cho \(\frac{a}{b}\)=\(\frac{4}{7}\); \(4b^2\)-\(6a^2\)=49. GTNN của 3a + 2b là ?
trình bày ra giúp mình nha
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a) a/4 = b/7
=> a/4 + 1 = b/7 + 1
=> \(\frac{a+4}{4}=\frac{b+7}{7}\left(đpcm\right)\)
b) \(\frac{a}{4}=\frac{b}{7}\\ \Leftrightarrow\frac{a}{4}+2=\frac{b}{7}+2\\ \Leftrightarrow\frac{a+8}{4}=\frac{b+14}{7}\\ \Leftrightarrow\frac{a+8}{b+14}=\frac{4}{7}\left(đpcm\right)\)
Chúc bạn học tốt!
a/ \(\frac{a}{4}=\frac{4}{4}+\frac{a}{4}+1\) và \(\frac{b}{7}+\frac{7}{7}=\frac{b}{7}+1\)
mà a/4 =b/7 và 1=1 suy ra ĐPCM
\(TA-CO':\)
\(A=\frac{4+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}{7+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}\)
\(A=\frac{4\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}{7\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(A=\frac{4}{7}\)
\(B=\frac{1+2+...+2^{2013}}{2^{2015}-2}\)
ĐẶT \(C=1+2+...+2^{2013}\)
\(\Rightarrow2C=2+2^2+...+2^{2014}\)
\(\Rightarrow2C-C=\left(2+2^2+...+2^{2014}\right)-\left(1+2+...+2^{2013}\right)\)
\(\Rightarrow C=2^{2014}-2\)
\(\Rightarrow B=\frac{2^{2014}-1}{2^{2015}-2}\)
\(B=\frac{2^{2014}-1}{2\left(2^{2014}-1\right)}\)
\(B=\frac{1}{2}\)
\(\Rightarrow A-B=\frac{3}{7}-\frac{1}{2}=\frac{6}{14}-\frac{7}{14}\)
\(A-B=\frac{6-7}{14}=\frac{-1}{14}\)
VẬY, \(A-B=\frac{-1}{14}\)
cho a, b, c thỏa mãn a+b+c=79, \(\frac{a+7}{a-7}=\frac{b+4}{b-4},\frac{b+5}{b-5}=\frac{c+6}{c-6}\)
Ta có: \(\frac{a+7}{a-7}=\frac{b+4}{b-4}\Rightarrow\left(a+7\right)\left(b-4\right)=\left(a-7\right)\left(b+4\right)\)
\(\Rightarrow ab-4a+7b-28=ab+4a-7b-28\)
\(\Rightarrow-4a-4a+7b+7b=0\Rightarrow-8a+14b=0\)
\(\Rightarrow8a=14b\Rightarrow4a=7b\Rightarrow\frac{a}{7}=\frac{b}{4}\Rightarrow\frac{a}{35}=\frac{b}{20}\left(1\right)\)
Lại có: \(\frac{b+5}{b-5}=\frac{c+6}{c-6}\Rightarrow\left(b+5\right)\left(c-6\right)=\left(b-5\right)\left(c+6\right)\)
\(\Rightarrow bc-6b+5c-30=bc+6b-5c-30\)
\(\Rightarrow6b+6b=5c+5c\) => 12b = 10c
=>\(6b=5c\Rightarrow\frac{b}{5}=\frac{c}{6}\Rightarrow\frac{b}{20}=\frac{c}{24}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\frac{a}{35}=\frac{b}{20}=\frac{c}{24}=\frac{a+b+c}{35+20+24}=\frac{79}{79}=1\)
=>a=35,b=20,c=24
tìm n N để \(\frac{n}{n+1}\) + \(\frac{n}{n+2}\) là số tự nhiên
giúp mik với sắp thi r
a) Ta có :
\(A=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}\)
\(A=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\)
\(A=\frac{2}{7}\)
b) Ta có :
\(B=\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(B=\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(B=\frac{2}{7}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{2}{7}}{\frac{2}{7}}=1\)
Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\left(x;y>0\right)\) (tự c/m ha)
\(\frac{7}{a}+\frac{5}{b}+\frac{4}{c}=\left(\frac{4}{a}+\frac{4}{b}\right)+\left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{3}{a}+\frac{3}{c}\right)\)
\(=4\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{b}+\frac{1}{c}\right)+3\left(\frac{1}{a}+\frac{1}{c}\right)\)
\(\ge4.\frac{4}{a+b}+\frac{4}{b+c}+3.\frac{4}{a+c}=4\left(\frac{4}{a+b}+\frac{1}{b+c}+\frac{3}{c+a}\right)\)
Dấu "=" <=> a = b = c
3)
3/5 + 3/7-3/11 / 4/5 + 4/7- 4/11
= 3.( 1/5 + 1/7 - 1/11)/4.(1/5+1/7-1/11)
= 3/4
1,
ta có B = 196+197/197+198 = 196/(197+198) + 197/(197+198)
196/197 > 196/197+198
197/198 > 197/197+198
=> A>B
\(\frac{a}{b}=\frac{4}{7}\Rightarrow\frac{a}{4}=\frac{b}{7}\)
Gọi k là giá trị chung của các tỉ số
\(\frac{a}{4}=k\Rightarrow a=4k\) ; \(\frac{b}{7}=k\Rightarrow b=7k\)
Từ đó
\(4b^2-6a^2=49\)
\(\Rightarrow4.49k^2-6.16k^2=49\)
\(\Rightarrow196k^2-96k^2=49\)
\(\Rightarrow\left(196-96\right)k^2=49\)
\(\Rightarrow100k^2=49\)
\(\Rightarrow k^2=\frac{49}{100}\)
\(\Rightarrow k=-\frac{7}{10}\)hoặc \(k=\frac{7}{10}\)
với \(k=-\frac{7}{10}\) thì \(\frac{a}{4}=k\Rightarrow\frac{a}{4}=-\frac{7}{10}\Rightarrow a=-\frac{14}{5}\)
\(\frac{b}{7}=k\Rightarrow\frac{b}{7}=-\frac{7}{10}\Rightarrow b=-\frac{49}{10}\)
Với \(k=\frac{7}{10}\)thì \(\frac{a}{4}=k\Rightarrow\frac{a}{4}=\frac{7}{10}\Rightarrow a=\frac{14}{5}\)
\(\frac{b}{7}=k\Rightarrow\frac{b}{7}=\frac{7}{10}\Rightarrow b=\frac{49}{10}\)
Vậy \(a=-\frac{14}{5};b=-\frac{49}{10}\)và \(a=\frac{14}{5};b=\frac{49}{10}\)
Suy ra GTNN của \(3a+2b=3.\left(\frac{-14}{5}\right)+2.\left(-\frac{49}{10}\right)=-\frac{42}{5}+-\frac{49}{5}=-\frac{91}{5}\)