-13/6 nha bạn

TL

-13/6

HT nha bn

\(A=\left(3+\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}-\frac{4}{5}\right)\)

\(A=\left(3-5-6\right)+\left(\frac{1}{4}+\frac{7}{4}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)+\left(\frac{6}{5}+\frac{4}{5}\right)\)

\(A=-8+2+1+3\)

\(A=-3\)

\(a,\dfrac{1}{2}-\dfrac{1}{3}-\left(-\dfrac{5}{4}\right)=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{5}{4}=\dfrac{1\times6-1\times4+5\times3}{12}=\dfrac{6-4+15}{12}=\dfrac{17}{12}\\ b,\dfrac{5}{4}-\dfrac{1}{2}-\dfrac{7}{8}=\dfrac{5\times2-1\times4-7}{8}=\dfrac{10-4-7}{8}=-\dfrac{1}{8}\\ c,\dfrac{1}{5}-\dfrac{1}{2}+\dfrac{9}{10}=\dfrac{1\times2-1\times5+9}{10}=\dfrac{2-5+9}{10}=\dfrac{6}{10}=\dfrac{3}{5}\\ d,\dfrac{5}{4}-\dfrac{1}{3}+\dfrac{7}{6}=\dfrac{5\times3-1\times4+7\times2}{12}=\dfrac{15-4+14}{12}=\dfrac{25}{12}\)

`1/2+(-1/3)-(-5/4)`

`=1/2-1/3+5/4`

`=3/6-2/6+5/4`

`=1/6+5/4`

`=2/12+15/12`

`=17/12`

__

`5/4-1/2+(-7/8)`

`=5/4-1/2-7/8`

`=10/8-4/8-7/8`

`=6/8-7/8`

`=-1/8`

__

`1/5-1/2+9/10`

`=2/10-5/10+9/10`

`=-3/10+9/10`

`=6/10`

`=3/5`

__

`5/4-1/3+7/6`

`=15/12-4/12+14/12`

`=11/12+14/12`

`=25/12`

`#lv`

Lời giải:

$A=1+4+4^2+4^3+...+4^{2023}$

$A=1+4+(4^2+4^3+4^4)+(4^5+4^6+4^7)+...+(4^{2021}+4^{2022}+4^{2023})$

$=5+4^2(1+4+4^2)+4^5(1+4+4^2)+....+4^{2021}(1+4+4^2)$

$=5+(1+4+4^2)(4^2+4^5+...+4^{2021})$

$=5+21(4^2+4^5+....+4^{2021})$ 

Do đó biểu thức chia 21 dư 5

Mik quên là bài này là bài tìm x nha mọi người ❤

chưa ai trả lời à!!!

a) \(\frac{3}{4}x-\frac{1}{4}=2\left(x-3\right)+\frac{1}{4}x\)

\(\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)

\(\frac{3}{4}x-2x-\frac{1}{4}x=\frac{1}{4}-6\)

\(x\left(\frac{3}{4}-2-\frac{1}{4}\right)=-\frac{23}{4}\)

\(-\frac{3}{2}x=-\frac{23}{4}\)

\(x=-\frac{23}{4}\div\left(-\frac{3}{2}\right)\)

\(x=\frac{23}{6}\)

b) \(30\%-x+\frac{5}{6}=\frac{1}{3}\)

\(\frac{3}{10}-x+\frac{5}{6}=\frac{1}{3}\)

\(\frac{3}{10}-x=\frac{1}{3}-\frac{5}{6}\)

\(\frac{3}{10}-x=-\frac{1}{2}\)

\(x=\frac{3}{10}-\left(-\frac{1}{2}\right)\)

\(x=\frac{4}{5}\)

c) \(\left(2x-1\right).\left(3x+12\right)=0\)

\(\hept{\begin{cases}2x-1=0\\3x+12=0\end{cases}\Rightarrow\hept{\begin{cases}2x=1\\3x=-12\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\x=-4\end{cases}}}\)

Vậy \(x\in\left\{\frac{1}{2};-4\right\}\)

\(\dfrac{-3}{-5}=\dfrac{3}{5}\)

\(\dfrac{4}{6}=\dfrac{2}{3}< \dfrac{3}{2}\)

\(\dfrac{-3}{-21}=\dfrac{1}{7}>\dfrac{-1}{7}\)

\(\dfrac{-9}{6}=\dfrac{-3}{2}< \dfrac{-2}{3}\)

\(=>\) Chọn đáp án A.