3}{2-\alpha}}\\b=\sqrt{\dfrac{9}{6-4\alpha}}\\c=\sqrt{\dfrac{4}{4-3\alpha}}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)
Ta có: \(a+2b+3c\ge20\). Xác định điểm rơi: \(a+2b+3c=20\)
\(\Rightarrow\sqrt{\dfrac{3}{2-\alpha}}+2\sqrt{\dfrac{9}{6-4\alpha}}+3\sqrt{\dfrac{4}{4-3\alpha}}=20\)
Giải ra ta có \(\alpha=\dfrac{5}{4}\Rightarrow\beta=\dfrac{5}{2};\gamma=\dfrac{15}{4}\)
Lời giải:
Ta có: \(A=2a+3b+4c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
\(=\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\left(\dfrac{5a}{4}+\dfrac{5b}{2}+\dfrac{15c}{4}\right)\)
\(\ge^{Cauchy}2\sqrt{\dfrac{3a}{4}.\dfrac{3}{a}}+2\sqrt{\dfrac{b}{2}.\dfrac{9}{2b}}+2\sqrt{\dfrac{c}{4}.\dfrac{4}{c}}+\dfrac{5}{4}\left(a+2b+3c\right)\)
\(=3+3+2+\dfrac{5}{4}\left(a+2b+3c\right)\)
\(\ge8+\dfrac{5}{4}.20=33\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3a}{4}=\dfrac{3}{a}\\\dfrac{b}{2}=\dfrac{9}{2b}\\\dfrac{c}{4}=\dfrac{4}{c}\\a+2b+3c=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
Vậy \(MinA=33\), đạt được khi \(a=2;b=3;c=4\)
a) a:b:c:d=2:3:4:5
<=>a/2=b/3=c/4=d/5=a+b+c+d/2+3+4+5=-42/14=-3
a/2=-3<=>a=-6
b/3=-3<=>b=-9
c/4=-3<=>c=-12
d/5=-3<=>d=-15
b)a/b=7/20<=>a/7=b/20
b/c=5/8 <=>b/5=c/8<=>b/20=c/32
<=>a/7=b/20=c/32=2a/14=5b/100=2c/64=2a+5b-2c/14+100+64=100/178=50/89
minh ko chac dung dau nha
1. 2a = 3b ; 5b =7c
Từ giả thiết 2a = 3b => \(\frac{a}{3}=\frac{b}{2}\)=>\(\frac{a}{3}.\frac{1}{7}=\frac{b}{2}.\frac{1}{7}=>\frac{a}{21}=\frac{b}{14}\)
5b = 7c => \(\frac{b}{7}=\frac{c}{5}=>\frac{b}{7}.\frac{1}{2}=\frac{c}{5}.\frac{1}{2}=>\frac{b}{14}=\frac{c}{10}\)
Do đó: \(\frac{a}{21}=\frac{b}{14}=\frac{c}{10}\) và 3a + 5c -7b = 30
Ta đặt \(\frac{a}{21}=\frac{b}{14}=\frac{c}{10}=k\)
Suy ra a= 21k, b= 14k, c= 10k
Theo giả thiết: 3a + 5c - 7b = 30 =>3.21k + 5.10k - 7.14k = 30
=>63k + 50k - 98k= 30 => 15k = 30=> k= 2
Vậy a = 21.2=42
b = 14.2= 28
c = 10.2=20.
2. Bạn giải như bài trên nha!