Cho x-y=7.Tinh B=3x-7/2x-y - 3y+7/2y+x
(các bn giải nhanh dùm mik nha.cảm ơn các bn)
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a. Ta có: \(x^2-10x+26+y^2+2y=0\Leftrightarrow\left(x^2-10x+25\right)+\left(y^2+2y+1\right)=0\\ \)
\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\Rightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}}\)
b. \(\left(2x+5\right)^2-\left(x-7\right)^2=0\Leftrightarrow\left(2x+5+x-7\right).\left(2x+5-x+7\right)=0\)
\(\Leftrightarrow\left(3x-2\right).\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-12\end{cases}}}\)
c. \(25.\left(x-3\right)^2=49.\left(1-2x\right)^2\Leftrightarrow\left(5x-15\right)^2=\left(7-14x\right)^2\Leftrightarrow\left(5x-15\right)^2-\left(7-14x\right)^2=0\)
\(\Leftrightarrow\left(5x-15-7+14x\right).\left(5x-15+7-14x\right)=0\Leftrightarrow\left(19x-22\right).\left(-9x-8\right)=0\)
\(\Leftrightarrow\left(19x-22\right).\left(9x+8\right)=0\Leftrightarrow\orbr{\begin{cases}19x-22=0\\9x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{22}{19}\\x=-\frac{8}{9}\end{cases}}}\)
d. \(\left(x+2\right)^2=\left(3x-5\right)^2\Leftrightarrow\left(x+2\right)^2-\left(3x-5\right)^2=0\Leftrightarrow\left(x+2+3x-5\right).\left(x+3-3x+5\right)=0\)
\(\Leftrightarrow\left(4x-3\right).\left(8-2x\right)=0\Leftrightarrow\orbr{\begin{cases}4x-3=0\\8-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=4\end{cases}}}\)
e. \(x^2-2x+1=16\Leftrightarrow\left(x-1\right)^2-16=0\Leftrightarrow\left(x-1-4\right).\left(x-1+4\right)=0\)
\(\Leftrightarrow\left(x-5\right).\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
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 \(x=y=z=2\)
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
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