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Từ \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
\(\Rightarrow\frac{y+z-x}{x}+2=\frac{z+x-y}{y}+2=\frac{x+y-z}{z}+2\)
\(\Rightarrow\frac{x+y+z}{x}=\frac{x+y+z}{y}=\frac{x+y+z}{z}\left(1\right)\)
*)Xét \(x+y+z\ne0\left(2\right)\). Từ (1) và (2)
\(\Rightarrow x=y=z\). Khi đó \(B=\frac{x+y}{y}\cdot\frac{y+z}{z}\cdot\frac{x+z}{x}=2\cdot2\cdot2=8\)
*)Xét \(x+y+z=0\)\(\Rightarrow\left\{\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
Khi đó \(B=\frac{x+y}{y}\cdot\frac{y+z}{z}\cdot\frac{x+z}{x}=\frac{-z}{y}\cdot\frac{-x}{z}\cdot\frac{-y}{x}=-1\)
a)
Ta có \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)
\(\Rightarrow\left\{\begin{matrix}\frac{y+z-x}{x}=1\\\frac{z+x-y}{y}=1\\\frac{x+y-z}{z}=1\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}y+z-x=x\\z+x-y=y\\x+y-z=z\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\) (1)
Ta có \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(\Rightarrow B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\)
Thế (1) vào biểu thức B
\(\Rightarrow B=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}\)
\(\Rightarrow B=2.2.2=8\)
Vậy biểu thức \(B=8\)
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)
\(=\frac{a+ab+1}{a+ab+1}=1\)
PT đã cho suy ra thành
\(\left(\frac{x^{2010}}{a^2+b^2+c^2+d^2}-\frac{x^{2010}}{a^2}\right)+\left(\frac{y^{2010}}{a^2+b^2+c^2+d^2}-\frac{y^{2010}}{b^2}\right)+\left(\frac{z^{2010}}{a^2+b^2+c^2+d^2}-\frac{z^{2010}}{c^2}\right)\)
\(+\left(\frac{t^{2010}}{a^2+b^2+c^2+d^2}-\frac{t^{2010}}{d^2}\right)=0\)
\(=>x^{2010}\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\right)+\left(tương\right)Tựnha=0\)
Do
\(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\ne0\)
máy cái bạn tự suy ra cx thế
\(=>x^{2010}=y^{2010}=z^{2010}=t^{2010}=0=>x=y=z=t=0\)
ta có
\(T=x^{2011}+y^{2011}+z^{2011}+t^{2011}=0+0+0+0=0\)
Ta có:
\(\frac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)
<=> \(x^{2010}\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+y^{2010}\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\)
\(+z^{2010}\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+t^{2010}\left(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)=0\)(1)
Lại có: \(x^{2010};y^{2010};z^{2010};t^{2010}\ge0;\forall x,y,z,t\)
và với mọi a; b ; c ; d khác 0 có:
\(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\)
\(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\);
\(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\);
\(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\)
=> \(x^{2010}\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
\(y^{2010}\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
\(z^{2010}\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
\(t^{2010}\left(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
=> \(x^{2010}\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+y^{2010}\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\)
\(+z^{2010}\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+t^{2010}\left(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
Như vậy (1) xảy ra<=> \(x^{2010}=y^{2010}=z^{2010}=t^{2010}=0\)
<=> x = y = z = t = 0
Thay vào T ta có : T = 0
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó : \(\frac{ac}{bd}=\frac{b.d.k^2}{b.d}=k^2\left(1\right);\)
\(\frac{2010a^2+2011c^2}{2010b^2+2011d^2}=\frac{2010b^2.k^2+2011d^2.k^2}{2010b^2+2011d^2}=\frac{k^2.\left(2010b^2+2011d^2\right)}{2010b^2+2011d^2}=k^2\left(2\right)\)
Từ (1)(2) => \(\frac{ac}{bd}=\frac{2010a^2+2011c^2}{2010b^2+2001d^2}\left(\text{đpcm}\right)\)
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) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)