\(38:\left(2x-1\right)=-2\)

\(\Rightarrow2x-1=38:\left(-2\right)\)

\(\Rightarrow2x-1=-19\)

\(\Rightarrow2x=-19+1\)

\(\Rightarrow2x=-18\)

\(\Rightarrow x=-18:2\)

\(\Rightarrow x=-9\)

38 : (2x - 1) = -2

       (2x - 1) = 38 : (-2)

       (2x - 1) = - 19

        2x - 1  = -19

        2x       = -19 + 1

        2x       = -18

          x       = -18 : 2

          x       = -9 

\(\dfrac{45\%}{x-1,25}-2\dfrac{3}{8}=-1\dfrac{31}{40}\)

\(\Rightarrow\dfrac{0,45}{x-1,25}-\dfrac{19}{8}=-\dfrac{71}{40}\)

\(\Rightarrow\dfrac{0,45}{x-1,25}=-\dfrac{71}{40}+\dfrac{19}{8}\)

\(\Rightarrow\dfrac{0,45}{x-1,25}=\dfrac{3}{5}\)

\(\Rightarrow3\left(x-1,25\right)=5\cdot0,45\)

\(\Rightarrow3\left(x-1,25\right)=\dfrac{9}{4}\)

\(\Rightarrow x-1,25=\dfrac{9}{4}:3\)

\(\Rightarrow x-\dfrac{5}{4}=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}+\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{8}{4}=2\)

Bạn xem lại dòng thứ 3 nhé!

100;200;300;400;500;600;700;...;99900

10000 > B(100) = {0; 100; 200;...; 9900}

6,45>0

-3,273<0

Do đó: 6,45>-3,273

\(x-\dfrac{4}{5}=\dfrac{6}{20}+\dfrac{-7}{3}\)

=>\(x-\dfrac{4}{5}=\dfrac{3}{10}-\dfrac{7}{3}=\dfrac{9-70}{30}=\dfrac{-61}{30}\)

=>\(x=-\dfrac{61}{30}+\dfrac{4}{5}=\dfrac{-61+24}{30}=\dfrac{-37}{30}\)

\(x\)  - \(\dfrac{4}{5}\) = \(\dfrac{6}{20}\) + \(\dfrac{-7}{3}\)

\(x\) - \(\dfrac{4}{5}\) = - \(\dfrac{61}{30}\)

\(x\)       = - \(\dfrac{61}{30}\) + \(\dfrac{4}{5}\)

\(x\)      = - \(\dfrac{37}{30}\)

Vậy \(x\) = - \(\dfrac{37}{30}\)

C là trung điểm của AB

=>\(CA=CB=\dfrac{AB}{2}=5\left(cm\right)\)

O là trung điểm của AC

=>\(AO=\dfrac{AC}{2}=\dfrac{5}{2}=2,5\left(cm\right)\)

AC=5 cm ;  CB=5 cm ;  AO=2,5 cm

Lời giải:
a.

$A=32.7^2-22.7^2+90.7^2+25.4.51$

$=7^2(32-22+90)+100.51=49.100+100.51=100(49+51)=100.100=10000$

b.

\(X=\frac{1}{2.6}+\frac{1}{4.9}+\frac{1}{6.12}+...+\frac{1}{36.57}+\frac{1}{438.60}\\ =\frac{1}{(1.2).(2.3)}+\frac{1}{(2.2).(3.3)}+\frac{1}{(3.2)(4.3)}+...+\frac{1}{(18.2)(19.3)}+\frac{1}{(19.2).(20.3)}\)

\(=\frac{1}{2.3}(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20})\)

$=\frac{1}{2.3}(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20})$

$=\frac{1}{6}(1-\frac{1}{20})=\frac{19}{120}$

$B=2023-X=2023-\frac{19}{120}=2022\frac{101}{120}$

c/

$C=1+2023+2023^2+2023^3+...+2023^{2022}+2023^{2023}$

$2023C=2023+2023^2+2023^3+2023^4+...+2023^{2023}+2023^{2024}$

$\Rightarrow 2023C-C=2023^{2024}-1$

$\Rightarrow C=\frac{2023^{2024}-1}{2023}< 2023^{2024}-1$

$\Rightarrow C< D$

= \(7^2\left(32.22+90\right)+\left(25.4\right).51\)

= \(7^2.100+100.51\)

= \(49.100+100.51\)

= \(100.\left(49+51\right)\)

= 100.100

= 10000