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:
$0.05\cdot 2(\dfrac{x-1}{2009}+\dfrac{x}{2010}+\dfrac{x+1}{2011})=3.3-(\dfrac{x-1}{2009}+\dfrac{x}{2010}+\dfrac{x+1}{2011})$
$\to 0.05\cdot 2u=3.3-u$
$\to 0.1u=3.3-u$
$\to 1.1u=3.3$
$\to u=3$
$\to \dfrac{x-1}{2009}+\dfrac{x}{2010}+\dfrac{x+1}{2011}=3$
$\to \dfrac{x-1}{2009}-1+\dfrac{x}{2010}-1+\dfrac{x+1}{2011}-1=0$
$\to \dfrac{x-1-2009}{2009}+\dfrac{x-2010}{2010}+\dfrac{x+1-2011}{2011}=0$
$\to \dfrac{x-2010}{2009}+\dfrac{x-2010}{2010}+\dfrac{x-2010}{2011}=0$
$\to (x-2010)(\dfrac1{2009}+\dfrac1{2010}+\dfrac1{2011})=0$
$\to x-2010=0$
$\to x=2010$
