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+, Nếu x+y+z=0 => B = x+y/y. y+z/z . z+x/x = (-z/y).(-x/z).(-y/x) = -xyz/xyz = -1
+, Nếu x+y+z khác o thì :
Áp dụng tính chất dãy tỉ số bằng nhau ta có : y+z-x/x = z+x-y/y = x+y-z/z = y+z-x+z+x-y+x+y-z/x+y+z = 1
=> y+z-x=x ; z+X-y=y ; x+y-z=z
=> x=y=z
=> B = (1+1).(1+1).(1+!) = 8
Vậy .............
Tk mk nha
ADTCDTSBN
\(\frac{y+z-x}{x}\)=\(\frac{z+x-y}{y}\)=\(\frac{x+y-z}{z}\)=\(\frac{y+z-x+z+x-y+x+y-z}{x+y+z}\)=1
\(\Rightarrow\)\(\hept{\begin{cases}y=-z\\z=-x\\x=-y\end{cases}}\)
Khi đó B=\(\left(1+\frac{-y}{y}\right)\)\(\left(1+\frac{-z}{z}\right)\)\(\left(1+\frac{-x}{x}\right)\)=0
Vậy B=0 ........... hjhjh
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\\ \Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\\ \Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\\ \Rightarrow x=y=z\\ \Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{x}{y}=\frac{y}{z}=\frac{z}{x}=\frac{x+y+z}{y+z+x}=1\)
=> x = y = z
Ta có: \(A=\frac{2013x^2+y^2+z^2}{x^2+2013y^2+z^2}=\frac{2013x^2+x^2+x^2}{x^2+2013x^2+x^2}=\frac{2015x^2}{2015x^2}=1\)
từ điều kiện suy ra \(\frac{y+z}{x}-1=\frac{x+z}{y}-1=\frac{x+y}{z}-1\)1\(\Rightarrow\frac{y+z}{x}=\frac{x+z}{y}=\frac{x+y}{z}\)
\(\frac{y+z}{x}=\frac{x+z}{y}\Rightarrow\frac{y+z}{x}-\frac{x+z}{y}=0\)\(\Rightarrow\frac{y\left(y+z\right)-x\left(x+z\right)}{xy}=0\)
\(\Rightarrow y^2+yz-xz-x^2=0\Rightarrow y^2-x^2+yz-zx=0\)\(\Rightarrow\left(y+x\right)\left(y-x\right)+z\left(y+x\right)\)=0
\(\Rightarrow\left(y-x\right)\left(x+y+z\right)=0\)\(\Rightarrow\)hoặc y-x=0 hoặc x+y+z=0 \(\Rightarrow\)x=y hoặc x+y=-z
giải tương tự ta có hoặc x=y=z hoặc x+y=-z;y+z=-x;x+z=-y
*x=y=z thay vào biểu thức ta có bt=8
*x+y=-z;y+z=-x;x+z=-y ta có bt =\(\left(\frac{x+y}{y}\right)\left(\frac{z+y}{z}\right)\left(\frac{x+z}{x}\right)\)=-1
Ta có :
\(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{\left(y+x\right)\left(z+y\right)\left(x+z\right)}{xyz}\)
+ ) Nếu \(x+y+z\ne0\)
Theo tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
\(=\frac{\left(y+z-x\right)\left(z+x-y\right)\left(x+y-x\right)}{x+y+z}\)
\(=\frac{x+y+z}{x+y+z}\)
\(=1\)
\(\Rightarrow\hept{\begin{cases}y+z-x=x\\z+x-y=y\\x+y-z=z\end{cases}\Leftrightarrow\hept{\begin{cases}y+z=2x\\z+x=2y\\x+y=2z\end{cases}}}\)
Do đó , \(B=\frac{\left(y+x\right)\left(z+y\right)\left(x+z\right)}{xyz}=\frac{2z.2x.2y}{xyz}=8\)
+ ) Nếu \(x+y+z\ne0\text{thì}\hept{\begin{cases}x+y=-z\\x+z=-y\\y+z=-x\end{cases}}\)
Do đó , \(B=\frac{\left(-x\right).\left(-y\right).\left(-z\right)}{xyz}=-1\)
Vậy : \(B=-1\text{hoặc}B=8\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)
Do đó :
\(\frac{y+z-x}{x}=1\)\(\Rightarrow\)\(2x=y+z\)
\(\frac{z+x-y}{y}=1\)\(\Rightarrow\)\(2y=x+z\)
\(\frac{x+y-z}{z}=1\)\(\Rightarrow\)\(2z=x+y\)
Suy ra :
\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{x}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=\frac{8xyz}{xyz}=8\)
Vậy \(P=8\)
Đề hơi sai
ta có y+z-x/x=z+x-y/y=x+y-z/z=y+z-x+z+x-y+x+y-z/x+y+z=(2y-y)+(2x-x)+(2z-z)/x+y+z=y+x+z/x+y+z=1
=>y+z-x/x=1 =>z+x-y/y=1
z+x-y/y=1 x+y-z/z=1
=> y+z-x=x => z+x-y=y
z+x-y=y x+y-z=z
=>2y-2x=x-y =>2z-2y=y-z
3y-3x=0 3z-3y=0
y-x=0 z-y=0
=>x=y =>z=y
=>x=y=z
=> y+z-x/x+z+x-y/y+x+y-z/z= 0,(3)+0,(3)+0,(3)=1
=>x +y+z=0,(3)+0,(3)+0,(3)=1
thay vào b=(1+x/y). (1+y/z). (1+z/x)
b=(1+0,(3)/0,(3)).(1+0,(3)/0,(3)).(1+0,(3)/0,(3))
b=(1+1).(1+1).(1+1)
b=2.2.2
b=2^3
b=8
CÂU TRẢ LỜI TRƯỚC MK BẤM NHẦM
Từ \(\frac{y+x-z}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
\(\Rightarrow\frac{y+x-z}{x}+2=\frac{z+x-y}{y}+2=\frac{x+y-z}{z}+2\)
\(\Rightarrow\frac{x+y+x}{x}=\frac{x+y+z}{y}=\frac{x+y+z}{z}\)
* Xét \(x+y+z\ne0\)
\(\Rightarrow x=y=z\)
Khi đó \(B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}=2.2.2=8\)
* Xét \(x+y+z=0\)
\(\Rightarrow\left\{\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
Khi đó \(B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{-z}{y}.\frac{-x}{z}.\frac{-y}{x}=-1\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)(vì x + y + z khác 0)
=> \(\frac{1}{x+y+z}=2\) => x + y + z = 1/2
=> \(\hept{\begin{cases}\frac{y+z+1}{x}=2\\\frac{x+z+2}{y}=2\\\frac{x+y-3}{z}=2\end{cases}}\) => \(\hept{\begin{cases}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\end{cases}}\) => \(\hept{\begin{cases}3x=x+y+z+1\\3y=x+y+z+2\\3z=x+y+z-3\end{cases}}\)=> \(\hept{\begin{cases}3x=\frac{3}{2}\\3y=\frac{5}{2}\\3z=-\frac{5}{2}\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=-\frac{5}{6}\end{cases}}\)
Khi đó: A = \(2016\cdot\frac{1}{2}+\left(\frac{5}{6}\right)^{2017}-\left(\frac{5}{6}\right)^{2017}=1008\)
Ta có \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)
\(=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
Khi đó \(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)
Lại có \(\frac{y+z+1}{x}=2\Rightarrow y+z+1=2x\Rightarrow x+y+z+1=3x\Rightarrow\frac{1}{2}+1=3x\Rightarrow3x=\frac{3}{2}\)
=> x = 1/2
Lại có \(\frac{x+z+2}{y}=2\Rightarrow x+z+2=2y\Rightarrow x+y+z+2=3y\Rightarrow\frac{1}{2}+2=3y\Rightarrow3y=\frac{5}{2}\)
=> y = 5/6
Lại có x + y + z = 1/2
=> 1/2 + 5/6 + z = 1/2
=> 5/6 + z = 0
=> z = -5/6
Khi đó A = 2016X + y2017 + z2017
= 2016.1/2 + (5/6)2017 - (5/6)2017
= 1008
Vậy A = 1008
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) nha bạn!
ko hỉu thì ib
\(\left(x+y+z\right).\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{x}\right)\ge9\) với x,y,z dương hay jj đó chứ? (cái này t k bt -.-) VD: x=2, y=-2,z=4
=> \(\left(x+y+z\right).\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{x}\right)=\left(2-2+4\right).\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{4}\right)=1\)
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\(\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\(\Leftrightarrow\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-\frac{x+y+z}{x+y+z}=0\)
\(\Leftrightarrow\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
vì x+y+z khác 0 => \(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{xy+yz+xz}{xyz}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{\left(xy+yz+xz\right).\left(x+y+z\right)-xyz}{xzy.\left(x+y+z\right)}=0\)
\(\Leftrightarrow\frac{x^2y+xy^2+xyz+zyx+y^2z+yz^2+x^2z+xyz+xz^2-xzy}{xyz.\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x^2y+xyz\right)+\left(xy^2+y^2z\right)+\left(yz^2+xzy\right)+\left(x^2z+xz^2\right)=0\)
\(\Leftrightarrow xy.\left(x+z\right)+y^2.\left(x+z\right)+yz.\left(z+x\right)+xz.\left(x+z\right)=0\)
\(\Leftrightarrow\left(x+z\right).\left(xy+y^2+yz+xz\right)=0\)
\(\Leftrightarrow\left(x+z\right).\left[x.\left(y+z\right)+y.\left(y+z\right)\right]=0\)
\(\Leftrightarrow\left(x+y\right).\left(y+z\right).\left(x+z\right)=0\Leftrightarrow\orbr{\begin{cases}x=-y\\y=-z\end{cases}\text{hoặc }x=-z}\)
\(\Rightarrow P=\left(\frac{1}{x}-\frac{1}{y}\right).\left(\frac{1}{y}+\frac{1}{z}\right).\left(\frac{1}{z}+\frac{1}{x}\right)=0\)
ps: bài này t làm cách l8, ai có cách ez hơn giải vs ak :') morongtammat