P=\(\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}\)=\(\sqrt{2+5+7+2\sqrt{5.2}+2\sqrt{2.7}+2\sqrt{3.5}}\)

=\(\sqrt{\left(\sqrt{2}+\sqrt{5}+\sqrt{7}\right)^2}\)=\(\sqrt{2}+\sqrt{5}+\sqrt{7}\)=\(\sqrt{a}+\sqrt{b}+\sqrt{c}\)

Vậy a+b+c=14

b)\(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}=\dfrac{a+\sqrt{a}+1}{\sqrt{a}}\)

\(\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}=\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\)

\(\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)=\dfrac{a-1}{\sqrt{a}}\left(\dfrac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{a-1}\right)=\dfrac{2a+2}{\sqrt{a}}\)

A=2+\(\dfrac{2a+2}{\sqrt{a}}\)

a)ĐKXĐ:a\(\ge0\),a\(\ne1\)

c)Để A=7=>2+\(\dfrac{2a+2}{\sqrt{a}}\)=7=>2a+2=5\(\sqrt{a}\)=>a=2 và a=1/2

d)Để A>6=>2+\(\dfrac{2a+2}{\sqrt{a}}\)>6 =>\(\dfrac{2a+2}{\sqrt{a}}\)>4=>2a+2>4\(\sqrt{a}\)=>a>1

N=(2²-1²)+(4²-3²)+(6²-5²)+....+(100²-99²)

=(2-1)(2+1)+(4-3)(4+3)+....+(100-99)(100+99)

=3+7+11+15+....+195+199=(199+3)(\(\dfrac{199-3}{4}\)+1)=10100

Cho 4 cặp oxi hóa-khử: Fe²⁺/Fe; Fe³⁺/Fe; Ag⁺/Ag; Cu²⁺/Cu. Dãy cặp xếp theo chiều tăng dần tính oxi hóa và giảm dần tính khử là:

A. Ag⁺/Ag; Fe^3+/Fe²⁺; Cu²⁺/Cu; Fe²⁺/Fe.

B. Cu²⁺/Cu; Fe²⁺/Fe; Fe³⁺/Fe²⁺; Ag⁺/Ag.

C. Fe³⁺/Fe²⁺; Fe²⁺/Fe; Ag⁺/Ag; Cu²⁺/Cu.

D. Fe²⁺/Fe; Cu²⁺/Cu; Fe³⁺/Fe²⁺; Ag⁺/Ag.