(x-y)(3x-4y)=0

=>x=y hoặc 3x=4y

TH1: x=y

\(B=\dfrac{3y+4y}{5y-4y}+\dfrac{3y-8y}{5y+8y}=7+\dfrac{-5}{13}=\dfrac{86}{13}\)

TH2: 3x=4y

=>x/4=y/3=k

=>x=4k; y=3k

\(B=\dfrac{3x+4y}{5x-4y}+\dfrac{3x-8y}{5x+8y}\)

\(=\dfrac{12k+12k}{20k-12k}+\dfrac{12k-24k}{20k+24k}=\dfrac{24}{8}+\dfrac{-12}{44}=\dfrac{30}{11}\)

a) |x| = 1212

=> x = 1212 hoặc -1212

T_T các câu kia tườn tự vậy thôi bạn, dài quá @_@

học tốt -_-"

a) \(\left|x\right|=1212\)

\(\Rightarrow\orbr{\begin{cases}x=1212\\x=-1212\end{cases}}\)

b) \(\left|2x+1212\right|=3434\)

\(\Rightarrow\orbr{\begin{cases}2x+1212=3434\\2x+1212=-3434\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=2222\\2x=-4646\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1111\\x=-2323\end{cases}}\)

\(\left|x+\dfrac{1}{2}\right|+\left|x-y+z\right|+\left|y+\dfrac{1}{3}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\y+\dfrac{1}{3}=0\\x-y+z=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{1}{3}\\z=-x+y=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\end{matrix}\right.\)

\(A=2x+y+z=-1-\dfrac{1}{3}+\dfrac{1}{6}=-\dfrac{4}{3}+\dfrac{1}{6}=-\dfrac{7}{6}\)

a, x/3 = y/-4 = z/-5 

=> 2x/6 = 3y/-12 = 4z/-20

 theo đề bài áp dụng tính chất của dãy tỉ số bằng nhau ta có :

2x/6 = 3y/-12 = 4z/-20 = 2x + 3y - 4z/6 + (-12) - (20) = 70/14 = 5

=> x = 5.3 = 15

     y = 5.(-4) = -20

     z = 5.(-5) = -25  

Áp dụng dãy tỉ số bằng nhau là đc mà

Ta có: \(\left|\frac{1}{2}+x\right|\ge0;\left|x-y+z\right|\ge0;\left|\frac{1}{3}+y\right|\ge0\)

\(\Rightarrow\left|\frac{1}{2}+x\right|+\left|x-y+z\right|+\left|\frac{1}{3}+y\right|\ge0\)

Mà \(\left|\frac{1}{2}+x\right|+\left|x-y+z\right|+\left|\frac{1}{3}+y\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|\frac{1}{2}+x\right|=0\\\left|x-y+z\right|=0\\\left|\frac{1}{3}+y\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{-1}{2}\\z=\frac{1}{6}\\y=-\frac{1}{3}\end{cases}}}\)

\(\Rightarrow A=2\cdot\left(\frac{-1}{2}\right)+\left(\frac{-1}{3}\right)+\frac{1}{6}=-1-\frac{1}{3}+\frac{1}{6}=\frac{-1}{2}\)

Thấy cái đề mà thấy khiếp ...

Ta có : \(x^2-xy+y^2=\frac{3}{4}\left(x^2-2xy+y^2\right)+\frac{1}{4}\left(x^2+2xy+y^2\right)\)

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

\(\Rightarrow\sqrt{x^2-xy+y^2}\ge\frac{x+y}{2}\)

Tương tự \(\sqrt{y^2-yz+z^2}\ge\frac{y+z}{2}\)

                \(\sqrt{z^2-zx+x^2}\ge\frac{x+z}{2}\)

Do đó : \(2S\ge\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{x+z}{x+z+2y}\)

\(\Rightarrow2S+3\ge\left(1+\frac{x+y}{x+y+2z}\right)+\left(1+\frac{y+z}{y+z+2x}\right)+\left(1+\frac{x+z}{x+z+2y}\right)\)

                       \(=2\left(x+y+z\right)\left(\frac{1}{x+y+2z}+\frac{1}{y+z+2x}+\frac{1}{x+z+2y}\right)\)

                                                         \(\ge2\left(x+y+z\right).\frac{9}{4\left(x+y+z\right)}\)\(=\frac{9}{2}\)

                                                          (Áp dụng bđt Cô-si dạng engel cho 3 số)

\(\Rightarrow2S+3\ge\frac{9}{2}\)

\(\Rightarrow S\ge\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)

Vậy ..............