Lời giải:
a. Các đơn thức: $\frac{4\pi r^3}{3}; \frac{p}{2\pi}; 0; \frac{1}{\sqrt{2}}$

b. Đa thức:

$\frac{4\pi r^3}{3}$ có 1 hạng tử

$\frac{p}{2\pi}$ có 1 hạng tử

$0$ có 1 hạng tử

$\frac{1}{\sqrt{2}}$ có 1 hạng tử 

$ab-\pi r^2$ có 2 hạng tử

 $x^3-x+1$ có 3 hạng tử

h

pt đã cho \(\Leftrightarrow\dfrac{2x-50}{50}-1+\dfrac{2x-51}{49}-1+\dfrac{2x-52}{48}-1+\dfrac{2x-53}{47}-1+\dfrac{2x-200}{25}+4=0\)

\(\Leftrightarrow\dfrac{2x-50-50}{50}+\dfrac{2x-51-49}{49}+\dfrac{2x-52-48}{48}+\dfrac{2x-53-47}{47}+\dfrac{2x-200+100}{25}=0\)

\(\Leftrightarrow\dfrac{2x-100}{50}+\dfrac{2x-100}{49}+\dfrac{2x-100}{48}+\dfrac{2x-100}{47}+\dfrac{2x-100}{25}=0\)

\(\Leftrightarrow\left(2x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)

\(\Leftrightarrow2x-100=0\) (vì \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}>0\))

\(\Leftrightarrow x=50\)

Vậy pt đã cho có tập nghiệm \(S=\left\{50\right\}\)

Do AB // DE (gt)

Theo hệ quả của định lý Thalès, ta có:

AB/DE = BC/CD

x = BC = AB.CD : DE

x = BC = 5.7,2 : 15 = 2,4

Do AB // DE (gt)

Theo hệ quả của định lý Thalès, ta có:

AB/DE = AC/CE

y = CE = AC.DE : AB

= 3.15 : 7,2

= 6,25

=> (x+1) . 5 = (2x + 5) . 3 

     5x + 5  = 6x + 15

     5x + 6x = 15-5

           11x =10

              x = 11\10

câu a

\(\left(\dfrac{2x}{3x+1}-1\right):\left(1-\dfrac{8x^2}{9x^2-1}\right)\\ =\left(\dfrac{2x}{3x+1}-\dfrac{3x+1}{3x+1}\right):\left(\dfrac{9x^2-1}{9x^2-1}-\dfrac{8x^2}{9x^2-1}\right)\\ =\left(\dfrac{2x}{3x+1}-\dfrac{3x+1}{3x+1}\right):\left(\dfrac{9x^2-1}{\left(3x-1\right)\left(3x+1\right)}-\dfrac{8x^2}{\left(3x-1\right)\left(3x+1\right)}\right)\\ =\left(\dfrac{2x-3x-1}{3x+1}\right):\left(\dfrac{9x^2-1-8x^2}{\left(3x-1\right)\left(3x+1\right)}\right)\)

\(=\left(\dfrac{-x-1}{3x+1}\right):\left(\dfrac{x^2-1}{\left(3x-1\right)\left(3x+1\right)}\right)\\ =\dfrac{-x-1}{3x+1}\cdot\dfrac{\left(3x-1\right)\left(3x+1\right)}{x^2-1}\)

\(=\dfrac{-\left(x+1\right)\cdot\left(3x-1\right)\cdot\left(3x+1\right)}{\left(3x+1\right)\cdot\left(x-1\right)\cdot\left(x+1\right)}\\ =\dfrac{-3x+1}{x-1}\)

câu b

thay \(x=2\) vào P ta được

\(\dfrac{-3\cdot2+1}{2-1}=\dfrac{-6+1}{1}=-5\)

vậy \(P=5\) khi \(x=2\)

câu a) 

\(\dfrac{2y-1}{y}-\dfrac{2x+1}{x}\\ =\dfrac{2xy-x}{xy}-\dfrac{2xy+y}{xy}\\ =\dfrac{2xy-x-2xy-y}{xy}\\ =\dfrac{-x-y}{xy}\)

câu b) 

\(\dfrac{2x}{3}:\dfrac{5}{6x^2}\\ =\dfrac{2x}{3}\cdot\dfrac{6x^2}{5}\\ =\dfrac{2x\cdot6x^2}{3\cdot5}\\ =\dfrac{12x^3}{15}=\dfrac{4x^3}{5}\)

Trục hoành (trục Ox) nằm ngang

Trục tung (trục Oy) thẳng đứng

Thế cx k bt :)

\(A=4x^2+16x-9=4\left(x^2+4x+4\right)-25=4\left(x+2\right)^2-25\ge-25\)

\(A_{min}=-25\) khi \(x=-2\)

\(B=-5x^2-29x-20=-5\left(x^2+\dfrac{29}{5}x+\dfrac{841}{100}\right)+\dfrac{441}{20}\)

\(B=-5\left(x+\dfrac{29}{10}\right)^2+\dfrac{441}{20}\le\dfrac{441}{20}\)

\(B_{max}=\dfrac{441}{20}\) khi \(x=-\dfrac{29}{10}\)