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1) Ta có : Đặt M = 3x + 1 + 3x + 2 + ... + 3x + 100
= 3x(3 + 32 + ... + 3100)
= 3x[(3 + 32 + 33 + 34) + (35 + 36 + 37 + 38) + ... + (397 398 + 399 + 3100)]
= 3x[(3 + 32 + 33 + 34) + 34.(3 + 32 + 33 + 34) + ... + 396.(3 + 32 + 33 + 34)]
= 3x(120 + 34.120 + .... + 396.120)
= 3x.120.(1 + 34 + .... + 396)
=> \(M⋮120\)(ĐPCM)
2) Ta có \(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}\)
\(\Rightarrow\frac{3a+b+c}{a}-2=\frac{a+3b+c}{b}-2=\frac{a+b+3c}{c}-2\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
Nếu a + b + c = 0
=> a + b = - c
b + c = -a
c + a = -b
Khi đó P = \(\frac{-c}{c}+\frac{-a}{a}+\frac{-b}{b}=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)
Nếu a + b + c \(\ne\)0
=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
Khi đó P = \(\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}=2+2+2=6\)
Vậy nếu a + b + c = 0 thì P = -3
nếu a + b + c \(\ne\)0 thì P = 6
Ta có :
\(3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}\)
\(=\left(3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4}\right)+...\)\(+\left(3^{x+97}+3^{x+98}+3^{x+99}+3^{x+100}\right)\)
\(=3^x\left(3+3^2+3^3+3^4\right)+...+3^{x+96}\left(3+3^2+3^3+3^4\right)\)
\(=3^x.120+3^{x+4}.120+...+3^{x+96}.120\)
\(=120.\left(3^x+3^{x+4}+...+3^{x+96}\right)\)
Vì \(120⋮120\)
\(\Rightarrow120.\left(3^x+3^{x+4}+...+3^{x+96}\right)⋮120\)
\(\Rightarrow3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}⋮120\left(\forall x\inℕ\right)\left(đpcm\right)\)
a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)
Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)
b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)
\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)
\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)
\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)
\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)
Do đó: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)
\(\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
\(\Rightarrow a=b=c\)
\(\Rightarrow\frac{b}{a}=1;\frac{a}{c}=1;\frac{c}{b}=1\)
\(\Rightarrow B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
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) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)
Bài 1:
\(\left(x-1\right).\left(xy-5\right)=5\)
Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-1\in Z\\xy-5\in Z\end{matrix}\right.\)
\(\Rightarrow x-1\inƯC\left(5\right);xy-5\inƯC\left(5\right)\)
\(\Rightarrow x-1\in\left\{\pm1;\pm5\right\};xy-5\in\left\{\pm1;\pm5\right\}.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=1\\xy-5=5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=5\\xy-5=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-1\\xy-5=-5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-5\\xy-5=-1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\2y=10\end{matrix}\right.\\\left\{{}\begin{matrix}x=6\\6y=6\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\0y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\-4y=4\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\left(TM\right)\\\left\{{}\begin{matrix}x=6\\y=1\end{matrix}\right.\left(TM\right)\\\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\left(TM\right)\\\left\{{}\begin{matrix}x=-4\\y=-1\end{matrix}\right.\left(TM\right)\end{matrix}\right.\)
Vậy cặp số nguyên \(\left(x;y\right)\) thỏa mãn đề bài là: \(\left(2;5\right),\left(6;1\right),\left(0;0\right),\left(-4;-1\right).\)
Chúc bạn học tốt!
Bài 1:
Ta có bảng sau:
| \(x-1\) | -5 | -1 | 1 | 5 |
| \(xy-5\) | -1 | -5 | 5 | 1 |
| x | -4 | 0 | 2 | 6 |
| y | -1 | mọi y∈Z | 5 | 1 |
Áp dụng ta đc:
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}=\frac{5a+5b+5c}{a+b+c}=5\left(\text{vì: a,b,c khác 0}\right)\)
\(\Rightarrow\hept{\begin{cases}b+c=2a\\c+a=2b\\a+b=2c\end{cases}}\Rightarrow a=b=c\)
\(\Rightarrow P=6\)
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}\)
\(\Rightarrow\frac{3a+b+c}{a}-2=\frac{a+3b+c}{b}-2=\frac{a+b+3c}{c}-2\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
Xét \(a+b+c\ne0\)
\(\Rightarrow a=b=c\)
Thay vào P ta được P=6
Xét \(a+b+c=0\)
\(\Rightarrow a+b=-c;b+c=-a;a+c=-b\)
Thay vào P ta được P= -3
Vậy P có 2 gtri là ...........
\(\Rightarrow3+\frac{y+z-2x}{x}=3+\frac{x+z-2y}{y}=3+\frac{x+y-2z}{z}\)
\(\Rightarrow\frac{x+y+z}{x}=\frac{x+y+z}{y}=\frac{x+y+z}{z}\)
\(TH1:x+y+z=0\)
\(\Rightarrow x=-\left(y+z\right),y=-\left(x+z\right),z=-\left(x+y\right)\)
\(A=\left(1+\frac{-y-z}{y}\right).\left(1+\frac{-x-z}{z}\right).\left(1+\frac{-x-y}{x}\right)\)
\(A=-\left(\frac{z}{y}\cdot\frac{x}{z}\cdot\frac{y}{x}\right)=-1\)
\(TH2:x+y+z\ne0\)
\(\Rightarrow x=y=z\Rightarrow A=2^3=8\)
sai đề ròi: tớ làm 2 trường hợp luôn vì trường hợp x+y+z khác 0 thì A mới t/m thuộc N
mà đề là x+y+z khác 0 -.-
Bài 1:
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}.\)
\(\Rightarrow\frac{3a}{a}+\frac{b+c}{a}=\frac{3b}{b}+\frac{a+c}{b}=\frac{3c}{c}+\frac{a+b}{c}\)
\(\Rightarrow3+\frac{b+c}{a}=3+\frac{a+c}{b}=3+\frac{a+b}{c}\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}.\)
+ TH1: \(a+b+c=0.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Ta có: \(P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
\(\Rightarrow P=\frac{-c}{c}+\frac{-a}{a}+\frac{-b}{b}\)
\(\Rightarrow P=\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(\Rightarrow P=-3.\)
+ TH2: \(a+b+c\ne0.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=\frac{b+c+a+c+a+b}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=\frac{2.\left(a+b+c\right)}{a+b+c}=2.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{b+c}{a}=2\\\frac{a+c}{b}=2\\\frac{a+b}{c}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\a+b=2c\end{matrix}\right.\)
Lại có: \(P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
\(\Rightarrow P=\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}\)
\(\Rightarrow P=2+2+2\)
\(\Rightarrow P=6.\)
Vậy \(P=-3\) hoặc \(P=6.\)
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Cảm ơn bạn ♥