Có a+b=3(b+c)=4(c+a)

\(\Rightarrow\frac{a+b}{12}=\frac{3\left(b+c\right)}{12}=\frac{4\left(c+a\right)}{12}\Leftrightarrow\frac{a+b}{12}=\frac{b+c}{4}=\frac{c+a}{3}\)

Áp dụng tc của dãy tỉ số bằng nhau,ta có:

\(\frac{a+b}{12}=\frac{b+c}{4}=\frac{c+a}{3}=\frac{a+b-b-c+c+a}{12-4+3}=\frac{2a}{11}\)

=>,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

\(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)=\frac{3}{4}\)

\(\frac{2}{3}:\left(x-1\right)=\frac{1}{4}\)

\(x-1=\frac{8}{3}\)

\(x=\frac{11}{3}\)

1/2+2/3 : (x-1) =3/4

=> 2/3 : (x-1) = 3/4 - 1/2

=> 2/3 : (x-1) = 3/4 - 2/4

=> 2/3 : (x-1) = 1/4

=> (x-1) = 2/3 : 1/4

=> (x-1) = 2/3 . 4

=> (x-1) = 8/3

=> x = 8/3 + 1

=> x = 8/3 + 3/3

=> x = 11/3

\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{x^2}\right)=\frac{1007}{2012}\)

\(\Rightarrow\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{\left(x^2-1\right)}{x^2}=\frac{1007}{2012}\)

\(\Rightarrow\frac{1.3.2.4.3.5...\left(x-1\right)\left(x+1\right)}{\left(2.3.4..x\right)^2}=\frac{1007}{2012}\)

\(\Rightarrow\frac{\left[2.3.4...\left(x-1\right)\right].\left[3.4.5...\left(x+1\right)\right]}{\left(2.3.4....x\right)\left(2.3.4....x\right)}=\frac{1007}{2012}\)

\(\Rightarrow\frac{\left(x+1\right)}{2x}=\frac{1007}{2012}\)

\(\Rightarrow2002.\left(x+1\right)=1007.2x\)

\(\Rightarrow2012x+2012=2014x\)

\(\Rightarrow2x=2012\)

\(\Rightarrow x=1006\)

Vậy x = 1006

bình phương 3 vế :

\(\left(\frac{x}{2}\right)^2=\left(\frac{y}{3}\right)^2=\left(\frac{z}{5}\right)^2\)<=> \(\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}\)<=> \(\frac{2}{2}.\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}\)<=> \(\frac{2x^2}{8}=\frac{y^2}{9}=\frac{z^2}{25}\)

áp dụng t/c dãy tỉ số bằng nhau 

\(\frac{2x^2}{8}=\frac{y^2}{9}=\frac{z^2}{25}=\frac{2x^2+y^2+z^2}{8+9+25}\)\(=\)\(\frac{-72}{42}=\frac{-12}{7}\)

\(\frac{x}{2}=\frac{-12}{7}\Leftrightarrow x=\frac{-24}{7}\)

\(\frac{y}{3}=\frac{-12}{7}\Leftrightarrow y=\frac{-36}{7}\)

\(\frac{z}{5}=\frac{-12}{7}\Leftrightarrow z=\frac{-60}{7}\)

Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=K\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=K\\\frac{y}{3}=K\\\frac{z}{5}=K\end{cases}}\Rightarrow\hept{\begin{cases}x=2K\\y=3K\\z=5K\end{cases}}\)

Theo đề bài có : 2x² + y² + z² = ( -72 )

Mà \(x^2\ge0\forall x\)nên \(2x^2\ge0\forall x\); \(y^2\ge0\forall y\); \(z^2\ge0\forall z\)

 => \(2x^2+y^2+z^2\ge0\forall x,y,z\)

=> Ko có giá trị thỏa mãn x,y,z

ai trả lời giúp em câu này với