a) 0.1+0.5+0.9+1.3+...+9.7+10.1

=(10.1+0.1)+...+(4.9+5.3)

=10.2+...+10.2

=10.2x13

=132.6

b) 2.5x4+20-5x2.5+7.5

=2.5x4+2.5x8-5x2.5+2.5x3

=2.5(4+8-5+3)

=2.5x10

=25

a) dãy số trên có 26 số hạng

tổng mỗi cặp số đầu và số cuối là 10.1+0.1 = 10.2

vậy tổng dãy trên là 10.2 x 26 : 2 = 81,6

b) 2.5 x 4 + 20 - 5 x 2,5 + 7,5

  10 + 20 - 12,5 + 7,5

30 - 5

25

2x2:(7.5-2.5)

=4:(7.5-2.5)

=4:5

=0.8

\(\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\)

\(\left|2x\right|-2,5=7,5\)

|2x| = 7,5 + 2,5

|2x | = 10

TH1: 2x = 10

x=5

TH2: 2x= -10

x = -5

KL: x = 5 hoặc x = -5

\(\left|2x\right|-\left|-2\cdot5\right|=\left|-7\cdot5\right|\)

\(\Rightarrow\left|2x\right|-\left|-10\right|=\left|-35\right|\)

\(\Rightarrow\left|2x\right|-10=35\)

\(\Rightarrow\left|2x\right|=45\)

\(\Rightarrow\orbr{\begin{cases}2x=45\\2x=-45\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=22,5\\x=-22,5\end{cases}}\)

\(\left|2,5-x\right|=7,5\)

\(\orbr{\begin{cases}2,5-x=7,5\\2,5-x=-7,5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2,5-7,5\\x=2,5+7,5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-5\\x=10\end{cases}}\)

\(Vayx\in\left\{-5;10\right\}\)

\(1,6-\left|x-0,2\right|=0\)

\(\left|x-0,2\right|=1,6\)

\(\orbr{\begin{cases}x-0,2=1,6\\x-0,2=-1,6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1,6+0,2\\x=-1,6+0,2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1,8\\x=-1,4\end{cases}}\)

\(A=\dfrac{6^{12}.3^3.5-7.9^7.2^{13}}{2.4^7.5-2^{14}.3^2}\)

\(=\dfrac{2^{12}.3^{12}.3^3.5-7.3^{14}.2^{12}}{2.2^{14}.5-2^{14}.3^2}\)

\(=\dfrac{2^{12}.3^{15}.5-7.3^{14}.2^{13}}{2.3^{14}.5-2^{14}.3^2}\)

\(=\dfrac{2^{12}.3^{14}.\left(15-14\right)}{2^{14}.\left(2.5-3\right)}\)

\(=\dfrac{3^{14}.1}{2^2}\)

\(=\dfrac{314}{4}\)

Đáp án là 2016

<7.5A+2.5A>+<6.9B+3.1B>+<5.6C+4.4C>

10A+10B+10C

10<A+B+C>

10<201.6>

2016

Giải:

a) \(\left(\dfrac{1}{x}-3\right)\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\\dfrac{2}{3}x+\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=3\\\dfrac{2}{3}x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{4}\end{matrix}\right.\)

Vậy ...

b) \(\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\)

\(\Leftrightarrow\left|2x\right|-2,5=7,5\)

\(\Leftrightarrow\left|2x\right|=10\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

Vây ...

c) \(x-7\ge0\Leftrightarrow x\ge7\)

\(\left|1-3x\right|=x-7\)

\(\Leftrightarrow\left[{}\begin{matrix}1-3x=x-7\\1-3x=7-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x-x=-7-1\\-3x+x=7-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-4x=-8\\-2x=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=-3\left(l\right)\end{matrix}\right.\)

Vậy ...

{ x² - [ 6² - ( 8² - 9.7)³ - 7.5]³ - 5.3 }³ = 1

{ x² + [ 36 - (64 - 63)³ - 35]³ - 15}³ = 1

[ x² - ( 36 - 1³ - 35 ) - 15 ]³ = 1

[ x² - ( 36 - 1 - 35 ) - 15]³ = 1

[ x² - ( 35 - 35 ) - 15]³ = 1

[ x² - 0 - 15]³ = 1

( x² - 15 )³ = 1

<=> ( x² - 15)³ = 1³

=> x² - 15 = 1

<=> x² = 16

=> x = 4