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1) Phân số đầu nhân 2.
_ Phân số thứ 2 nhân 3, p/s thứ 3 giữ nguyên.
_ Lấy phân số đầu + p/s thứ 2 - p/s thứ 3.
_ Dựa vào dãy tỉ số bằng nhau tìm x, y, z.
2) \(x-y-z=0\Rightarrow x=y+z\)
Khi đó thay vào B được:
\(B=\left(1-\dfrac{z}{y+z}\right)\left(1-\dfrac{y+z}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\dfrac{y}{y+z}.\dfrac{z}{y}.\dfrac{y+z}{z}\)
\(=1\)
Vậy B = 1.
Ta có :
\(\dfrac{x+y-z}{z}=\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}\\ \Leftrightarrow\dfrac{x+y+z}{z}=\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\left(cùngcộngthêm2\right)\)
TH1: \(x+y+z\ne0\)
\(\Rightarrow x=y=z\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)\\ =2\cdot2\cdot2=8\)
TH2: \(x+y+z=0\Rightarrow\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(y+x\right)\end{matrix}\right.\)(*)
\(\Rightarrow P=\left(1+\dfrac{-\left(y+z\right)}{y}\right)\left(1+\dfrac{-\left(z+x\right)}{z}\right)\left(1+\dfrac{-\left(x+y\right)}{z}\right)\\ =\left(1-1-\dfrac{z}{y}\right)\left(1-1-\dfrac{x}{z}\right)\left(1-1-\dfrac{y}{z}\right)\\ =\left(-\dfrac{z}{y}\right)\left(-\dfrac{x}{z}\right)\left(-\dfrac{y}{z}\right)\\ =-1\)
Vậy P=8 hoặc P=-1
\(\dfrac{y+z+t-nx}{x}=\dfrac{z+t+x-ny}{y}=\dfrac{t+x+y-nz}{z}=\dfrac{x+y+z-nt}{t}\)
\(=\dfrac{y+z+t-nx+z+t+x-ny+t+x+y-nz+x+y+z-nt}{x+y+z+t}\)
\(=\dfrac{3x+3y+3z+3t-n\left(x+y+z+t\right)}{x+y+z+t}\)
\(=\dfrac{3\left(x+y+z+t\right)-n\left(x+y+z+t\right)}{x+y+z+t}=\dfrac{\left(3-n\right)\left(x+y+z+t\right)}{x+y+z+t}=3-n\)
Nên \(\left\{{}\begin{matrix}y+z+t-nx=3x-nx\\z+t+x-ny=3y-ny\\t+x+y-nz=3z-nz\\x+y+z-nt=3t-nt\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y+z+t=3x\\z+t+x=3y\\t+x+y=3z\\x+y+z=3t\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{y+z+t}{3}\\y=\dfrac{z+t+x}{3}\\z=\dfrac{t+x+y}{3}\\t=\dfrac{x+y+z}{3}\end{matrix}\right.\)
Thay vào \(P\) ta có:
\(P=x+2y-3z+t\)
\(P=\dfrac{y+z+t}{3}+\dfrac{2\left(z+t+x\right)}{3}-\dfrac{3\left(t+x+y\right)}{3}+\dfrac{x+y+z}{3}\)
\(P=\dfrac{y+z+t+2z+t+x-3t-3x-3y+x+y+z}{3}\)
\(P=\dfrac{\left(x+x-3x\right)+\left(y+y-3y\right)+\left(z+z+2z\right)+\left(t+t-3t\right)}{3}\)
\(P=\dfrac{-x-y-z+4t}{3}\)
\(P=\dfrac{-\left(x+y+z+t\right)+5t}{3}\)
\(P=\dfrac{-2012+5t}{3}\)
Tốn sức quá T^T
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z\\ =\frac{x+y+z}{z+y+x+z+1+x+y-2}\\ =\frac{x+y+z}{\left(x+x\right)+\left(y+y\right)+\left(z+z\right)+\left(1+1-2\right)}\\ =\frac{x+y+z}{2x+2y+2z}\\ =\frac{x+y+z}{2\left(x+y+z\right)}\\ =\frac{1}{2}\)
Ta có:
\(\frac{z}{x+y-2}=\frac{1}{2}\\ \Rightarrow2z=x+y-2\\\Rightarrow x+y=2z+2 \)
Thay \(x+y=2z+2\) vào \(x+y+z=\frac{1}{2}\), ta có:
\(2z+2+z=\frac{1}{2}\\ \Rightarrow3z=\frac{1}{2}-2\\ \Rightarrow3z=\frac{1}{2}-\frac{4}{2}\\ \Rightarrow3z=-\frac{3}{2}\\ \Rightarrow z=-\frac{\frac{3}{2}}{3}\\ \Rightarrow z=-\frac{3}{2}\cdot\frac{1}{3}\\ \Rightarrow z=-\frac{1}{2}\)
Ta có:
\(x+y+z=\frac{1}{2}\)
hay \(x+y-\frac{1}{2}=\frac{1}{2}\\ x+y=\frac{1}{2}+\frac{1}{2}\\ x+y=1\\ \Rightarrow x=1-y\)
Lại có:\(\frac{x}{y+z+1}=\frac{1}{2}\)
hay \(\frac{1-y}{y-\frac{1}{2}+1}=\frac{1}{2}\\ \Rightarrow2\left(1-y\right)=y-\frac{1}{2}+1\\ \Rightarrow2-2y=y-\frac{1}{2}+\frac{2}{2}\\ \Rightarrow2-2y=y+\frac{1}{2}\\ \Rightarrow2-\frac{1}{2}=y+2y\\ \Rightarrow\frac{4}{2}-\frac{1}{2}=3y\\ \Rightarrow\frac{3}{2}=3y\\ \Rightarrow y=\frac{3}{\frac{2}{3}}\\ \Rightarrow y=\frac{3}{2}\cdot\frac{1}{3}\\ \Rightarrow y=\frac{1}{2}\)
Lại có:\(x=1-y\)
hay \(x=1-\frac{1}{2}\\ \Rightarrow x=\frac{2}{2}-\frac{1}{2}\\ \Rightarrow x=\frac{1}{2}\)
Vậy: \(\left(x;y;z\right)=\left(\frac{1}{2};\frac{1}{2};-\frac{1}{2}\right)\)
\(\Rightarrow\dfrac{z+y+1}{x}=\dfrac{x+z+1}{y}=\dfrac{x+y-2}{z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2=x+y+z\\ \Rightarrow\left\{{}\begin{matrix}z+y+1=2x\\x+z+1=2y\\x+y-2=2z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z=2x-1\\x+z=2y-1\\x+y=2z+2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x-1=2-x\\2y-1=2-y\\2z+2=2-z\end{matrix}\right.\Rightarrow\left(x,y,z\right)=\left(1;1;0\right)\)
ta có x-y-z=0
->x=y+z
y=x-z
z=x-y
B=\(\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1-\dfrac{y}{z}\right)\)
B=\(\left(\dfrac{x-z}{x}\right)\left(\dfrac{y-x}{y}\right)\left(\dfrac{z+y}{z}\right)\)
B=\(\dfrac{y}{x}.\left(-\dfrac{z}{y}\right)\left(\dfrac{x}{z}\right)\)
B=\(\dfrac{-\left(xyz\right)}{xyz}\)
B=-1
\(2x=4z\Rightarrow z=\dfrac{x}{2}\)
\(2x=-3y\Rightarrow y=\dfrac{-2}{3}x\)
Thay vào \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{\dfrac{-2}{3}x}+\dfrac{1}{\dfrac{x}{2}}=3\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{\dfrac{-3}{2}}{\dfrac{-2}{3}.\dfrac{-3}{2}.x}+\dfrac{2}{2\dfrac{x}{2}}=3\)
\(\dfrac{1}{x}+\dfrac{\dfrac{-3}{2}}{x}+\dfrac{2}{x}\)
\(\Rightarrow\dfrac{\left(1+\dfrac{-3}{2}+2\right)}{x}=3\)
\(\Rightarrow\dfrac{\dfrac{3}{2}}{x}=3\)
\(\Rightarrow x=\dfrac{1}{2}\)
\(z=\dfrac{x}{2}=\dfrac{\dfrac{1}{2}}{2}=\dfrac{1}{4}\)
\(y=\dfrac{-2}{3}x=\dfrac{-2}{3}.\dfrac{1}{4}=\dfrac{-1}{6}\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{4}\\z=\dfrac{-1}{6}\end{matrix}\right.\)
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-2y+4+3z-9}{2-6+12}\)
\(=\dfrac{\left(x-2y+3z\right)+\left(-1+4-9\right)}{8}=\dfrac{14-6}{8}=\dfrac{8}{8}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\\\dfrac{y-2}{3}=1\\\dfrac{z-3}{4}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=2\\y-2=3\\z-3=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)