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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) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)
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
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
TH1: Nếu a+b+c+d\(\ne\)0 thì theo tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2d}{a+b+c+d}\)\(=\frac{5a+5b+5c+5d}{a+b+c+d}=\frac{5\left(a+b+c+d\right)}{a+b+c+d}=5\)
<=> \(2a+b+c+d=5a;a+2b+c+d=5b;a+b+2c+d=5c;a+b+c+2d=5d\)
<=>\(b+c+d=3a;a+c+d=3b;a+b+d=3c;a+b+c=3d\)
=>\(b+c+d+a+c+d=3a+3b\Leftrightarrow a+b+2c+2d=3a+3b\)
<=>\(2c+2d=2a+2b\Leftrightarrow2\left(c+d\right)=2\left(a+b\right)\Leftrightarrow c+d=a+b\)
Chứng minh tương tự ta được b+c=d+a ; c+d=a+b ; d+a=b+c
=>\(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
TH2: a+b+c+d=0
\(\Leftrightarrow a+b=-\left(c+d\right);b+c=-\left(a+b\right);c+d=-\left(a+b\right);d+a=-\left(b+c\right)\)
\(\Rightarrow M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(d+a\right)}{d+a}+\frac{-\left(a+b\right)}{a+b}+\frac{-\left(b+c\right)}{b+c}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Vậy ........................
\(Giai\)
\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)
\(=\frac{a+b+c+d}{3\left(a+b+c+d\right)}=\frac{1}{3}\Rightarrow a=b=c=d\)
\(\Rightarrow M=\frac{a+b}{c+d}=\frac{b+c}{a+d}=\frac{c+d}{a+b}=\frac{d+a}{b+c}=\frac{a+b+b+c+c+d+d+a}{c+d+a+d+a+b+b+c}\)
\(=1?!?.Mknghĩ:M=a+b+c+d\left(chứ\right)\)
Ta có \(\frac{A}{B+C+D}\)=\(\frac{B}{A+C+D}\)=\(\frac{C}{D+B+A}\)=\(\frac{D}{B+C+A}\)
=>\(\frac{A}{B+C+D}\)+1=\(\frac{B}{A+C+D}\)+1=\(\frac{C}{D+B+A}\)+1=\(\frac{D}{B+C+A}\)+1
=>\(\frac{A+B+C+D}{B+C+D}\)=\(\frac{A+B+C+D}{A+C+D}\)=\(\frac{A+B+C+D}{D+B+A}\)=\(\frac{A+B+C+D}{A+B+C}\)
Nếu A+B+C+D=0
=>\(\hept{\begin{cases}A+B=-\left(C+D\right)\\B+C=-\left(A+D\right)\\D+A=-\left(C+B\right)\end{cases}}\)
=>M=(-1)+(-1)+(-1)+(-1)
=>M= -4
Nếu A+B+C+D khác 0
=>B+C+D=A+C+D=A+B+D=A+B+C
=>A=B=C=D
=>M=1+1+1+1=4
Vậy M= -4 hoặc M=4
Học tốt nha bạn
Đặt dãy tỷ số bằng nhau là (1)
\(\Rightarrow\left(1\right)=\frac{5\left(a+b+c+d\right)}{a+b+c+d}=5\)
\(\Rightarrow\left(1\right)=\frac{2\left(a+b\right)+3\left(c+d\right)}{c+d}=\frac{2\left(a+b\right)}{c+d}+3=5\Rightarrow\frac{\left(a+b\right)}{c+d}=1\)
Chứng minh tương tự ta tính và suy ra
\(\frac{b+c}{d+a}=\frac{c+d}{a+b}=\frac{d+a}{b+c}=1\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
Từ giả thiết suy ra:
\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
* Nếu a + b + c + d = 0 thì a + b = - ( c + d ); b + c = - ( d + a ); c + d = - ( a + b ); d + a = - ( b + c )
Khi đó M = ( - 1 ) + ( - 1 ) + ( - 1 ) + ( - 1 ) = - 4
* Nếu a + b + c + d \(\ne0\) thì \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\)nên a = b = c = d
Khi đó M = 1 + 1 + 1 + 1 = 4
\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}=\frac{a+b+c+d}{3\left(a+b+c+d\right)}=\frac{1}{3}\)
=>3a=b+c+d
3b=a+c+d
3c=a+b+d
3d=a+b+c
=>a=b=c=d
=>\(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
a) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\)
\(\Rightarrow\frac{a}{b}+\frac{b}{b}=\frac{c}{d}+\frac{d}{d}\)
\(\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\left(đpcm1\right).\)
b) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\)
\(\Rightarrow\frac{a}{b}-\frac{b}{b}=\frac{c}{d}-\frac{d}{d}\)
\(\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\left(đpcm2\right).\)
c) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{b}{a}=\frac{d}{c}.\)
\(\Rightarrow\frac{b}{a}+1=\frac{d}{c}+1\)
\(\Rightarrow\frac{b}{a}+\frac{a}{a}=\frac{d}{c}+\frac{c}{c}\)
\(\Rightarrow\frac{b+a}{a}=\frac{d+c}{c}\left(đpcm3\right).\)
d) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{b}{a}=\frac{d}{c}.\)
\(\Rightarrow\frac{b}{a}-1=\frac{d}{c}-1\)
\(\Rightarrow\frac{b}{a}-\frac{a}{a}=\frac{d}{c}-\frac{c}{c}\)
\(\Rightarrow\frac{b-a}{a}=\frac{d-c}{c}\left(đpcm4\right).\)
Còn 2 câu kia tí nữa mình làm sau nhé.
Chúc bạn học tốt!
Ta có:\(\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{d+a+b}{c}=\frac{a+b+c}{d}\)
=>\(\frac{b+c+d}{a}+1=\frac{c+d+a}{b}+1=\frac{d+a+b}{c}+1=\frac{a+b+c}{d}+1\)
=>\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Vì các phân số trên có cùng tử. Nên các mẫu của phân số đó bằng nhau.
=>a=b=c=d
=>M=\(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)=\(\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}\)=1+1+1+1=4
Vậy M=4
\(\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{d+a+b}{c}=\frac{a+b+c}{d}=\frac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)
Vậy 3a= b+c+d 3b=c+d+a 3c=d+a+b 3d=a+b+c
Suy ra a=b=c=d
Thay vào ta có M=1+1+1+1=4
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