Cho a-b = 5. Tính \(\dfrac{4a-b}{3a+5}+\dfrac{3b-a}{2b-5}\)
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11 tháng 11 2023
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: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
30 tháng 8 2017
a, Theo bài ta có :
\(\dfrac{a}{b}=\dfrac{10}{3}\Leftrightarrow\dfrac{a}{10}=\dfrac{b}{3}\)
Đặt :
\(\dfrac{a}{10}=\dfrac{b}{3}=k\left(k\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=10k\\b=3k\end{matrix}\right.\)
Ta có :
\(Q=\dfrac{3a-2b}{a-3b}=\dfrac{3.10k-2.3k}{10k-3.3k}=\dfrac{30k-6k}{10k-9k}=\dfrac{24k}{1k}=24\)
Vậy ...........
30 tháng 8 2017
a-b=3=>a=b+3 Thay a=b+3 vào B
\(\Rightarrow B=\dfrac{b+3-8}{b-5}-\dfrac{4\left(b+3\right)-b}{3\left(b+3\right)+3}\)
\(\Rightarrow B=1-\dfrac{4b-b+12}{3b+9+3}=1-1=0\)
HT
1
7 tháng 7 2022
4a-b=6 nên b=4a-6
\(\dfrac{6a-b}{3a+5}-\dfrac{4a-4b}{3b-5}\)
\(=\dfrac{6a-\left(4a-6\right)}{3a+5}-\dfrac{4a-4\left(4a-6\right)}{3\left(4a-6\right)-5}\)
\(=\dfrac{6a-4a+6}{3a+5}-\dfrac{4a-16a+24}{12a-18-5}\)
\(=\dfrac{2a+6}{3a+5}-\dfrac{-12a+24}{12a-23}\)
\(=\dfrac{2a+6}{3a+5}+\dfrac{12a-24}{12a-23}\)
\(=\dfrac{\left(2a+6\right)\left(12a-23\right)+\left(12a-24\right)\left(3a+5\right)}{\left(3a+5\right)\left(12a-23\right)}\)
\(=\dfrac{24a^2-46a+72a-138+36a^2+60a-72a-120}{\left(3a+5\right)\left(12a-23\right)}\)
\(=\dfrac{60a^2+14a-258}{\left(3a+5\right)\left(12a-23\right)}\)
HH
1
YN
8 tháng 4 2022
`Answer:`
a. Ta có: \(\frac{a}{b}=\frac{1}{3}\Rightarrow\frac{a}{1}=\frac{b}{3}\)
Đặt \(k=\frac{a}{1}=\frac{b}{3}\Rightarrow\hept{\begin{cases}a=k\\b=3k\end{cases}}\)
\(E=\frac{3a+2b}{4a-3b}\)
\(=\frac{3k+2.3k}{4k-3.3k}\)
\(=\frac{3k+6k}{4k-9k}\)
\(=\frac{9k}{-5k}\)
\(=-\frac{9}{5}\)
b. Thay `a-b=5` vào biểu thức `F`, ta được:
\(F=\frac{3a-\left(a-b\right)}{2a+b}-\frac{4b+\left(a-b\right)}{a+3b}\)
\(=\frac{3a-a+b}{2a+b}-\frac{4b+a-b}{a+3b}\)
\(=\frac{2a+b}{2a+b}-\frac{3b+a}{a+3b}\)
\(=1+1\)
\(=0\)
24 tháng 12 2021
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Leftrightarrow\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-2b=c\\3b-2c=a\\3c-2a=b\end{matrix}\right.\text{ và }\left\{{}\begin{matrix}3a-c=2b\\3b-a=2c\\3c-b=2a\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{a\cdot b\cdot c}{2a\cdot2b\cdot3c}=\dfrac{1}{8}\)
B
1
LA
4 tháng 2 2023
\(đk:a;b\ne\dfrac{5}{3}\)
\(\dfrac{3b-28}{3a-5}-\dfrac{38-3a}{5-3b}=\dfrac{3b-28}{3\left(11+b\right)-5}-\dfrac{38-3\left(11+b\right)}{5-3b}=1-1=0\)
5 tháng 11 2021
Vì \(a,b,c>0\Rightarrow a+b+c\ne0\)
Áp dụng tc dtsbn:
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-2b=c\\3b-2c=a\\3c-2a=b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-c=2b\\3b-a=2c\\3c-b=2a\end{matrix}\right.\\ \Rightarrow P=\dfrac{abc}{2a\cdot2b\cdot2c}=\dfrac{1}{8}\)
AH
Akai Haruma
Giáo viên
9 tháng 2 2018
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{a^3}{2b+3c}+\frac{b^3}{2c+3a}+\frac{c^3}{2a+3b}=\frac{a^4}{2ab+3ac}+\frac{b^4}{2bc+3ba}+\frac{c^4}{2ac+3bc}\)
\(\geq \frac{(a^2+b^2+c^2)^2}{2ab+3ac+2bc+3ba+2ac+3bc}=\frac{(a^2+b^2+c^2)^2}{5(ab+bc+ac)}\)
Theo hệ quả của BĐT AM-GM ta có:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)(ab+bc+ac)}{5(ab+bc+ac)}=\frac{a^2+b^2+c^2}{5}\)
Ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c\)
HP
3
18 tháng 6 2017
Ta có : \(\frac{4a-b}{3a+5}=\frac{3a+\left(a-b\right)}{3a+5}=\frac{3a+5}{3a+5}=1\)
\(\frac{3b-a}{2b-5}=\frac{2b+b-a}{2b-5}=\frac{2b-a+b}{2b-5}=\frac{2b-\left(a-b\right)}{2b-5}=\frac{2b-5}{2b-5}=1\)
Nên : \(\frac{4a-b}{3a+5}+\frac{3b-a}{2b-5}=1+1=2\)
Giải:
Ta có: \(a-b=5\Leftrightarrow a=b+5\)
\(\dfrac{4a-b}{3a+5}+\dfrac{3b-a}{2b-5}=\dfrac{4b+20-b}{3b+15+5}+\dfrac{3b-b-5}{2b-5}\)
\(=\dfrac{3b+20}{3b+20}+\dfrac{2b-5}{2b-5}=1+1=2\)
Vậy...
ta có : a-b=5 => a=b+5 khi đó pt trên trở thành:
\(\dfrac{3a+a-b}{3a+5}+\dfrac{2b+b-a}{2b+5}=\dfrac{3a+5}{3a+5}+\dfrac{2b+5}{2b+5}=1+1=2\)
vậy ......