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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) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)
\(P=\frac{n-7+9}{n-7}=1+\frac{9}{n-7}\)
\(\left(\text{Để P}\right)max\Rightarrow\left(\frac{9}{n-7}\right)max\Rightarrow\left(n-7\right)min\text{ và }n-7>0\left(\text{vì }9>0\right)\)
n-7 min và n-7>0 => n-7=1 => n=8. Vậy MaxP=10
\(\hept{\begin{cases}b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\\c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\end{cases}}\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{abc}{bcd}=\frac{a}{d}\)
áp dụng t.c dtsbn:
\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{abc}{bcd}=\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(đpcm\right)\)
Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1
c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1
=> A = 1+bc+b/bc+b+1 = 1
Tk mk nha
BÀI 1:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\) (thay abc = 1)
\(=\frac{a+ab+1}{a+ab+1}=1\)
1) \(D=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+....+\frac{10}{1400}\)
\(D=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+.....+\frac{5}{700}\)
\(D=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+......+\frac{5}{25.28}\)
\(D=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+.....+\frac{3}{25.28}\right)\)
\(D=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{25}-\frac{1}{28}\right)\)
\(D=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}.\frac{6}{28}=\frac{5}{14}\)
\(E=\frac{1}{1+2}+\frac{1}{1+2+3}+.......+\frac{1}{1+2+3+....+24}\)
Ta có: \(1+2=\)\(\frac{2.\left(2+1\right)}{2}=3\);\(1+2+3=\frac{3.\left(3+1\right)}{2}=6\);\(1+2+3+...+24=\frac{24.\left(24+1\right)}{2}=300\)
\(E=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{300}\)
=>\(\frac{1}{2}E=\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{600}=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{24.25}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{24}-\frac{1}{25}=\frac{1}{2}-\frac{1}{25}=\frac{23}{50}\)
=>\(E=\frac{46}{50}\)
Vậy \(\frac{D}{E}=\frac{5}{14}:\frac{46}{50}=\frac{250}{644}=\frac{125}{322}\)
2) Theo t/c dãy tỉ số=nhau:
\(\frac{a+b}{a+c}=\frac{a-b}{a-c}=\frac{a+b-\left(a-b\right)}{a+c-\left(a-c\right)}=\frac{a+b-a+b}{a+c-a+c}=\frac{2b}{2c}=1\)
=>b=c
do đó \(A=\frac{10b^2+9bc+c^2}{2b^2+bc+2c^2}=\frac{10b^2+9b^2+b^2}{2b^2+b^2+2b^2}=\frac{\left(10+9+1\right).b^2}{\left(2+1+2\right).b^2}=4\)
Từ x-y=7
=>x=y+7
Thay x=y+7 vào B ta được:
\(B=\frac{3.\left(y+7\right)-7}{2.\left(y+7\right)+y}-\frac{3y+7}{2y+\left(y+7\right)}\)\(=\frac{3y+21-7}{2y+14+y}-\frac{3y+7}{3y+7}=\frac{3y+14}{3y+14}-\frac{3y+7}{3y+7}=1-1=0\)
Vậy B=0 khi x-y=7
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{b+c+d}{a}=\frac{a+c+d}{b}=\frac{a+b+d}{c}=\frac{a+b+c}{d}=k\)
\(=\frac{b+c+d+a+c+d+a+b+d+a+b+c}{a+b+c+d}\)
= \(\frac{3b+3c+3a+3d}{a+b+c+d}=\frac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)(Do a + b + c + d \(\ne\)0)
=> k = 3
Với k = 3 => M = (3 - 3)2019 = 0
ADTCCDTSBN Ta có
\(\frac{b+c+d}{a}+\frac{a+c+d}{b}+\frac{a+b+d}{c}+\frac{a+b+c}{d}\)
\(=\frac{b+c+d+a+c+d+a+b+d+a+b+c}{a+b+c+d}=3\)
\(=>k=3\)
Thay vào M Ta có:
\(M=\left(k-3\right)^{2019}=\left(3-3\right)^{2019}=0\)
\(=>M=0\)
P/S:Ko chắc~!!
Áp dụng TC của dãy tỉ số bằng nhau , ta có :
\(\frac{b+c+d}{a}=\frac{a+c+d}{b}=\frac{a+b+d}{c}=\frac{a+b+c}{d}\left(1\right)\)
\(=\frac{b+c+d+a+c+d+a+b+d+a+b+c}{a+b+c+d}\)
\(=\frac{3a+3b+3c+3d}{a+b+c+d}=\frac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)
Mà \(\left(1\right)=k\Rightarrow k=3\)
Ta có : \(M=\left(k-3\right)^{2019}\)
\(\Leftrightarrow M=\left(3-3\right)^{2019}\)
\(\Leftrightarrow M=0\)
còn ai nữa à =='
đk a,b,c,d khác 0
áp dugnj tc dãy tỉ số = nhau \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(=\frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2d}{a+b+c+d}=\frac{5\left(a+b+c+d\right)}{a+b+c+d}\)
+> nếu a+b+c+d =0\(\Rightarrow\hept{\begin{cases}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\end{cases}\hept{\begin{cases}d+a=-\left(b+c\right)\\\end{cases}}}\)\(\Rightarrow M=-4\)
+> a+b+c+d khác 0 \(\Rightarrow\frac{2a+b+c+d}{a}=5\Rightarrow b+c+d=3a\)
Tương tự ta có \(\hept{\begin{cases}a+b+c=3d\\a+c+d=3b\\a+b+d=3c\end{cases}}\)\(\Rightarrow a=b=c=d\)
Khi đó M=4
Vậy M=4 hoặc M=-4