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\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)

\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)

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

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)

Do đó: x=16; y=24; z=30

B nha em

Chọn B

Ta có:

`x/10=y/5 -> x/20=y/10` `(1)`

`y/2=z/3 -> y/10=z/15` `(2)`

Từ `(1)` và `(2)`

`-> x/20=y/10=z/15` `-> x/20=y/10=(4z)/60`

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

`x/20=y/10=(4z)/60=(x+4z)/(20+60)=320/80=4`

`-> x/20=y/10=z/15=4`

`-> x=20*4=80, y=10*4=40, z=15*4=60`.

Ta có:

\(\left\{{}\begin{matrix}\dfrac{x}{10}=\dfrac{y}{5}\Rightarrow\dfrac{x}{20}=\dfrac{y}{10}\\\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{y}{10}=\dfrac{z}{15}\end{matrix}\right.\Rightarrow\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\)

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

\(\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}=\dfrac{x+4z}{20+4.15}=\dfrac{320}{80}=4\)

Do đó:

\(\dfrac{x}{20}=4\Rightarrow x=80\)

\(\dfrac{y}{10}=4\Rightarrow y=40\)

\(\dfrac{z}{15}=4\Rightarrow z=60\)

a) Ta có: \(\dfrac{x}{y}=\dfrac{10}{9}\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}\)

               \(\dfrac{y}{z}=\dfrac{3}{4}\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{9}=\dfrac{z}{12}\)

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{12}=\dfrac{x-y+z}{10-9+12}=\dfrac{78}{13}=6\)

\(\Rightarrow\left\{{}\begin{matrix}x=6.10=60\\y=6.9=54\\z=6.12=72\end{matrix}\right.\)

b)Ta có:  \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)

               \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)

c) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{3}\)

\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{9+16+9}=\dfrac{200}{34}=\dfrac{100}{17}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{900}{17}\\y^2=\dfrac{1600}{17}\\z^2=\dfrac{900}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{30\sqrt{17}}{17}\\y=\pm\dfrac{40\sqrt{17}}{17}\\z=\pm\dfrac{30\sqrt{17}}{17}\end{matrix}\right.\)

Vậy\(\left(x;y;z\right)\in\left\{\left(\dfrac{30\sqrt{17}}{17};\dfrac{40\sqrt{17}}{17};\dfrac{30\sqrt{17}}{17}\right),\left(-\dfrac{30\sqrt{17}}{17};-\dfrac{40\sqrt{17}}{17};-\dfrac{30\sqrt{17}}{17}\right)\right\}\)

1/2-2y=9/20

=>2y=1/2-9/20=1/20

=>y=1/20:2=1/40

b,3/5:4/3:y=2+7/10=9/20:y=27/10

=>y=9/20:27/10=1/6

c,y+y*3/2-y*1/2=1/10

=>y(1+3/2-1/2)=1/10

=>2y=1/10

=>y=1/10:2=1/20

b:

ĐKXĐ: x<>0

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

=>\(\dfrac{6+xy}{3x}=\dfrac{1}{6}\)

=>\(6\left(6+xy\right)=3x\)

=>\(x=2\left(6+xy\right)=12+2xy\)

=>\(x\left(1-2y\right)=12\)

mà x,y là các số nguyên

nên \(\left(x;1-2y\right)\in\left\{\left(12;1\right);\left(-12;-1\right);\left(4;3\right);\left(-4;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(12;0\right);\left(-12;1\right);\left(4;-1\right);\left(-4;2\right)\right\}\)

c: ĐKXĐ: y<>-1

\(\dfrac{x}{3}+\dfrac{1}{y+1}=\dfrac{1}{6}\)

=>\(\dfrac{xy+x+3}{3\left(y+1\right)}=\dfrac{1}{6}\)

=>\(\dfrac{2\left(xy+x+3\right)}{6\left(y+1\right)}=\dfrac{y+1}{6\left(y+1\right)}\)

=>\(2xy+2x+6=y+1\)

=>\(2x\left(y+1\right)-\left(y+1\right)=-6\)

=>\(\left(2x-1\right)\left(y+1\right)=-6\)

mà x,y là các số nguyên

nên \(\left(2x-1;y+1\right)\in\left\{\left(1;-6\right);\left(-1;6\right);\left(3;-2\right);\left(-3;2\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(0;5\right);\left(2;-3\right);\left(-1;1\right)\right\}\)

Áp dụng t/c dtsbn ta có:

\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\dfrac{1}{x+y+z}=2\Rightarrow2x+2y+2z=1\Rightarrow x+y+z=0,5\Rightarrow\left\{{}\begin{matrix}x+y=0,5-z\\y+z=0,5-x\\x+z=0,5-y\end{matrix}\right.\\ \dfrac{y+z+1}{x}=2\Rightarrow y+z+1=2x\Rightarrow0,5-x+1=2x\Rightarrow x=0,5\\ \dfrac{x+z+2}{y}=2\Rightarrow x+z+2=2y\Rightarrow0,5-y+2=2y\Rightarrow y=\dfrac{5}{6}\\ \dfrac{x+y-3}{z}=2\Rightarrow x+y-3=2z\Rightarrow0,5-z-3=2z\Rightarrow z=-\dfrac{5}{6}\)