Cho 2 số a và b thỏa mãn a+b=38 và \(\dfrac{2a}{3}\)=\(\dfrac{5b}{2}\). Vậy 3a-2b bằng...
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Ta có :\(\dfrac{2a}{3}=\dfrac{5b}{2}\Rightarrow\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{2}{5}}\) và \(a+b=38\)
Áp dụng t/c của dãy tỉ số = nhau ta có:
\(\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{2}{5}}=\dfrac{a+b}{\dfrac{3}{2}+\dfrac{2}{5}}=\dfrac{38}{\dfrac{19}{10}}=20\)
\(\Rightarrow\left[{}\begin{matrix}a=20.\dfrac{3}{2}\\b=20.\dfrac{2}{5}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=30\\b=8\end{matrix}\right.\)
\(\Rightarrow3a-2b=3.30-2.8=74\)
Vậy...........................
Ta có : \(\dfrac{2a}{3}=\dfrac{5b}{2}\)
\(\Rightarrow4a=15b\\ \Rightarrow\dfrac{a}{15}=\dfrac{b}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có :
\(\dfrac{a}{15}=\dfrac{b}{4}=\dfrac{a+b}{15+4}=\dfrac{38}{19}=2\\ \Rightarrow\left\{{}\begin{matrix}a=2\cdot15=30\\b=2\cdot4=8\end{matrix}\right.\)
Vậy \(3a-2b=3\cdot30-2\cdot8=90-16=74\)
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`a^2+4ab-5b^2=0`
`<=>a^2+4ab+4b^2-9b^2=0`
`<=>(a+2b)^2-9b^2=0`
`<=>(a+2b-3b)(a+2b+3b)=0`
`<=>(a-b)(a+5b)=0`
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=-5b\end{matrix}\right.\)
`Q={2a-b}/{a-b}+{3a-2b}/{a+b}`
Với `a=b` `=>` giá trị vô nghĩa
Với `a=-5b`
`Q={-10b-b}/{-5b-b}+{-15b-2b}/{-5b+b}`
`Q={-11b}/{-6b}+{-17b}/{-4b}`
`Q=11/6+17/4`
`Q=73/12`
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bđt Schwarz ta có:
\(P=\dfrac{a^4}{2ab+3ac}+\dfrac{b^4}{2cb+3ab}+\dfrac{c^4}{2ac+3bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(ab+bc+ca\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(a^2+b^2+c^2\right)}=\dfrac{1}{5}\).
Đẳng thức xảy ra khi và chỉ khi \(a=b=c=\dfrac{\sqrt{3}}{3}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{ab}{2b}\right)\)
\(=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
Tương tự:
\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{b}{2}\right)\)
\(\dfrac{ac}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ac}{b+c}+\dfrac{ac}{a+b}+\dfrac{c}{2}\right)\)
Cộng vế:
\(P\le\dfrac{1}{9}\left(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{a+b+c}{2}\right)\)
\(P\le\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
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\(\frac{2a}{a+b}+\frac{b}{a-b}=2< =>2\left(a-b\right)a+b\left(a+b\right)=2\left(a-b\right)\left(a+b\right).\)
\(< =>2a^2-2ab+ab+b^2=2a^2-2b^2\)
\(< =>3b^2-ab=0< =>b\left(3b-a\right)=0=>\orbr{\begin{cases}b=0\\3b-a=0\end{cases}}\)\(< =>\orbr{\begin{cases}b=0\\a=3b\end{cases}=>\orbr{\begin{cases}A=3\\A=1\end{cases}}}\)
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1) \(\left\{{}\begin{matrix}a^3+b^3+c^3=3abc\\a+b+c\ne0\end{matrix}\right.\) \(\left(a;b;c\in R\right)\)
Ta có :
\(a^3+b^3+c^3\ge3abc\) (Bất đẳng thức Cauchy)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=1\left(a^3+b^3+c^3=3abc\right)\)
Thay \(a=b=c\) vào \(P=\dfrac{a^2+2b^2+3c^2}{3a^2+2b^2+c^2}\) ta được
\(\Leftrightarrow P=\dfrac{6a^2}{6a^2}=1\)
\(3^x=y^2+2y\left(x;y>0\right)\)
\(\Leftrightarrow3^x+1=y^2+2y+1\)
\(\Leftrightarrow3^x+1=\left(y+1\right)^2\left(1\right)\)
- Với \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
\(pt\left(1\right)\Leftrightarrow3^0+1=\left(0+1\right)^2\Leftrightarrow2=1\left(vô.lý\right)\)
- Với \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
\(pt\left(1\right)\Leftrightarrow3^1+1=\left(1+1\right)^2=4\left(luôn.luôn.đúng\right)\)
- Với \(x>1;y>1\)
\(\left(y+1\right)^2\) là 1 số chính phương
\(3^x+1=\overline{.....1}+1=\overline{.....2}\) không phải là số chính phương
\(\Rightarrow\left(1\right)\) không thỏa với \(x>1;y>1\)
Vậy với \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\) thỏa mãn đề bài
74
Ta có : \(\dfrac{2a}{3}=\dfrac{5b}{2}=\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{2}{5}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{2}{5}}=\dfrac{a+b}{\dfrac{3}{2}+\dfrac{2}{5}}=\dfrac{38}{\dfrac{19}{10}}=20\)
(vì a+b=38 )
Với \(\dfrac{a}{\dfrac{3}{2}}=20\) thì a=30
Với \(\dfrac{b}{\dfrac{2}{5}}=20\) thì b=8
Vậy b=8 ;a=30