Giải thích các bước giải:

Đặt như hướng dẫn ta có:

a.$6u-9=3u+7$

$\to  3u=16$

$\to u=\dfrac{16}{7}$

$\to \dfrac{16x+3}{7}=\dfrac{16}{7}$

$\to 16x+3=16$

$\to 16x=13$

$\to x=\dfrac{13}{16}$

b.Ta có:

$(\sqrt{2}+2)(x\sqrt{2}-1)=2x\sqrt{2}-\sqrt{2}$

$\to (\sqrt{2}+2)(x\sqrt{2}-1)=2x\sqrt{2}-2+2-\sqrt{2}$

$\to (\sqrt{2}+2)(x\sqrt{2}-1)=2(x\sqrt{2}-1)+2-\sqrt{2}$

$\to (\sqrt{2}+2)(x\sqrt{2}-1)-2(x\sqrt{2}-1)=2-\sqrt{2}$

$\to (\sqrt{2}+2-2)(x\sqrt{2}-1)=2-\sqrt{2}$

$\to (\sqrt{2}+2-2)u=2-\sqrt{2}$

$\to \sqrt{2}u=2-\sqrt{2}$

$\to u= \sqrt{2}-1$

$\to x\sqrt{2}-1= \sqrt{2}-1$

$\to x\sqrt{2}=\sqrt{2}$

$\to x=1$

c.Ta có:

$0.05\cdot 2(\dfrac{x-1}{2009}+\dfrac{x}{2010}+\dfrac{x+1}{2011})=3.3-(\dfrac{x-1}{2009}+\dfrac{x}{2010}+\dfrac{x+1}{2011})$

$\to 0.05\cdot 2u=3.3-u$

$\to 0.1u=3.3-u$

$\to 1.1u=3.3$

$\to u=3$

$\to \dfrac{x-1}{2009}+\dfrac{x}{2010}+\dfrac{x+1}{2011}=3$

$\to \dfrac{x-1}{2009}-1+\dfrac{x}{2010}-1+\dfrac{x+1}{2011}-1=0$

$\to \dfrac{x-1-2009}{2009}+\dfrac{x-2010}{2010}+\dfrac{x+1-2011}{2011}=0$

$\to \dfrac{x-2010}{2009}+\dfrac{x-2010}{2010}+\dfrac{x-2010}{2011}=0$

$\to (x-2010)(\dfrac1{2009}+\dfrac1{2010}+\dfrac1{2011})=0$

$\to x-2010=0$

$\to x=2010$