x-y=7

nên x=y+7

\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y-7}{2y+x}\)

\(=\dfrac{3\left(y+7\right)-7}{2\cdot\left(y+7\right)+y}-\dfrac{3y-7}{2y+y+7}\)

\(=\dfrac{3y+21-7}{2y+14+y}-\dfrac{3y-7}{3y+7}\)

\(=\dfrac{3y+14}{3y+14}-\dfrac{3y-7}{3y+7}\)

\(=1-\dfrac{3y-7}{3y+7}=\dfrac{3y+7-3y+7}{3y+7}=\dfrac{14}{3y+7}\)

B=\(\frac{3x-7}{2x+y}-\frac{3y+7}{2y+x}\)

=\(\frac{3x-\left(x-y\right)}{2x+y}-\frac{3y+\left(x-y\right)}{2y+x}\)

=\(\frac{3x-x+y}{2x+y}-\frac{3y+x-y}{2y+x}\)

=\(\frac{2x+y}{2x+y}-\frac{2x+x}{2x+x}\)

=1-1

=0

\(x-y=7\Rightarrow x=7+y\)

B=\(\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2x+y}\)

=\(\dfrac{3\left(7+y\right)-7}{2\left(7+y\right)+y}-\dfrac{3y+7}{2y+7+y}\)

=\(\dfrac{21+3y-7}{14+2y+y}-\dfrac{3y+7}{3y+7}\)

=\(\dfrac{14+3y}{14+3y}-\dfrac{3y+7}{3y+7}\)

=1-1=2

Vậy B=2

mk nhầm

B=1-1=0

Vậy B=0

Với mọi a;b;c không âm ta có:

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)

\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}\)

Áp dụng:

a.

\(VT\le\sqrt{3\left(x+7+y+7+z+7\right)}=\sqrt{3\left(6+21\right)}=9\)

Dấu "=" xảy ra khi \(x=y=z=2\)

b.

\(VT\le\sqrt{3\left(3x+2y+3y+2z+3z+2x\right)}=\sqrt{15\left(x+y+z\right)}=\sqrt{15.6}=3\sqrt{10}\)

Dấu "=" xảy ra khi \(x=y=z=2\)

c.

\(VT\le\sqrt{3\left(2x+5+2y+5+2z+5\right)}=\sqrt{3\left(2.6+15\right)}=9\)

Dấu "=" xảy ra khi \(x=y=z=2\)

\(\frac{4m-2n}{4m+5n}\) với \(\frac{m}{n}=\frac{1}{5}\)

Ta có : \(\frac{m}{n}=\frac{1}{5}\)hay \(\frac{m}{1}=\frac{n}{5}\)

Đặt \(\frac{m}{1}=\frac{n}{5}=k\Rightarrow\hept{\begin{cases}m=k\\n=5k\end{cases}}\)

Do đó \(\frac{4m-2n}{4m+5n}=\frac{4k-2\cdot5k}{4k+5\cdot5k}=\frac{4k-10k}{4k+25k}=\frac{-6k}{29k}=-\frac{6}{29}\)

b. \(\frac{2x+7}{3x-y}+\frac{2y-7}{3y-x}\)

Ta có : x - y = 7 => x = 7 + y

Do đó \(\frac{2x+7}{3x-y}+\frac{2y-7}{3y-x}=\frac{2\left(7+y\right)+7}{3\left(7+y\right)-y}+\frac{2y-7}{3y-\left(7+y\right)}\)

\(=\frac{14+2y+7}{21+3y-y}+\frac{2y-7}{3y-7-y}\)

\(=\frac{21+2y}{21+2y}+\frac{2y-7}{2y-7}=1+1=2\)

a) \(\frac{m}{n}=\frac{1}{5}\Rightarrow\frac{m}{1}=\frac{n}{5}\)

Đặt \(\frac{m}{1}=\frac{n}{5}=k\Rightarrow\hept{\begin{cases}m=k\\n=5k\end{cases}}\)

Thế vào ta được :

\(\frac{4m-2n}{4m+5n}=\frac{4k-2.5k}{4k+5.5k}=\frac{4k-10k}{4k+25k}=\frac{-6k}{29k}=-\frac{6}{29}\)

b) x - y = 7 => x = 7 + y

Thế vào ta được :

\(\frac{2x+7}{3x-y}+\frac{2y-7}{3y-x}=\frac{2\left(7+y\right)+7}{3\left(7+y\right)-y}+\frac{2y-7}{3y-\left(7+y\right)}\)

\(=\frac{21+2y}{21+2y}+\frac{2y-7}{3y-7-y}\)

\(=\frac{21+2y}{21+2y}+\frac{2y-7}{2y-7}=1+1=2\)