(x+162,9)x0,02=4,078

x+162,9           = 4,078 : 0,02 

x + 162,9         = 203,9

x                      = 203,9 - 162,9

x                      = 41

( x + 162,9 ) × 0,02 = 4,078

 x + 162,9 = 4,078 : 0,02

 x + 162,9 = 203,9

 x = 203,9 - 162,9

 x = 41

\(x^2+x-6=0\)

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

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

\(=>\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy

=>(x+3)(x-2)=0

=>x=-3 hoặc x=2

a) \(=(\dfrac{1}{6}+\dfrac{24}{6})\times\dfrac{3}{10}\)

    = \(\dfrac{25}{6}\times\dfrac{3}{10}\)

    = \(\dfrac{75}{60}=\dfrac{5}{4}\)

tổng số túi là

945:(15+12)=35(túi)

số kg gạo tẻ là

35.15=525(kg)

số kg gạo nếp là

945-525=420(kg)

\(=\left(\dfrac{1}{21}+\dfrac{1}{210}+\dfrac{1}{2010}\right)\cdot\dfrac{10-6-3-1}{30}=0\)

\(\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{10}-\dfrac{1}{30}\right)\times\left(\dfrac{1}{21}+\dfrac{1}{210}+\dfrac{1}{2010}\right)\)

\(=\left(\dfrac{10}{30}-\dfrac{6}{30}-\dfrac{3}{30}-\dfrac{1}{30}\right)\times\left(\dfrac{1}{21}+\dfrac{1}{210}+\dfrac{1}{2010}\right)\)

\(=0\times\left(\dfrac{1}{21}+\dfrac{1}{210}+\dfrac{1}{2010}\right)\)

\(=0\)

\(\left(\dfrac{1}{6}+4\right)\times\dfrac{3}{10}=\left(\dfrac{1}{6}+\dfrac{24}{6}\right)=\dfrac{25}{6}\times\dfrac{3}{10}=\dfrac{5}{4}\)

\(\left(\dfrac{1}{8}-\dfrac{1}{9}\right)\times\dfrac{3}{2}=\left(\dfrac{9}{72}-\dfrac{8}{72}\right)\times\dfrac{3}{2}=\dfrac{1}{72}\times\dfrac{3}{2}=\dfrac{1}{48}\)

a) = 25/6 x 3/10

= 5/4

b) = 1/72 x 3/2 

= 1/48

3081 l 237

  711 l 13

      0

7824 l 326

1304 l 24

      0

35467 l 678

  1567 l 52

     211

Đáp án C , 3 giờ nhé

c

\(B=x+1+\dfrac{25}{x+2}=x+2+\dfrac{25}{x+2}-1\)

Áp dụng bđt cauchy, ta có \(x+2+\dfrac{25}{x+2}\ge2.\sqrt{\left(x+2\right).\dfrac{25}{x+2}}=2.5=10\)

Do đó B ≥ 10 - 1 = 9

Dấu = xảy ra <=> x = 3

a: góc OBA+góc OCA=180 độ

=>OBAC nội tiếp

b: Xét ΔABE và ΔADB có

góc ABE=góc ADB

góc BAE chung

=>ΔABE đồng dạng với ΔADB

=>AB^2=AE*AD