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a: \(\Leftrightarrow\left\{{}\begin{matrix}2x+2y+4z=8\\2x-y+3z=6\\2x-6y+8z=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y+z=2\\8y-4z=1\\x+y+2z=4\end{matrix}\right.\)
=>y=9/20; z=13/20; x=4-y-2z=9/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}z=23-x-y\\z=31-y-t\\z=27-t-x\\x+y+t=33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x-y+23=-y-t+31\\-y-t-31=-x-t+27\\x+y+t=33\\z=23-x-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x+t=8\\x-y=58\\x+y+t=33\\z=23-x-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=x+8\\y=x-58\\x-58+x+8+x=33\\z=23-x-y\end{matrix}\right.\)
=>x=83/3; t=107/3; y=-91/3; z=23-83/3+91/3=77/3
a: Sửa đề:
\(\left\{{}\begin{matrix}3xy=2\left(x+y\right)\\4yz=3\left(y+z\right)\\5xz=6\left(z+x\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{3}{2}\\\dfrac{y+z}{yz}=\dfrac{4}{3}\\\dfrac{x+z}{xz}=\dfrac{5}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{2}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{4}{3}\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{3}{2}\\\dfrac{1}{y}=1\\\dfrac{1}{z}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{2}{3};y=1;z=3\)
b: Áp dụng tính chất của dãy tỉ số bằng nhau,ta được:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}=\dfrac{7x-3y+2z}{7\cdot4-3\cdot3+2\cdot9}=\dfrac{37}{37}=1\)
=>x=4; y=3; z=9
a) ĐK xác định : x≠0;y≠0
ta có : \(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{6}{y}=9\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{7}{x}=16\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{7}{16}\\y=-\dfrac{42}{17}\end{matrix}\right.\)
Vậy S = {(\(\dfrac{7}{16};-\dfrac{42}{17}\))}
b) Đk xác định : x≠0;y≠0
ta có : \(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{1}{y}=14\\\dfrac{8}{x}-\dfrac{1}{y}=-8\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{13}{x}=6\\\dfrac{5}{x}+\dfrac{1}{y}=14\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{13}{6}\\y=\dfrac{13}{152}\end{matrix}\right.\)
Vậy S={(\(\dfrac{13}{6};\dfrac{13}{152}\))}
c) ĐK xác định : x≠0;y≠0
ta có : \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{7}{y}=21\\-\dfrac{2}{x}-\dfrac{5}{y}=-11\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{2}{y}=10\\\dfrac{2}{x}+\dfrac{7}{y}=21\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{1}{5}\\x=-\dfrac{1}{7}\end{matrix}\right.\)
Vậy S={(\(-\dfrac{1}{7};\dfrac{1}{5}\))}
d) ĐK xác định : x≠0;y≠0
ta có : \(\left\{{}\begin{matrix}\dfrac{9}{x}+\dfrac{2}{y}=22\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{14}{x}=35\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)
Vậy S={(0,4;-4)}
e) ĐKXĐ : x≠0;y≠0
ta có : \(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{5}{y}=10\\-\dfrac{3}{x}-\dfrac{7}{y}=8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-\dfrac{2}{y}=18\\\dfrac{3}{x}+\dfrac{5}{y}=10\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{1}{9}\\x=\dfrac{3}{55}\end{matrix}\right.\) 'Vậy....
hpt \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{1}{2}\\\dfrac{y+z}{yz}=\dfrac{1}{4}\\\dfrac{z+x}{xz}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{4}\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{1}{3}\end{matrix}\right.\) ( đk : x , y , z # 0 )
Cộng từng vế của các pt lại với nhau , ta có :
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{13}{12}\)
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{13}{24}-\left(\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{13}{24}-\dfrac{1}{4}=\dfrac{7}{24}\)
\(\Leftrightarrow x=\dfrac{24}{7}\left(tm\right)\)
\(\Rightarrow y=\dfrac{24}{5}\left(tm\right);z=8\left(tm\right)\)
Lời giải:
Ta có: \(\left\{\begin{matrix} x+\frac{1}{y}=3\\ y+\frac{1}{z}=3\\ z+\frac{1}{x}=3\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+\frac{1}{y}=y+\frac{1}{z}\\ y+\frac{1}{z}=z+\frac{1}{x}\\ z+\frac{1}{x}=x+\frac{1}{y}\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x-y=\frac{y-z}{yz}\\ y-z=\frac{z-x}{xz}\\ z-x=\frac{x-y}{xy}\end{matrix}\right.(*)\) \(\Rightarrow (x-y)(y-z)(z-x)=\frac{(x-y)(y-z)(z-x)}{x^2y^2z^2}\)
\(\Leftrightarrow (x-y)(y-z)(z-x)\left(1-\frac{1}{x^2y^2z^2}\right)=0\)
Bây giờ ta xét các TH sau:
TH1: \(x-y=0\Rightarrow(*)\) kéo theo \(y-z=0\Rightarrow (*)\) kéo theo \(z-x=0\)
Do đó \(x=y=z\)
Thay vào pt ban đầu: \(x+\frac{1}{x}=3\Leftrightarrow x^2-3x+1=0\)
\(\Leftrightarrow x=\frac{3\pm \sqrt{5}}{2}\)
Ta có bộ nghiệm \((x,y,z)=\left(\frac{3+\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right);\left(\frac{3-\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}\right) \)
TH2: \(\left[\begin{matrix} y-z=0\\ z-x=0\end{matrix}\right.\) (hoàn toàn tương tự TH1)
TH3: \(1-\frac{1}{x^2y^2z^2}=0\Leftrightarrow xyz=\pm 1\)
\(\bullet\)Nếu \(xyz=1\):
\(\left\{\begin{matrix} x+\frac{1}{y}=3\\ y+\frac{1}{z}=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=3\\ y+xy=3\end{matrix}\right.\)
\(\Rightarrow x+\frac{1}{y}=y+xy\Leftrightarrow x(y-1)+\frac{y^2-1}{y}=0\)
\(\Leftrightarrow (y-1)(x+\frac{y+1}{y})=0\)
+) \(y=1\Rightarrow x=2; z=\frac{1}{2}\), thử vào pt số 3 thấy không thỏa mãn (loại)
\(+) x+\frac{y+1}{y}=0\Leftrightarrow x+1+\frac{1}{y}=0\Leftrightarrow 3+1=0\) (vô lý- loại )
\(\bullet xyz=-1\)
\(\left\{\begin{matrix} x+\frac{1}{y}=3\\ y-xy=3\\ \end{matrix}\right.\) \(\Rightarrow x+\frac{1}{y}=y-xy\Leftrightarrow (y+1)(x+\frac{1-y}{y})=0\)
+) Nếu \(y+1=0\Leftrightarrow y=-1\Rightarrow x=4; z=\frac{1}{4}\)
Thử lại vào pt thứ 3 thấy không đúng (loại )
+ Nếu \(x+\frac{1-y}{y}=0\Leftrightarrow x+\frac{1}{y}-1=0\Leftrightarrow 3-1=0\) (vô lý- loại )
Vậy \((x,y,z)=\left(\frac{3+\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right);\left(\frac{3-\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}\right) \)
\(\left\{{}\begin{matrix}x+\dfrac{1}{y}=3\left(1\right)\\y+\dfrac{1}{z}=3\left(2\right)\\z+\dfrac{1}{x}=3\left(3\right)\end{matrix}\right.\) đk : x,y,z khác 0
từ (1) \(x=3-\dfrac{1}{y};x\ne0\Rightarrow y\ne\dfrac{1}{3}\) (4)
từ (3) và (4) => \(z=3-\dfrac{1}{x}=3-\dfrac{1}{3-\dfrac{1}{y}}=3-\dfrac{y}{3y-1}=\dfrac{8y-3}{3y-1};z\ne0\Rightarrow y\ne\dfrac{3}{8}\) (5)
từ (5) và (2) => \(y+\dfrac{3y-1}{8y-3}=3\Leftrightarrow8y^2-3y+3y-1=3\left(8y-3\right)\)
\(\Leftrightarrow y^2-3y+1=0\) \(\Delta_y=9-4=5\)
=>\(\left[{}\begin{matrix}y_1=\dfrac{3-\sqrt{5}}{2}\\y_2=\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\) thỏa mãn đk y nhận
thế vào (4)=> \(\left[{}\begin{matrix}x_1=\dfrac{3-\sqrt{5}}{2}\\x_2=\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\) thế vào (3) \(\Rightarrow\left[{}\begin{matrix}z_1=\dfrac{3-\sqrt{5}}{2}\\z_2=\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\)
Đơn giản vậy thôi cần gì biết đổi hầm hố phân ra nhiều các trường hợp rắc rối
\(\left\{{}\begin{matrix}\dfrac{xy}{x+y}=\dfrac{12}{5}\\\dfrac{yz}{y+z}=\dfrac{18}{5}\\\dfrac{zx}{z+x}=\dfrac{36}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{5}{12}\\\dfrac{y+z}{yz}=\dfrac{5}{18}\\\dfrac{z+x}{zx}=\dfrac{13}{36}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{12}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{5}{18}\\\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{13}{36}\end{matrix}\right.\)
Cộng vế theo vế ta thu được :
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{19}{18}\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{19}{36}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{4}\\\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{1}{z}=\dfrac{1}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=9\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(4;6;9\right)\)
Lời giải:
Từ \(\left\{\begin{matrix} \frac{x}{105}=\frac{y}{90}\\ \frac{y}{24}=\frac{z}{21}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{x}{420}=\frac{y}{360}\\ \frac{y}{360}=\frac{z}{315}\end{matrix}\right.\) \(\Rightarrow \frac{x}{420}=\frac{y}{360}=\frac{z}{315}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{420}=\frac{y}{360}=\frac{z}{315}=\frac{x+y+z}{420+360+315}=\frac{292}{1095}=\frac{4}{15}\)
\(\Rightarrow \left\{\begin{matrix} x=\frac{4}{15}.420=112\\ y=\frac{4}{15}.360=96\\ z=\frac{4}{15}.315=84\end{matrix}\right.\)