3,1 . 28 - 72 . 3,1-21 + (-3100)

= 3,1 . 28 - 72 . 3,1 - 21 - 3,1 * 1000

= 3,1 (28 - 72 - 1000) -21

= 3,1 (-1044) -21

=

.....

(thấy kì kì)

Ta có:

\(3,1\left(21\right)=3,1+0,0\left(21\right)=\frac{31}{10}+0,0\left(01\right).21=\frac{31}{10}+\frac{0,\left(01\right)}{10}.21=\frac{31}{10}+\frac{\frac{1}{99}}{10}.21\)

\(=\frac{31}{10}+\frac{21}{9900}=\frac{3069}{9900}+\frac{21}{9900}=\frac{3090}{9900}=\frac{103}{330}\)

\(A=4,8.\left(3,1-1,5\right)+1,5.\left(4,8-3,1\right)\)

\(A=4,8.3,1-4,8.1,5+1,5.4,8-1,5.3,1\)

\(A=3,1.\left(4,8-1,5\right)-4,8\left(1,5+1,5\right)\)

\(A=3,1.3,3-4,8.3\)

\(A=10,23-14,4=-4,17\)

\(B=\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\dfrac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{2.3^9.2^{10}+3^{10}.\left(2^2\right)^{10}}=\dfrac{2^{19}.3^9+3^9.2^{18}.5}{2^{11}.3^9+3^{10}.2^{20}}=\dfrac{2^{18}.3^9\left(2+5\right)}{2^{11}.3^9\left(1+3.2^9\right)}=\dfrac{2^7.7}{1+3.2^9}\)

\(C=-1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-\dfrac{1}{28}-\dfrac{1}{36}-\dfrac{1}{45}\)

\(\Rightarrow C=\dfrac{-1}{2}\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(\Rightarrow C=\dfrac{-1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right)\)

\(\Rightarrow C=\dfrac{-1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(\Rightarrow C=\dfrac{-1}{2}\left(1-\dfrac{1}{10}\right)\)

\(\Rightarrow C=\dfrac{-1}{2}.\dfrac{9}{10}\)

\(\Rightarrow C=\dfrac{-9}{20}\)

a, | 0,2 . x - 3,1 | = 6,3

\(\Rightarrow0,2.x-3,1=\pm6,3\)

+) \(0,2.x-3,1=6,3\Rightarrow x=47\)

+) \(0,2.x-3,1=-6,3\Rightarrow x=16\)

Vậy x = 47 hoặc x = 16

b, | 12,1 . x - 3,1 | = 12,1

\(\Rightarrow12,1.x-3,1=\pm12,1\)

+) \(12,1.x-3,1=12,1\Rightarrow x=\frac{151}{121}\)

+) \(12,1.x-3,1=-12,1\Rightarrow x=\frac{90}{121}\)

Vậy \(x=\frac{151}{121}\) hoặc \(x=\frac{90}{121}\)

c, | 0,2 . x - 3,1 | + | 0,2 . x + 3,1 | = 0

\(\Rightarrow0,2.x-3,1=0\) hoặc \(0,2.x+3,1=0\)

+) \(0,2.x-3,1=0\Rightarrow x=15,5\)

+) \(0,2.x+3,1=0\Rightarrow x=-15,5\)

Vậy x = 15,5 hoặc x = -15,5

c) ( sửa lại )

| 0,2 . x - 3,1 | + | 0,2 . x + 3,1 | = 0

\(\Rightarrow\) 0,2 . x - 3,1 = 0 hoặc 0,2 . x + 3,1 = 0

+) \(0,2.x-3,1=0\Rightarrow x=15,5\)

+) \(0,2.x+3,1=0\Rightarrow x=-15,5\)

Vì x không thể bằng 15,5 và -15,5 ( 15,5 khác -15,5 ) nên x không có giá trị thỏa mãn đề bài.

Vậy x không có giá trị thỏa mãn đề bài

a) \(\left|0,2x-3,1\right|=6,3\)

\(\Leftrightarrow\begin{cases}x\ge15,5\\0,2x-3,1=6,3\end{cases}\)  hoặc \(\begin{cases}x< 15,5\\0,2x-3,1=-6,3\end{cases}\)

\(\Leftrightarrow\begin{cases}x\ge15,5\\x=47\left(tm\right)\end{cases}\) hoạc \(\begin{cases}x< 15,5\\x=-47\left(tm\right)\end{cases}\)

Vậy x={47;-47}

b) \(\left|12,1x-3,1\right|=12,1\)

\(\Leftrightarrow\begin{cases}x\ge\frac{31}{121}\\12,1x-3,1=12,1\end{cases}\)  hoạc \(\begin{cases}x< \frac{31}{121}\\12,1x-3,1=-12,1\end{cases}\)

\(\Leftrightarrow\begin{cases}x\ge\frac{31}{121}\\x=\frac{152}{121}\left(tm\right)\end{cases}\)  hoặc \(\begin{cases}x< \frac{31}{121}\\x=-\frac{90}{121}\left(tm\right)\end{cases}\)

Vaayj x={152/121;-90/121}

c) \(\left|0,2x-3,1\right|+\left|0,2x+3,1\right|=0\)

\(\Leftrightarrow\begin{cases}0,2x-3,1=0\\0,2x+3,1=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=15,5\\x=-15,5\end{cases}\)(vô lý)

Vậy k có x thỏa mãn

A=0

\(A=\left(3.1-2.5\right)-\left(-2.5+3.1\right)\)

\(=3.1-2.5+2.5-3.1\)

=0

(-20) . 3,1 - 7,2 : 4 + 3,2 . (1,2 - 2,4)

= (-20) . 3,1 - 7,2 : 4 + 3,2 . (-1,2)

= -62 - 1,8 + (-3,84)

= - 63,8 + (-3,84)

= -67,64

(  3,1 - 2,5 ) - ( -2,5 +3,1)

= 3,1 -2,5 + 2,5 -3,1

= ( 3,1 - 3,1) - ( 2,5 -2,5)

=   0 - 0

=      0

2/5 : 2/1