a: ĐKXĐ: \(\begin{cases}x\ge0\\ x<>4\end{cases}\)

\(P=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{4\sqrt{x}-4}{4-x}\right):\left(1+\frac{5}{\sqrt{x}-2}\right)\)

\(=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{\sqrt{x}-2+5}{\sqrt{x}-2}\)

\(=\frac{x+5\sqrt{x}+6+x-3\sqrt{x}+2-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

\(=\frac{2x-2\sqrt{x}+12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

b: \(P>\frac12\)

=>\(\frac{2x-2\sqrt{x}+12}{x+5\sqrt{x}+6}-\frac12>0\)

=>\(\frac{4x-4\sqrt{x}+24-x-5\sqrt{x}-6}{2\left(x+5\sqrt{x}+6\right)}>0\)

=>\(3x-9\sqrt{x}+18>0\)

=>\(x-3\sqrt{x}+6>0\)

=>\(\left(\sqrt{x}-\frac32\right)^2+\frac{15}{4}>0\) (luôn đúng với mọi x thỏa mãn ĐKXĐ)

=>x>=0 và x<>4

b. nhé

mn giúp e với ạ

a: 2+5/6=12/6+5/6=17/6

b: 5/12+3/4+1/3=5/12+9/12+4/12=18/12=3/2

c: 2/3+3/4=8/12+9/12=17/12

\(=>\dfrac{R1}{R2}=\dfrac{l1}{l2}=>\dfrac{20}{R2}=\dfrac{3}{6}=>R2=\dfrac{20.6}{3}=40\Omega\)

a) \(2\left(x+3\right)=4x-\left(2+x\right)\)

\(2x+6=3x-2\)

\(-x=-8\)=>x=8

b) \(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\) đk x khác 2 và -2

\(\dfrac{\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}\)

=>\(x-2-5x-10=2x-3\)

\(-6x=9=>x=\dfrac{3}{2}tm\)