cho x-y=7. tính B= 3x-7/2x+y- 3y-7/2y+x
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x-y=7
nên x=y+7
\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y-7}{2y+x}\)
\(=\dfrac{3\left(y+7\right)-7}{2\cdot\left(y+7\right)+y}-\dfrac{3y-7}{2y+y+7}\)
\(=\dfrac{3y+21-7}{2y+14+y}-\dfrac{3y-7}{3y+7}\)
\(=\dfrac{3y+14}{3y+14}-\dfrac{3y-7}{3y+7}\)
\(=1-\dfrac{3y-7}{3y+7}=\dfrac{3y+7-3y+7}{3y+7}=\dfrac{14}{3y+7}\)
\(x-y=7\Rightarrow x=7+y\)
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
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 \(x=y=z=2\)
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
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