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

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

\(\Rightarrow\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\) 

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

\(\dfrac{z+y-3}{z}=2\Rightarrow2z=z+y-3\Rightarrow z=y-3\) 

Thay `y+1=x` và `z=y-3` vào `x+y+z=1/2` ta có:

\(y+1+y-3+y=\dfrac{1}{2}\\ \Rightarrow3y-2=\dfrac{1}{2}\Rightarrow3y=\dfrac{5}{2}\\ \Rightarrow y=\dfrac{5}{2}:3\Rightarrow y=\dfrac{5}{6}\)

\(\Rightarrow\left\{{}\begin{matrix}x=y+1=\dfrac{5}{6}+1=\dfrac{11}{6}\\z=y-3=\dfrac{5}{6}-3=\dfrac{-13}{6}\end{matrix}\right.\)

a:

2x=3y=5z

=>\(\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}\)

=>\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

Đặt \(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=k\)

=>x=15k; y=10k; z=6k

|x+y-z|=95

=>|15k+10k-6k|=95

=>|19k|=95

=>|k|=5

=>\(\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\)

TH1: k=5

=>\(x=15\cdot5=75;y=10\cdot5=50;z=6\cdot5=30\)

TH2: k=-5

=>\(x=15\cdot\left(-5\right)=-75;y=10\cdot\left(-5\right)=-50;z=6\cdot\left(-5\right)=-30\)

b: \(\dfrac{6}{11}x=\dfrac{9}{2}y=\dfrac{18}{5}z\)

=>\(\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}\)

mà -x+z=-196

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

\(\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}=\dfrac{-x+z}{-\dfrac{11}{6}+\dfrac{5}{18}}=\dfrac{-196}{-\dfrac{14}{9}}=126\)

=>\(x=126\cdot\dfrac{11}{6}=231;y=126\cdot\dfrac{2}{9}=28;z=126\cdot\dfrac{5}{18}=35\)

a: Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\)

=>a=2k; b=3k; c=4k

\(a^2-b^2+2c^2=108\)

=>\(\left(2k\right)^2-\left(3k\right)^2+2\cdot\left(4k\right)^2=108\)

=>\(4k^2-9k^2+32k^2=108\)

=>\(27k^2=108\)

=>\(k^2=4\)

=>\(\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

TH1: k=2

=>\(a=2\cdot2=4;b=3\cdot2=6;c=4\cdot2=8\)

TH2: k=-2

=>\(a=2\cdot\left(-2\right)=-4;b=3\cdot\left(-2\right)=-6;c=4\cdot\left(-2\right)=-8\)

b: Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)

=>x=3k; y=4k; z=5k

\(-3x^2-2y^2+5z^2=594\)

=>\(-3\cdot\left(3k\right)^2-2\cdot\left(4k\right)^2+5\cdot\left(5k\right)^2=594\)

=>\(-27k^2-32k^2+125k^2=594\)

=>\(k^2=9\)

=>\(\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)

TH1: k=3

=>\(x=3\cdot3=9;y=4\cdot3=12;z=5\cdot3=15\)

TH2: k=-3

=>\(x=3\cdot\left(-3\right)=-9;y=4\cdot\left(-3\right)=-12;z=5\cdot\left(-3\right)=-15\)

a) Đặt: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)

\(a^2-b^2+2c^2=108\)

\(\Rightarrow\left(2k\right)^2-\left(3k\right)^2+2\cdot\left(4k\right)^2=108\\ \Rightarrow4k^2-9k^2+32k^2=108\\ \Rightarrow27k^2=108\\ \Rightarrow k^2=4\\ \Rightarrow k=\pm2\)

Với k=2 \(\Rightarrow\left\{{}\begin{matrix}a=2\cdot2=4\\b=3\cdot2=6\\c=4\cdot2=8\end{matrix}\right.\)

Với k=-2 \(\Rightarrow\left\{{}\begin{matrix}a=2\cdot-2=-4\\b=3\cdot-2=-6\\c=4\cdot-2=-8\end{matrix}\right.\)

Vậy: ...

b) \(x:y:z=3:4:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

Đặt: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\)

\(5z^2-3x^2-2y^2=594\\ \Rightarrow5\cdot\left(5k\right)^2-3\cdot\left(3k\right)^2-2\cdot\left(4k\right)^2=594\\ \Rightarrow125k^2-27k^2-32k^2=594\\ \Rightarrow66k^2=594\\ \Rightarrow k^2=\dfrac{594}{66}\\ \Rightarrow k^2=9\\ \Rightarrow k=\pm3\)

Với \(k=3\Rightarrow\left\{{}\begin{matrix}x=3\cdot3=9\\y=4\cdot3=12\\z=5\cdot3=15\end{matrix}\right.\)

Với \(k=-3\Rightarrow\left\{{}\begin{matrix}x=3\cdot-3=-9\\y=4\cdot-3=-12\\z=5\cdot-3=-15\end{matrix}\right.\)

Vậy: ... 

a: \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)

=>\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

mà x+y+z=49

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

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

=>\(x=12\cdot\dfrac{3}{2}=18;y=12\cdot\dfrac{4}{3}=16;z=12\cdot\dfrac{5}{4}=15\)

b: \(\dfrac{6}{11}x=\dfrac{9}{2}y=\dfrac{18}{5}z\)

=>\(\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}\)

mà -x+y+z=-120

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

\(\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}=\dfrac{-x+y+z}{-\dfrac{11}{6}+\dfrac{2}{9}+\dfrac{5}{18}}=\dfrac{-120}{-\dfrac{4}{3}}=90\)

=>\(x=90\cdot\dfrac{11}{6}=165;y=90\cdot\dfrac{2}{9}=20;z=90\cdot\dfrac{5}{18}=25\)

a) \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{x}{\dfrac{3}{2}}+\dfrac{y}{\dfrac{4}{3}}+\dfrac{z}{\dfrac{5}{4}}\)

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

\(\dfrac{x}{\dfrac{3}{2}}+\dfrac{y}{\dfrac{4}{3}}+\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\) 

\(\Rightarrow\dfrac{x}{\dfrac{3}{2}}=12\Rightarrow x=\dfrac{3}{2}\cdot12=18\)

\(\Rightarrow\dfrac{y}{\dfrac{4}{3}}=12\Rightarrow y=\dfrac{4}{3}\cdot12=16\)

\(\Rightarrow\dfrac{z}{\dfrac{5}{4}}=12\Rightarrow z=12\cdot\dfrac{5}{4}=15\)

Vậy: ... 

\(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}\)

=>\(\dfrac{x}{8}=\dfrac{y}{64}=\dfrac{z}{216}\)

=>\(\dfrac{x}{2}=\dfrac{y}{16}=\dfrac{z}{54}\)

=>\(\dfrac{x}{1}=\dfrac{y}{8}=\dfrac{z}{27}\)

Đặt \(\dfrac{x}{1}=\dfrac{y}{8}=\dfrac{z}{27}=k\)

=>x=k; y=8k; z=27k

\(2x^2+2y^2-z^2=1\)

=>\(2k^2+2\cdot\left(8k\right)^2-\left(27k\right)^2=1\)

=>\(2k^2+128k^2-729k^2=1\)

=>\(k^2=-\dfrac{1}{599}\)(vô lý)

Vậy: KHông có bộ số (x;y;z) nào thỏa mãn yêu cầu đề bài

TH1: p=2

\(p^2+8=2^2+8=12\) không là số nguyên tố

=>Loại

TH2: p=3

\(p^2+8=3^2+8=17\) là số nguyên tố

\(p^2+2=3^2+2=11\) là số nguyên tố

=>Nhận

TH3: p=3k+1

\(p^2+8=\left(3k+1\right)^2+8=9k^2+6k+9=3\left(3k^2+2k+3\right)⋮3\)

=>p^2+8 không là số nguyên tố

=>Loại

TH4: p=3k+2

\(p^2+8=\left(3k+2\right)^2+8\)

\(=9k^2+12k+4+8=9k^2+12k+12=3\left(3k^2+4k+4\right)⋮3\)

=>p^2+8 không là số nguyên tố

=>Loại

Vậy: p=3

\(3^{n-2}=81\\ =>3^{n-2}=3^4\\ =>n-2=4\\ =>n=4+2\\ =>n=6\)

Để pt có nghiệm duy nhất thì: \(-\dfrac{2}{m}\ne\dfrac{1}{1}\Leftrightarrow m\ne-2\)

\(\left\{{}\begin{matrix}-2x+y=-3m-1\\mx+y=m^2+m+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(m+2\right)y=m^2+m+3+3m+1\\-2x+y=-3m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m^2+4m+4}{m+2}\\-2x+y=-3m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\left(m+2\right)^2}{m+2}=m+2\\-2x+\left(m+2\right)=-3m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m+2\\2x=m+2+3m+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m+2\\2x=4m+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m+2\\x=\dfrac{4m+2}{2}\end{matrix}\right.\) 

Mà: \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4m+2}{3}>0\\m+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4m>-2\\m>-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>-\dfrac{1}{2}\\m>-2\end{matrix}\right.\Leftrightarrow m>-\dfrac{1}{2}\)

\(2x^4+ax^3+3x^2+4x+b⋮x^2-4x+4\)

=>\(2x^4-8x^3+8x^2+\left(a+8\right)x^3-\left(4a+32\right)x^2+\left(4a+32\right)x+\left(4a+27\right)x^2-4\cdot\left(4a+27\right)x+4\cdot\left(4a+27\right)+\left(12a+80\right)x+b-16a-108⋮x^2-4x+4\)

=>\(\left\{{}\begin{matrix}12a+80=0\\b-16a-108=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=-\dfrac{20}{3}\\b=16a+108=\dfrac{4}{3}\end{matrix}\right.\)