b) ĐKXĐ: x>-4

\(x^2+3x+4=x+4\)

\(x^2+2x=0\)

\(x\left(x+2\right)=0\)

\(\left[{}\begin{matrix}x=0\\x+2=0\Leftrightarrow x=-2\end{matrix}\right.\)

a) \(\dfrac{3x^2+1}{\sqrt{x-1}}=\dfrac{4}{\sqrt{x-1}}\)

ĐKXĐ: \(x>1\)

\(3x^2+1=4\)

\(3x^2=3\)

\(x^2=1\)

\(x=\pm1\)

=> Pt vô nghiệm

a) \(3x-2=0\Leftrightarrow x=\dfrac{2}{3}\)

Thay \(x=\dfrac{2}{3}\)

\(\left(m+3\right)\)\(\dfrac{2}{3}-m+4=0\)

\(\dfrac{2}{3}m+2-m+4=0\)

\(\dfrac{-1}{3}m+6=0\)

\(\dfrac{-1}{3}m=-6\)

\(m=18\)

2) \(\dfrac{2m-1}{x-1}=m-2\)(1)

ĐKXĐ: \(x\ne1\)

\(2m-1=\left(m-2\right)\left(x-1\right)\)

\(2m-1=xm-m-2x+2\)

\(0=xm-3x-2x+3\)

\(0=m\left(x-3\right)-2x+3\)

Để (1) có nghiệm duy nhất thì m(x-3)=0=>m=0

Vậy khi m=0 thì pt (1) có nghiệm duy nhất

3) \(\dfrac{x+a}{x+1}+\dfrac{x-2}{x}=2\)(1)

\(\dfrac{x+a}{x+1}+\dfrac{x-2}{x}-2=0\)

ĐKXĐ: \(x\ne1;x\ne0\)

\(\dfrac{\left(x+a\right)x}{x\left(x+1\right)}+\dfrac{\left(x-2\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{2\left(x+1\right)x}{\left(x+1\right)x}=0\)

\(\left(x+a\right)x+\left(x+1\right)\left(x+2\right)-2\left(x+1\right)x=0\)

\(x^2+ax+x^2-x-2-2x^2-2x=0\)

\(ax-3x-2=0\)

\(\left(a-3\right)x-2=0\)

Để pt vô nghiệm thì \(\left(a-3\right)=0\Leftrightarrow a=3\)

Vậy khi a=3 thì pt (1) vô nghiệm

5) 0<a<b, ta có: a<b

<=> a.a<a.b

<=>a²<a.b

<=>\(a< \sqrt{ab}\)(1)

- BĐT Cauchy:

\(\dfrac{a+b}{2}\ge\sqrt{ab}\) khi \(a\ge0;b\ge0\)

\(\Leftrightarrow\sqrt{ab}\le\dfrac{a+b}{2}\)

Dấu = xảy ra khi a=b=0 mà 0<a<b

=> \(\sqrt{ab}< \dfrac{a+b}{2}\)(2)

- 0<a<b, ta có: a<b<=> a+b<b+b

\(\Leftrightarrow\)\(\dfrac{a+b}{2}< \dfrac{b+b}{2}\)

\(\Leftrightarrow\dfrac{a+b}{2}< b\left(3\right)\)

Từ (1), (2), (3), ta có đpcm

Đậu phộng rang!!!! v~~ =='