Toán

4:

=>x^3+8-x^3-2x=15

=>-2x=7

=>x=-7/2

AC=AB*sin63=5,97(m)

 Giúp mũi với

a) Có: \(2^{1930}=2^4.2^4.....2^4.2^2\)

\(\Rightarrow2^{1930}=16.16.16.....16.4\)

\(\Rightarrow2^{1930}=\overline{...6}.4=\overline{...4}\)

Lại có: \(3^{1945}=3^4.3^4.3^4.....3^4.3\)

\(\Rightarrow3^{1945}=81.81.81.....81.3\)

\(\Rightarrow3^{1945}=\overline{...1}.3=\overline{...3}\)

Vậy \(2^{1930}.3^{1945}=\overline{...4}.\overline{...3}=\overline{...2}\) tận cùng là 2

b) Có: \(12^{2018}=12^4.12^4.12^4.....12^4.12^2\)

\(\Rightarrow12^{2018}=20736.20736.....20736.144\)

\(\Rightarrow12^{2018}=\overline{...6}.144=\overline{...4}\)

Lại có: \(11^{2015}=\overline{...1}\)

Vậy \(12^{2018}-11^{2015}=\overline{...4}-\overline{...1}=\overline{...3}\) tận cùng là 3

a: góc BAC=góc xCA
=>CD//AB

b: AD vuông góc AB

AB//CD

=>AD vuông góc Cx

c: Xét tứ giác ANCD có

AN//CD

AN=CD

=>ANCD là hình bình hành

=>I là trung điểm chung của AC và ND

\(a,x+\dfrac{2}{3}=\dfrac{-1}{5}\)

\(x=\dfrac{-1}{5}-\dfrac{2}{3}\)

\(x=\dfrac{-3-10}{15}=\dfrac{-13}{15}\)

\(b,x-\dfrac{7}{2}=\dfrac{-7}{3}\)

\(x=\dfrac{-7}{3}+\dfrac{7}{2}\)

\(x=\dfrac{-14+21}{6}=\dfrac{7}{6}\)

\(c,\dfrac{4}{7}-x=\dfrac{11}{2}\)

\(x=\dfrac{4}{7}-\dfrac{11}{2}\)

\(x=\dfrac{8-77}{14}=\dfrac{-69}{14}\)

\(d,-\dfrac{35}{14}-x=-1\)

\(x=-\dfrac{35}{14}-\left(-1\right)\)

\(x=\dfrac{\left(-35\right)-\left(-14\right)}{14}=\dfrac{-21}{14}=\dfrac{-3}{2}\)

\(e,3x-\dfrac{-22}{16}=1\)

\(3x=1+\dfrac{-22}{16}\)

\(3x=\dfrac{-3}{8}\)

\(x=\dfrac{-3}{8}:3\)

\(x=\dfrac{-3}{24}=\dfrac{-1}{8}\)

\(f,-\dfrac{12}{7}x+\dfrac{3}{14}=-\dfrac{1}{7}\)

\(-\dfrac{12}{7}x=-\dfrac{1}{7}-\dfrac{3}{14}\)

\(-\dfrac{12}{7}x=\dfrac{-5}{14}\)

\(x=\dfrac{-5}{14}:\dfrac{-12}{7}\)

\(x=\dfrac{-35}{-168}=\dfrac{-5}{-24}=\dfrac{5}{24}\)

\(g,\dfrac{-1}{4}+\dfrac{x}{24}=-2\)

\(\dfrac{x}{24}=-2-\dfrac{-1}{4}\)

\(\dfrac{x}{24}=\dfrac{-7}{4}\)

\(x:24=\dfrac{-7}{4}\)

\(x=\dfrac{-7}{4}.24\)

\(x=\dfrac{-168}{4}=-42\)

\(h,\dfrac{x}{28}+2=2x-\dfrac{1}{7}\)

\(x+56=56x-4\)

\(x+56x=-4-56\)

\(-55x=-60\)

\(x=-60:-55\)

\(x=\dfrac{-60}{-55}=\dfrac{12}{11}\)

\(h,\dfrac{x}{11}+\dfrac{2x}{3}=1\)

\(3x+22x=33\)

\(\left(3+22\right)x=33\)

\(25x=33\)

\(x=33:25\)

\(x=\dfrac{33}{25}\)

\(k,\dfrac{x}{2}+\dfrac{3x}{-4}=\dfrac{4x}{5}\)

\(\dfrac{x}{2}-\dfrac{3x}{4}=\dfrac{4x}{5}\)

\(10x-15x=16x\)

\(\left(10-15\right)x=16x\)

\(-5x=16x\)

\(-5x-16x=0\)

\(\left(-5-16\right)x=0\)

\(-21x=0\)

\(x=0:-21\)

\(x=0\)

\(m,\left|x-7\right|=7\)

\(x-7=7\) hoặc \(x-7=-7\)

\(x=14\) hoặc \(x=0\)

\(n,\left|x-7\right|=x-7\)

\(x-7\ge0\)

\(x\ge7\)

\(q,\left|2x+1\right|=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}\left|2x+1\right|=\dfrac{1}{2}\\\left|2x+1\right|=-\dfrac{1}{2}\end{matrix}\right.\)

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)

\(f,\left|3x-\dfrac{3}{4}\right|=-1\)

\(\Rightarrow\left\{{}\begin{matrix}\left|3x-\dfrac{3}{4}\right|=-1\\\left|3x-\dfrac{3}{4}\right|=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3x=-1+\dfrac{3}{4}\\3x=1+\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3x=-\dfrac{1}{4}\\3x=\dfrac{7}{4}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}:3\\x=\dfrac{7}{4}:3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{12}\\x=\dfrac{7}{12}\end{matrix}\right.\)

\(4,3=\dfrac{43}{10}=4\dfrac{3}{10}\)

\(8,49=\dfrac{849}{100}=8\dfrac{49}{100}\)

\(3,107=\dfrac{3107}{1000}=3\dfrac{107}{1000}\)

`4,3=43:10=43/10=4 3/10`

`8,49 = 849 : 100 =849/100 = 8 49/100`

`3,107 = 3107:1000 = 3107/1000 = 3 107/1000`

`4 1/5 + 25% - 3 7/2 + (9/10-1,25)`

`= 21/5 + 1/4 - 13/2 + 9/10 - 5/4`

`=(21/5-13/2+9/10)+(1/4-5/4)`

`=(-7)/5 - 1`

`= (-12)/5`

\(4\dfrac{1}{5}+25\%+3\dfrac{7}{2}+\left(\dfrac{9}{10}-1,25\right)\)

\(\dfrac{21}{5}+\dfrac{1}{4}-\dfrac{13}{2}+\left(\dfrac{9}{10}-\dfrac{5}{4}\right)\)

\(=\dfrac{21}{5}+\dfrac{1}{4}-\dfrac{13}{2}+\dfrac{-7}{20}\)

\(=\dfrac{84}{20}+\dfrac{5}{20}-\dfrac{130}{20}+\dfrac{-7}{20}\)

\(=\dfrac{84+5-130+\left(-7\right)}{20}=\dfrac{-48}{20}=\dfrac{-12}{5}\)

Đây 

ok

Bạn tham khảo nhé :

Gọi số tuổi của chị là a,tuổi của em là b.Suy ra hiệu số tuổi của 2 chị em là a-b

Theo bài ra ta có:

`b-(a-b)=12 (tuổi)      (1)`

`a xx 2-(a+b)=3( tuổi)`

`a xx 2-a-b=3 (tuổi)`

`a-b=3  (tuổi)`

Suy ra:  Hiệu số tuổi `2` chị em là `3` tuổi

Mà từ `(1)`  ta có: `b-(a-b)=12 (tuổi)`

Suy ra:   `b-3=12  (tuổi)`

`b=12+3 (tuổi)`

`b=15 (tuổi)`

`a=15+3=18(tuổi)`

Vậy chị là `18` tuổi,em là `15` tuổi