Coi đề lại câu a

b,

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\\ \dfrac{x-1}{2}=\dfrac{2\left(y-2\right)}{2\cdot3}=\dfrac{3\cdot\left(z-3\right)}{3\cdot4}\\ \dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-\left(2y-4\right)+3z-9}{2-6+12}=\dfrac{x-1-2y+4+3z-9}{8}=\dfrac{\left(x-2y+3z\right)+\left(4-1-9\right)}{8}=\dfrac{14+\left(-6\right)}{8}=\dfrac{8}{8}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\Rightarrow x-1=2\Rightarrow x=3\\\dfrac{2y-4}{6}=1\Rightarrow2y-4=6\Rightarrow2y=10\Rightarrow y=5\\\dfrac{3z-9}{12}=1\Rightarrow3z-9=12\Rightarrow3z=21\Rightarrow z=7\end{matrix}\right.\)

Vậy x = 3; y = 5; z = 7

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)

\(\Rightarrow\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

\(=\dfrac{x-1-2y+4+3z-9}{2-6+12}\)

\(=\dfrac{14-6}{14-6}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\\\dfrac{y-2}{3}=1\\\dfrac{z-3}{4}=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)

Bài này quá dễ mình làm được

áp dụng dãy tỉ số = nhau ta có \(\dfrac{1+x}{2}=\dfrac{4-2y}{6}=\dfrac{4+z}{5}=\dfrac{x-2y+z+1+4+4}{2+6+5}=\dfrac{11}{13}\)

\(\dfrac{1+x}{2}=\dfrac{11}{13}\Leftrightarrow13\left(1+x\right)=22\Leftrightarrow13x+13=22\Leftrightarrow x=\dfrac{9}{13}\)

\(\dfrac{2-y}{3}=\dfrac{11}{13}\Leftrightarrow13\left(2-y\right)=33\Leftrightarrow-13y+26=33\Leftrightarrow y=-\dfrac{7}{13}\)

\(\dfrac{4+z}{5}=\dfrac{11}{13}\Leftrightarrow13\left(4+z\right)=55\Leftrightarrow13z+52=55\Leftrightarrow z=\dfrac{3}{13}\)

vậy..................

\(\dfrac{3x-2y}{4}=\dfrac{4y-3z}{2}=\dfrac{2z-4x}{3}=\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\2z-4x=0\\4y-3z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{z}{4}=\dfrac{x}{2}\end{matrix}\right.\\ \Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x-2y+3z}{2-6+12}=\dfrac{8}{8}=1\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\\z=4\end{matrix}\right.\)

b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x-4y+5z+3-12-25}{-3\cdot2-4\cdot4+5\cdot6}=\dfrac{16}{8}=2\)

Do đó: x=5; y=5; z=17

\(a,\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)

Áp dụng t/c dtsbn:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+2y^2-3z^2}{4+18-48}=\dfrac{-650}{-26}=25\\ \Rightarrow\left\{{}\begin{matrix}x^2=100\\y^2=225\\z^2=400\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm10\\y=\pm15\\z=\pm20\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)\) có giá trị là hoán vị của \(\left(\pm10;\pm15;\pm20\right)\)

\(VT=\dfrac{2y+3z+5}{1+x}+1+\dfrac{3z+x+5}{2y+1}+1+\dfrac{x+2y+5}{1+3z}+1-3\)

\(VT=\dfrac{x+2y+3z+6}{1+x}+\dfrac{x+2y+3z+6}{1+2y}+\dfrac{x+2y+3z+6}{1+3z}-3\)

\(VT=24\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)-3\ge\dfrac{24.9}{1+x+1+2y+1+3z}-3=\dfrac{216}{21}-3=\dfrac{51}{7}\)

1: Ta có: \(\dfrac{x}{3}=\dfrac{y}{6}\)

mà 4x-y=42

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{6}=\dfrac{4x-y}{4\cdot3-6}=\dfrac{42}{12-6}=\dfrac{42}{6}=7\)

=>\(x=7\cdot3=21;y=6\cdot7=42\)

2: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)

mà x-2y+3z=33

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-2y+3z}{2-2\cdot3+3\cdot5}=\dfrac{33}{2-6+15}=\dfrac{33}{11}=3\)

=>\(x=3\cdot2=6;y=3\cdot3=9;z=3\cdot5=15\)

3: \(\dfrac{x}{y}=\dfrac{6}{5}\)

=>\(\dfrac{x}{6}=\dfrac{y}{5}\)

mà x+y=121

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{x+y}{6+5}=\dfrac{121}{11}=11\)

=>\(x=11\cdot6=66;y=11\cdot5=55\)

1) Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).3=-3\\y=\left(-1\right).5=-5\\z=\left(-1\right).7=-7\end{matrix}\right.\)

2) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{19}.8=-\dfrac{224}{19}\\y=-\dfrac{28}{19}.12=-\dfrac{336}{19}\\z=-\dfrac{28}{19}.15=-\dfrac{420}{19}\end{matrix}\right.\)

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{3\cdot2+5\cdot3-7}=\dfrac{-14}{14}=-1\\ \Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)

b, \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{224}{19}\\y=-\dfrac{336}{19}\\z=-\dfrac{420}{19}\end{matrix}\right.\)

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)

⇒\(\dfrac{2x}{10}=\dfrac{3y}{15}=\dfrac{z}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{10}=\dfrac{3y}{15}=\dfrac{z}{7}=\dfrac{2x+3y-z}{10+15-7}=\dfrac{-14}{18}=\dfrac{-7}{9}\)

⇒\(\left\{{}\begin{matrix}x=\dfrac{-7}{9}.3=\dfrac{-7}{3}\\y=\dfrac{-7}{9}.5=\dfrac{-35}{9}\\z=\dfrac{-7}{9}.7=\dfrac{-49}{9}\end{matrix}\right.\)