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\(\frac{\overline{ab}}{b}=\frac{\overline{bc}}{c}=\frac{\overline{ca}}{a}=\frac{10a+b}{b}=\frac{10b+c}{c}=\frac{10c+a}{a}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{\overline{ab}}{b}=\frac{\overline{bc}}{c}=\frac{\overline{ca}}{a}=\frac{10a+b}{b}=\frac{10b+c}{c}=\frac{10c+a}{a}=\frac{11.\left(a+b+c\right)}{a+b+c}=11\)
\(\frac{10a+b}{b}=11\Rightarrow10a+b=11b\Rightarrow10a=10b\Rightarrow a=b\)(1)
\(\frac{10b+c}{c}=11\Rightarrow10b+c=11c\Rightarrow10b=10c\Rightarrow b=c\)(2)
\(\frac{10c+a}{a}=11\Rightarrow10c+a=11a\Rightarrow10c=10a\Rightarrow c=a\)(3)
từ (1), (2), (3) => a=b=c (đpcm)
1) Ta có : \(\frac{2016a+b+c+d}{a}=\frac{a+2016b+c+d}{b}=\frac{a+b+2016c+d}{c}=\frac{a+b+c+2016d}{d}\)
Trừ 4 vế với 2015 ta được : \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Nếu a + b + c + d = 0
=> a + b = -(c + d)
=> b + c = (-a + d)
=> c + d = -(a + b)
=> d + a = (-b + c)
Khi đó M = (-1) + (-1) + (-1) + (-1) = - 4
Nếu a + b + c + d\(\ne0\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\Rightarrow a=b=c=d\)
Khi đó M = 1 + 1 + 1 + 1 = 4
2) a) Ta có : \(\hept{\begin{cases}\left|x+2013\right|\ge0\forall x\\\left(3x-7\right)^{2004}\ge0\forall y\end{cases}\Rightarrow\left|x+2013\right|+\left(3x-7\right)^{2014}\ge0}\)
Dấu "=" xảy ra \(\hept{\begin{cases}x+2013=0\\3y-7=0\end{cases}\Rightarrow\hept{\begin{cases}x=-2013\\y=\frac{7}{3}\end{cases}}}\)
b) 72x + 72x + 3 = 344
=> 72x + 72x.73 = 344
=> 72x.(1 + 73) = 344
=> 72x = 1
=> 72x = 70
=> 2x = 0 => x = 0
c) Ta có :
\(\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{5}{x+4}\Leftrightarrow\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{10}{2x+8}=\frac{7-10}{2x+2-2x-8}=\frac{1}{2}\)(dãy tỉ số bằng nhau)
=> 2x + 2 = 14 => x = 6 ;
2y - 4 = 6 => y = 5 ;
6 + 5 + z = 17 => z = 6
Vậy x = 6 ; y = 5 ; z = 6
3) a) Ta có : \(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)(dãy ti số bằng nhau)
=> a + b + c = a + b - c => a + b + c - a - b + c = 0 => 2c = 0 => c = 0;
Lại có : \(\frac{a+b+c}{a+b-c}-1=\frac{a-b+c}{a-b-c}-1\Leftrightarrow\frac{2c}{a+b-c}=\frac{2c}{a-b-c}\Rightarrow a+b-c=a-b-c\) => b = 0
Vậy c = 0 hoặc b = 0
c) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b+b+c+a+c}{c+a+b}=2\)(dãy tỉ số bằng nhau)
=> \(\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}\)
Khi đó P = \(\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{b}{a}\right)=\frac{b+c}{b}.\frac{c+a}{c}=\frac{a+b}{a}=\frac{2a.2b.2c}{abc}=8\)
Vậy P = 8
2. b) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)
Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)
\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\frac{a}{2009}=\frac{b}{2011}=\frac{a-b}{2009-2011}=\frac{a-b}{-2}\)
\(\frac{b}{2011}=\frac{c}{2013}=\frac{b-c}{2011-2013}=\frac{b-c}{-2}\)
\(\frac{a}{2009}=\frac{c}{2013}=\frac{a-c}{2009-2013}=\frac{a-c}{-4}\)
=> \(\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
=> \(\frac{a-b}{-2}.\frac{b-c}{-2}=\left(\frac{a-c}{4}\right)^2\)
=> \(\frac{\left(a-c\right)^2}{4^2}=\frac{\left(a-b\right)\left(b-c\right)}{4}\)
=> \(\frac{\left(a-c\right)^2}{4}=\left(a-c\right)\left(b-c\right)\)
Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)
\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
Ta gọi 3 số lần lượt là a , b , c
Theo đề bài ta có :
\(\left\{\begin{matrix}\frac{a}{\frac{2}{5}}=\frac{b}{\frac{3}{4}}=\frac{c}{\frac{1}{6}}\\a^2+b^2+c^2=24309\end{matrix}\right.\)
Ta có \(\frac{a}{\frac{2}{5}}=\frac{b}{\frac{3}{4}}=\frac{c}{\frac{1}{6}}\)
\(\Leftrightarrow\frac{a^2}{\left(\frac{2}{5}\right)^2}=\frac{b^2}{\left(\frac{3}{4}\right)^2}=\frac{c^2}{\left(\frac{1}{6}\right)^2}\)
\(\Leftrightarrow\frac{a^2}{\frac{4}{25}}=\frac{b^2}{\frac{9}{16}}=\frac{c^2}{\frac{1}{36}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a^2}{\frac{4}{25}}=\frac{b^2}{\frac{9}{16}}=\frac{c^2}{\frac{1}{36}}=\frac{a^2+b^2+c^2}{\frac{2701}{3600}}=\frac{24309}{\frac{2701}{3600}}=32400\)
\(\Rightarrow\left\{\begin{matrix}\frac{a}{\frac{2}{5}}=32400\\\frac{b}{\frac{3}{4}}=32400\\\frac{c}{\frac{1}{6}}=32400\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}a=32400.\frac{2}{5}=12960\\b=32400.\frac{3}{4}=24300\\c=32400.\frac{1}{6}=5400\end{matrix}\right.\)
\(\Rightarrow A=12960+24300+5400=42660\)
Vậy số A = 42660
Ta có : a/c=c/b
=> c^2=a.b (1)
Cm:a/b=a^2+c^2/b^2+c^2 (2)
Từ (1),(2) suy ra :
a^2+c^2/b^2+c^2=a^2+a.b/b^2+a.b=a(a+b)/b(b+a)=a/b
Vậy a/b = a^2+c^2/b^2+c^2 (đpcm)
\(\frac{1a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}\)
Không xác định vì không thể chia cho 0
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Leftrightarrow a=\frac{1}{2}\left(b+c\right);b=\frac{1}{2}\left(c+a\right);c=\frac{1}{2}\left(a+b\right)\)
\(\Leftrightarrow a+b=c+\frac{1}{2}b+\frac{1}{2}a\Leftrightarrow\frac{1}{2}a+\frac{1}{2}b=c\left(1\right)\)
\(b+c=a+\frac{1}{2}c+\frac{1}{2}b\Leftrightarrow\frac{1}{2}b+\frac{1}{2}c=a\left(2\right)\)
\(c+a=b+\frac{1}{2}a+\frac{1}{2}c\Leftrightarrow\frac{1}{2}a+\frac{1}{2}c=b\left(3\right)\)
Từ (1);(2) và (3)
=> a=b=c (đpcm)