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\(\dfrac{x-1}{2000}+\dfrac{x-3}{1998}+\dfrac{x-5}{1996}+\dfrac{x}{667}=6\\ \Leftrightarrow\left(\dfrac{x-1}{2000}-1\right)+\left(\dfrac{x-3}{1998}-1\right)+\left(\dfrac{x-5}{1996}-1\right)+\left(\dfrac{x}{667}-3\right)=0\\ \Leftrightarrow\dfrac{x-2001}{2000}+\dfrac{x-2001}{1998}+\dfrac{x-2001}{1996}+\dfrac{x-2001}{667}=0\\ \Leftrightarrow\left(x-2001\right)\left(\dfrac{1}{2000}+\dfrac{1}{1998}+\dfrac{1}{1996}+\dfrac{1}{667}\right)=0\\ \Leftrightarrow x-2001=0\\ \Leftrightarrow x=2001\)

ĐK: \(2\le x\le10\)

\(VP=x^2-12x+40=x^2-12x+36+4=\left(x-6\right)^2+4\ge4\forall x\)

\(VT=\sqrt{x-2}+\sqrt{10-x}\le\dfrac{x-2+10-x}{2}=4\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-6\right)^2=0\\\sqrt{x-2}=\sqrt{10-x}\end{matrix}\right.\Leftrightarrow x=6\left(tmđk\right)\)

15D 16A

a) Số học sinh giỏi lớp 6B: \(28\cdot\dfrac{5}{14}=10\) (học sinh)

b) Số học sinh lớp 6B: \(28:70\cdot100=40\) (học sinh)

\(\left(x^2-xy+y^2\right)\left(x+y\right)=x^3+y^3\)

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\(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}}\)

a) \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{b}+1=\dfrac{c}{d}+1\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)

b) \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\). Áp dụng dãy tỉ số bằng nhau:

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\\ \Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

c) Áp dụng dãy tỉ số bằng nhau:

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)