Ta có: \(3x\left(x+5\right)-\left(3x+18\right)\left(x-1\right)\)

\(=3x^2+15x-\left(3x^2-3x+18x-18\right)\)

\(=3x^2+15x-3x^2-15x+18\)

=18

ta có :

3x.(x+5)-(3x+18)(x-1)

=(\(3x^2+15x\))-(\(3x^2-3x+18x-17\))

=\(3x^2+15x-3x^2+3x-18x+17\)

=18x-18x+17

=17

Vậy biểu thức trên không thuộc biến x

địt con mẹ mày lên

\(\dfrac{1}{3}\)-\(\dfrac{2}{15}\)+\(\dfrac{14}{15}\)=\(\dfrac{3}{15}\)-\(\dfrac{2}{15}\)+\(\dfrac{14}{15}\)

=\(\dfrac{13}{15}\)

\(\dfrac{3}{5}\)+4-\(\dfrac{6}{7}\)=\(\dfrac{3}{5}\)+\(\dfrac{20}{5}\)-\(\dfrac{6}{7}\)=\(\dfrac{23}{5}\)-\(\dfrac{6}{7}\)=\(\dfrac{161}{35}\)-\(\dfrac{30}{35}\)=\(\dfrac{131}{35}\)

Ta có : \(\left\{{}\begin{matrix}\dfrac{1}{14}< \dfrac{1}{10}\\\dfrac{1}{23}< \dfrac{1}{10}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{14}+\dfrac{1}{23}< \dfrac{1}{10}+\dfrac{1}{10}=\dfrac{1}{5}\)

Lại có : \(\left\{{}\begin{matrix}\dfrac{1}{62}< \dfrac{1}{60}\\\dfrac{1}{83}< \dfrac{1}{60}\\\dfrac{1}{117}< \dfrac{1}{60}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{62}+\dfrac{1}{83}+\dfrac{1}{117}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

\(\Rightarrow\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{23}+\dfrac{1}{62}+\dfrac{1}{83}+\dfrac{1}{117}< \dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{20}=\dfrac{9}{20}< \dfrac{10}{20}=\dfrac{1}{2}\)
 

a) 4 . ( - 2 ) + 14  <  4 . ( - 1 ) + 14

Ta có : 4 . ( - 2 ) + 14

          = -8  + 14

          = 6

Lại có : 4 . ( - 1 ) + 14

          =    - 4   + 14

          =      10

Vậy  6 < 10 hay 4 . ( - 2 ) + 14 < 4 . ( -1 ) + 14

b) ( - 3 ) . 2 + 5 < ( - 3 ) .( - 5 ) + 5

Ta có : ( - 3 ) . 2 + 5

          = - 6 + 5 

          = -1

Lại có : ( -3 ) .( - 5 ) + 5

           = 15 + 5 

           = 20

Vậy ( - 1 ) < 20 hay ( - 3 ) . 2 + 5 < ( - 3 ) .( -5 ) + 5

ap dung bdt am gm

\(\sqrt{1+8a^3}=\sqrt{\left(1+2a\right)\left(4a^2-4a+1\right)}\)\(\le\frac{1+2a+4a^2-2a+1}{2}=\frac{4a^2+2}{2}=2a^2+1\)

\(\Rightarrow\frac{1}{\sqrt{1+8a^3}}\ge\frac{1}{2a^2+1}\)

tuongtu ta cung co \(\frac{1}{\sqrt{1+8b^3}}\ge\frac{1}{2b^2+1};\frac{1}{\sqrt{1+8c^3}}\ge\frac{1}{2c^2+1}\)

\(\Rightarrow\)VT\(\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\)

tiep tuc ap dung bat cauchy-schwarz dang engel ta co

\(VT\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\ge\frac{\left(1+1+1\right)^2}{2\left(a^2+b^2+c^2\right)+3}=\frac{3^2}{6+3}=1\)(dpcm)

dau = xay ra \(\Leftrightarrow a=b=c=1\)

Ta thấy cả ba phân số trong biểu thức đều bé hơn 3 nên cộng vào nó cũng bé hơn 3.Vậy:

1/5 + 1/14 + 1/28 < 1/3

Ta có :\(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}=\frac{1}{5}+\frac{3}{28}< \frac{1}{5}+\frac{3}{27}=\frac{1}{5}+\frac{1}{9}=\frac{14}{15}< \frac{15}{45}=\frac{1}{3}\left(\text{đpcm}\right)\)

Mình nhầm.

Ta thấy cả ba phân số trong biểu thức đều bé hơn 1/3,nên cộng vào nó cũng bé hơn 1/3.Vậy:

1/5 + 1/14 + 1/28 < 1/3

 S=1/5+ 1/13+1/14+1/15+1/61+1/62+1/63< 1/2

S = 1/5 + ( 1/13 + 1/14 + 1/15 ) + ( 1/ 61 + 1/ 62 + 1/ 63 )

=> S < 1/5 + 1/12 . 3 + 1/ 60 . 3

=> S < 1/5 + 1/4 + 1/20

=> S < 1/2

Vậy S < 1/2

\(\Delta s=r\Delta\alpha\) 

=> \(\frac{\Delta s}{\Delta t}=r\frac{\Delta\alpha}{\Delta t}\)

mà \(\omega=\frac{\Delta\alpha}{\Delta t}\)

=> \(v=r\omega\)

a(ht)=(v^2)/r

       = ((rω)^2)/r

       = (r^2xω^2)/r

a(ht) = rω^2