3/10 + 5/8 = ? giúp em ạ, cần gấp ạ
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
CN
1
NA
20 tháng 4 2022
\(a,\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow4x-x+3x=30+5+3\\ \Leftrightarrow6x=38\\ \Leftrightarrow x=\dfrac{19}{3}\)
15 tháng 9 2021
a) \(\dfrac{1}{4}-3\left(\dfrac{1}{12}+\dfrac{3}{8}\right)=\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{9}{8}=-\dfrac{9}{8}\)
b) \(\left(-\dfrac{2}{3}+\dfrac{3}{5}\right):\dfrac{1}{50}-30=\left(-\dfrac{2}{3}+\dfrac{3}{5}\right).50-30=-\dfrac{100}{3}+30-30=-\dfrac{100}{3}\)
Y
11 tháng 4 2022
\(d,\dfrac{6}{5}-3:\dfrac{15}{4}=\dfrac{6}{5}-3\times\dfrac{4}{15}=\dfrac{6}{5}-\dfrac{4}{5}=\dfrac{2}{5}\)
\(b,\dfrac{2}{5}+\dfrac{4}{5}:4=\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{1}{4}=\dfrac{2}{5}+\dfrac{1}{5}=\dfrac{3}{5}\)
\(a,\dfrac{5}{8}+\dfrac{4}{3}=\dfrac{15}{24}+\dfrac{32}{24}=\dfrac{47}{24}\)
24 tháng 8 2023
a) \(...=-\dfrac{1}{4}.\dfrac{4}{17}.\left(-\dfrac{63}{21}\right).\left(-\dfrac{7}{12}\right)\)
\(=-\dfrac{1}{17}.\dfrac{63}{21}.\dfrac{7}{12}\)
\(=-\dfrac{7}{68}\)
b) \(...=-\dfrac{2}{5}.\dfrac{4}{15}-\dfrac{3}{10}.\dfrac{4}{15}\)
\(=\dfrac{4}{15}\left(-\dfrac{2}{5}-\dfrac{3}{10}\right)\)
\(=\dfrac{4}{15}\left(-\dfrac{4}{10}-\dfrac{3}{10}\right)\)
\(=\dfrac{4}{15}.\left(-\dfrac{7}{10}\right)=-\dfrac{14}{75}\)
c) \(...=21-\dfrac{15}{4}:\left(\dfrac{9}{24}-\dfrac{4}{24}\right)\)
\(=21-\dfrac{15}{4}:\dfrac{5}{24}\)
\(=21-\dfrac{15}{4}.\dfrac{24}{5}\)
\(=21-3.6=3\)
d) \(...=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right).\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right).\dfrac{7}{3}\)
\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)
\(=\dfrac{7}{3}\left(-\dfrac{3}{4}-\dfrac{1}{4}+\dfrac{2}{5}+\dfrac{3}{5}\right)\)
\(=\dfrac{7}{3}\left(-1+1\right)=0\)
NX
16 tháng 8 2023
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
NX
16 tháng 8 2023
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
22 tháng 2 2022
uses crt;
var i,n:integer;
s:real;
begin
clrscr;
n:=1;
s:=0;
while (n<=10000) do
begin
n:=n+2;
s:=s+1/n;
end;
writeln(s:4:2);
readln;
end.
NN
14
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 4 2022
10.
\(\dfrac{sin3x-cos3x}{sinx+cosx}=\dfrac{3sinx-4sin^3x-\left(4cos^3x-3cosx\right)}{sinx+cosx}\)
\(=\dfrac{3\left(sinx+cosx\right)-4\left(sin^3x+cos^3x\right)}{sinx+cosx}\)
\(=\dfrac{3\left(sinx+cosx\right)-4\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)}{sinx+cosx}\)
\(=\dfrac{3\left(sinx+cosx\right)-4\left(sinx+cosx\right)\left(1-sinx.cosx\right)}{sinx+cosx}\)
\(=\dfrac{\left(sinx+cosx\right)\left(3-4+4sinx.cosx\right)}{sinx+cosx}\)
\(=-1+4sinx.cosx\)
\(=2sin2x-1\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 4 2022
11.
\(tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\dfrac{1+cos\left(\dfrac{\pi}{2}+x\right)}{sin\left(\dfrac{\pi}{2}+x\right)}=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{1+sin\left(-x\right)}{cos\left(-x\right)}\)
\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{1-sinx}{cosx}=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\dfrac{sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}-2sin\dfrac{x}{2}cos\dfrac{x}{2}}{cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}}\)
\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{\left(cos\dfrac{x}{2}-sin\dfrac{x}{2}\right)^2}{\left(cos\dfrac{x}{2}-sin\dfrac{x}{2}\right)\left(cos\dfrac{x}{2}+sin\dfrac{x}{2}\right)}\)
\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{cos\dfrac{x}{2}-sin\dfrac{x}{2}}{cos\dfrac{x}{2}+sin\dfrac{x}{2}}\)
\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{cos\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)}{sin\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)}\)
\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).cot\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\)
\(=1\)
\(\dfrac{3}{10}+\dfrac{5}{8}\)
\(=\dfrac{12}{40}+\dfrac{25}{40}=\dfrac{37}{40}\)
\(\dfrac{3}{10}+\dfrac{5}{8}=\dfrac{24}{80}+\dfrac{50}{80}=\dfrac{24+50}{80}=\dfrac{74}{80}=\dfrac{37}{40}\)