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Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=2018k\\b=2019k\\c=2020k\end{matrix}\right.\)
\(\Rightarrow\left(a-c\right)^3=\left(2018k-2020k\right)^3=\left(-2k\right)^3=-8k^3\) (1)
\(8\left(a-b\right)^2.\left(b-c\right)=8\left(2018k-2019k\right)^2.\left(2019k-2020k\right)=8k^2\left(-k\right)=8\left(-k\right)^3=-8k^3\left(2\right)\)
Từ (1) và (2) ⇒ \(\left(a-c\right)^3=8\left(a-b\right)^2.\left(b-c\right)\left(đpcm\right)\)
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) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)
a/ \(\left(3x-1\right)^6=\left(3x-1\right)^4\Rightarrow\left(3x-1\right)=\left\{-1;0;1\right\}\)
\(\Rightarrow x=\left\{0;\frac{1}{3};\frac{2}{3}\right\}\)
b/
\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=\frac{a+b-c+a-b+c-a+b+c}{a+b+c}=1\)
\(\Rightarrow\frac{a+b-c}{c}=1\Rightarrow a+b=2c\)
Tương tự
\(b+c=2a;a+c=2b\)
\(\Rightarrow M=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2c.2a.2b}{abc}=8\)
Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\)
\(\Rightarrow a=2018k\), \(b=2019k\), \(c=2020k\)
Ta có: \(4\left(a-b\right)\left(b-c\right)=4\left(2018k-2019k\right)\left(2019k-2020k\right)\)
\(=4.\left(-k\right).\left(-k\right)=4k^2=\left(2k\right)^2\)
Ta lại có: \(\left(a-c\right)^2=\left(2018k-2020k\right)^2=\left(-2k\right)^2=\left(2k\right)^2\)
Vậy \(4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)
Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\Rightarrow\hept{\begin{cases}a=2018k\\b=2019k\\c=2020k\end{cases}}\)
Thế vị trí tương ứng ta được :
VT = 4( a - b )( b - c )
= 4( 2018k - 2019k )( 2019k - 2020k )
= 4(-k)(-k)
= 4k2
VP = ( a - c )2
= ( 2018k - 2020k )2
= ( -2k )2
= 4k2
=> VT = VP
=> đpcm
a, b, c khác 0
Áp dụng tính chất dãy tỉ số bằng nhau. Ta có:
\(\frac{a+3b-c}{c}\)=\(\frac{-a+b+3c}{a}\) =\(\frac{c-b+3a}{b}\)=\(\frac{a+3b-c-a+b+3c+c-b+3a}{a+b+c}=\frac{3a+3b+3c}{a+b+c}=3\)
=> \(\frac{a+3b-c}{c}=3\Rightarrow\frac{a+3b}{c}-\frac{c}{c}=3\Rightarrow\frac{a+3b}{c}=4\)
\(\frac{-a+b+3c}{a}=3\Rightarrow-1+\frac{b+3c}{a}=3\Rightarrow\frac{b+3c}{a}=4\)
\(\frac{c-b+3a}{b}=3\Rightarrow\frac{c+3a}{b}-\frac{b}{b}=3\Rightarrow\frac{c+3a}{b}=4\)
=> P =\(\left(3+\frac{a}{b}\right).\left(3+\frac{b}{c}\right).\left(3+\frac{c}{a}\right)=\frac{3b+a}{b}.\frac{3c+b}{c}.\frac{3a+c}{a}\)
= \(\frac{a+3b}{c}.\frac{b+3c}{a}.\frac{c+3a}{b}=4.4.4=64\)
a,\(\frac{-2}{5}+\frac{7}{21}=\frac{-2}{5}+\frac{1}{3}=\frac{-6}{15}+\frac{5}{15}=\frac{-1}{15}\)
b,\(\left(\frac{1}{3}\right)^5.3^5-2020^0=\left(\frac{1}{3}.3\right)^5-1=1^5-1=1-1=0\)
c,\(\left(-\frac{1}{4}\right).6\frac{2}{11}+3\frac{9}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right).\left(6\frac{2}{11}+3\frac{9}{11}\right)=\left(-\frac{1}{4}\right).\left[\left(6+3\right)+\left(\frac{2}{11}+\frac{9}{11}\right)\right]\)
\(=\left(-\frac{1}{4}\right).\left[9+1\right]=\frac{-1}{4}.10=\frac{\left(-1\right).10}{4}=\frac{\left(-1\right).5}{2}=\frac{-5}{2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a-b}{3}=\frac{b+c}{6}=\frac{c-a}{7}=\frac{a-b+b+c+c-a}{3+6+7}=\frac{2c}{16}=\frac{c}{8}\)
mà \(\frac{b+c}{6}=\frac{c-a}{7}=\frac{\left(b+c\right)-\left(c-a\right)}{6-7}=\frac{b+c-c+a}{-1}=-\left(a+b\right)\)
\(\Rightarrow\frac{c}{8}=-\left(a+b\right)\)\(\Rightarrow c=-8\left(a+b\right)\)
Ta có: \(P=c+8\left(a+b\right)-2020=-8\left(a+b\right)+8\left(a+b\right)-2020=-2020\)
Ta có :\(\frac{a-b}{3}=\frac{b+c}{6}=\frac{c-a}{7}=\frac{a-b+b+c-c+a}{3+6-7}=\frac{2a}{2}=a\)(1)(dãy tỉ số bằng nhau)
\(\frac{a-b}{3}=\frac{b+c}{6}=\frac{c-a}{7}=\frac{a-b-b-c+c-a}{3-6+7}=\frac{-2b}{4}=-\frac{b}{2}\)(2)(dãy tỉ số bằng nhau)
\(\frac{a-b}{3}=\frac{b+c}{6}=\frac{c-a}{7}=\frac{a-b+b+c+c-a}{3+6+7}=\frac{2c}{16}=\frac{c}{8}\)(3)(dãy tỉ số bằng nhau)
Từ (1)(2)(3) => \(\frac{a}{1}=\frac{-b}{2}=\frac{c}{8}\)
Đựt \(\frac{a}{1}=\frac{-b}{2}=\frac{c}{8}=k\Rightarrow\hept{\begin{cases}a=k\\b=-2k\\c=8k\end{cases}}\)
Khi đó P = c + 8(a + b) - 2020 = 8k + 8(k - 2k) - 2020 = 8k - 8k - 2020 = -2020
Vậy P = -2020