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Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{2a-3b}{2a+3b}=\dfrac{2bk-3b}{2bk+3b}=\dfrac{2k-3}{2k+3}\)
\(\dfrac{2c-3d}{2c+3d}=\dfrac{2dk-3d}{2dk+3d}=\dfrac{2k-3}{2k+3}\)
=>\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
b: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
=>\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
c: \(\left(\dfrac{a-b}{c-d}\right)^4=\left(\dfrac{bk-b}{dk-d}\right)^4=\left(\dfrac{b}{d}\right)^4\)
\(\dfrac{a^4+b^4}{c^4+d^4}=\dfrac{b^4k^4+b^4}{d^4k^4+d^4}=\dfrac{b^4}{d^4}\)
Do đó: \(\left(\dfrac{a-b}{c-d}\right)^4=\dfrac{a^4+b^4}{c^4+d^4}\)
3.
Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Leftrightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}\) và \(a+2b-3c=-20\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)
+) \(\dfrac{a}{2}=5\Rightarrow a=5.2=10\)
+) \(\dfrac{2b}{6}=5\Rightarrow2b=5.6=30\Rightarrow b=30:2=15\)
+) \(\dfrac{3c}{12}=5\Rightarrow3c=5.12=60\Rightarrow c=60:3=20\)
Vậy ...
3.
ta có:\(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)=\(\dfrac{c}{4}\)=>\(\dfrac{a}{2}\)=\(\dfrac{2b}{6}\)=\(\dfrac{3c}{12}\) và a+2b-3c=-20
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{a}{2}\)=\(\dfrac{2b}{6}\)=\(\dfrac{3c}{12}\)=\(\dfrac{a+2b-3c}{2+6-12}\)\(\dfrac{-20}{-4}\)=5
vì\(\dfrac{a}{2}\)=5=>a=2.5=10
\(\dfrac{2b}{6}\)=5=>2b=5.6=30=>b=30:2=15
\(\dfrac{3c}{12}\)=5=>3c=5.12=60=>c=60:3=20
vậy a=10,b=15,c=20
chúc bạn hok tốt
Câu 1:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a^2}{c^2}=\dfrac{b^2k^2}{d^2k^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{2a^2+3b^2}{2c^2+3d^2}=\dfrac{2b^2k^2+3b^2}{2d^2k^2+3d^2}=\dfrac{b^2}{d^2}\)
=>\(\dfrac{a^2}{c^2}=\dfrac{2a^2+3b^2}{2c^2+3d^2}\)
b: \(\dfrac{2a-3c}{c}=\dfrac{2bk-3dk}{dk}=\dfrac{2b-3d}{d}\)
ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\Rightarrow\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}\)
\(\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}=\dfrac{2a^4}{2b^4}=\dfrac{3b^4}{3c^4}=\dfrac{4c^4}{4d^4}=\dfrac{4d^4}{4e^4}\\ =\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}\\ \dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)
Bài 1:
$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:
\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)
$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)
Từ $(1);(2)$ suy ra đpcm.
Bài 2:
Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:
$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)
Áp dụng dãy tỉ số bằng nhau ta có:
\(\frac{a}{2b+c}=\frac{b}{2c+a}=\frac{c}{2a+b} =\frac{a+b+c}{3(a+b+c)}=\frac{1}{3} \)
=>a=3(2b+c)
=>b=3(2c+a)
=>c=3(2a+b)
=> A=\(\frac{2b+c}{a}+\frac{2c+a}{b}+\frac{2a+b}{c}=\frac{2b+c}{3(2b+c)} +\frac{2c+a}{3(2c+a)}+\frac{2a+b}{3(2a+b)} \)=\(\frac{1}{3}+\frac{1}3{}+\frac{1}3{} \)=1
Tham khảo thêm thôi chứ mình không chắc nhé! dạng này mình chưa từng gặp (hay có gặp nhưng rất ít). Thôi không dài dòng nữa. Vào bài thôi.
Giải
Theo t/c dãy tỉ số bằng nhau: \(\dfrac{a}{2b+3c}=\dfrac{b}{2c+3a}=\dfrac{c}{2a+3b}=\dfrac{a+b+c}{2b+3c+2c+3a+2a+3b}\)
\(=\dfrac{a+b+c}{\left(2b+3b\right)+\left(2c+3c\right)+\left(2a+3a\right)}=\dfrac{a+b+c}{5b+5c+5a}\) (*)
Từ (*) ta có: \(\dfrac{a}{2b+3c}=\dfrac{b}{2c+3a}=\dfrac{c}{2a+3b}=\dfrac{a+b+c}{5b+5c+5a}=\dfrac{1}{5}\)
Vì: \(5.\dfrac{a}{2b+3c}=5.\dfrac{b}{2c+3a}=5.\dfrac{c}{2a+3b}=\dfrac{5a+5b+5c}{5b+5c+5a}=1\)
Mà \(1:5=\dfrac{1}{5}\)
\(\Leftrightarrow5a\left(2b+3c\right)=5b\left(2c+3a\right)=5c\left(2a+3b\right)\)
\(\Leftrightarrow10ab+15ac=10bc+15ba=10ca+15cb\Leftrightarrow a=b=c^{\left(đpcm\right)}\)
b/
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
* \(\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2b+c=3a\\2c+a=3b\\2a+b=3c\end{matrix}\right.\)
+)\(\Rightarrow\left\{{}\begin{matrix}c=3a-2b\\a=3b-2c\\b=3c-2a\end{matrix}\right.\)
\(\Rightarrow\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)=abc\left(1\right)\)
+) \(\Rightarrow\left\{{}\begin{matrix}2b=3c-a\\2c=3b-a\\2a=3c-b\end{matrix}\right.\)
\(\Rightarrow\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)=8abc\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{abc}{8abc}=\dfrac{1}{8}\)
\(\Rightarrow P=\dfrac{1}{8}\)
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\dfrac{1}{a}+\dfrac{1}{c}\ge\dfrac{4}{a+c}\)
\(\dfrac{4}{a+b}+\dfrac{4}{a+c}\ge4\left(\dfrac{4}{a+b+a+c}\right)=\dfrac{16}{2a+b+c}\)
\(\Rightarrow\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{16}{2a+b+c}\)
Tương tự ta có:
\(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}\ge\dfrac{16}{a+2b+c}\) ; \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\ge\dfrac{16}{a+b+2c}\)
Cộng vế:
\(4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{16}{2a+b+c}+\dfrac{16}{a+2b+c}+\dfrac{16}{a+b+2c}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{2a+b+c}+\dfrac{4}{a+2b+c}+\dfrac{4}{a+b+2c}\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có:
\(VP=\dfrac{4}{2a+b+c}+\dfrac{4}{2b+a+c}+\dfrac{4}{2c+a+b}\)
\(\le\dfrac{1}{2a}+\dfrac{1}{b+c}+\dfrac{1}{2b}+\dfrac{1}{c+a}+\dfrac{1}{2c}+\dfrac{1}{a+b}\)
\(=\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{4}{b+c}\right)+\dfrac{1}{2b}+\dfrac{1}{4}\left(\dfrac{4}{c+a}\right)+\dfrac{1}{2c}+\dfrac{1}{4}\left(\dfrac{4}{a+b}\right)\)
\(\le\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{1}{2b}+\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+\dfrac{1}{2c}+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(=\dfrac{1}{2a}+\dfrac{1}{4b}+\dfrac{1}{4c}+\dfrac{1}{2b}+\dfrac{1}{4c}+\dfrac{1}{4a}+\dfrac{1}{2c}+\dfrac{1}{4a}+\dfrac{1}{4b}\)
\(=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(=VT\)
Ta có đpcm. Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Chú ý: Trong bài ta đã sử dụng bất đẳng thức \(\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}\) với \(x,y>0\) hai lần