Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=1+2+3+4+5+...+2000\)
Số phần tử trong dãy: \(\dfrac{2000-1}{1}+1=2000\)
Tổng của dãy trên: \(A=(2000+1)\cdot2000:2=2001000\)
\(B=2+4+6+8+10+...+200\)
Số phần tử trong dãy: \(\dfrac{200-2}{2}+1=100\)
Tổng của dãy trên: \(B=(200+2)\cdot100:2=10100\)
\(C=1+3+5+7+9+...+499\)
Số phần tử trong dãy: \(\dfrac{499-1}{2}+1=250\)
Tổng của dãy trên: \(C=(499+1)\cdot250:2=62500\)
a\()\) 16/9 +3/5
=107/45
b\()\) 4/13--2/17
=51/221--26/221
=77/221
c\()\) -3/2+4/5
=-15/10+8/10
=-7/10
d\()\) 3/-4-1/4
=-1
e\()\) -1/5.5/7
=-1/7
f\()\) 7/8.64/49
=8/7
g\()\) 3/4.15/24
=15/32
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
1)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}\cdot\left(-2\right)^2\\ =\dfrac{6}{7}+\dfrac{5}{8}\cdot\dfrac{1}{5}-\dfrac{3}{16}\cdot4\\ =\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}\\ =\dfrac{55}{16}-\dfrac{3}{4}\\ =\dfrac{13}{56}\)
2)
\(-\dfrac{1}{6}+\dfrac{2}{3}\cdot\dfrac{-3}{4}+\dfrac{\left(-2\right)^2}{5}\\ =-\dfrac{1}{6}-\dfrac{1}{2}+\dfrac{4}{5}\\ =-\dfrac{2}{3}+\dfrac{4}{5}\\ =\dfrac{2}{15}\)
3)
\(2\dfrac{3}{4}\cdot\left(-0,4\right)+1\dfrac{3}{5}\cdot1,75+\left(-7,2\right):\dfrac{9}{11}\)
\(=\dfrac{11}{4}\cdot\dfrac{-2}{5}+\dfrac{8}{5}\cdot\dfrac{7}{4}-\dfrac{36}{5}\cdot\dfrac{11}{9}\\ =-\dfrac{11}{10}+\dfrac{14}{5}-\dfrac{44}{5}\\ =-\dfrac{11}{10}-6\\ =-\dfrac{71}{10}\)
4)
\(\dfrac{1}{8}\cdot15\dfrac{2}{5}+1\dfrac{4}{5}\cdot\dfrac{7}{8}-17\dfrac{1}{5}\cdot\dfrac{1}{8}\)
\(=\dfrac{1}{8}\cdot\dfrac{77}{5}+\dfrac{9}{5}\cdot\dfrac{7}{8}-\dfrac{86}{5}\cdot\dfrac{1}{8}\)
\(=\dfrac{1}{8}\cdot\dfrac{77}{5}+\dfrac{63}{5}\cdot\dfrac{1}{8}-\dfrac{86}{5}\cdot\dfrac{1}{8}\)
\(=\dfrac{1}{8}\left(\dfrac{77}{5}+\dfrac{63}{5}-\dfrac{86}{5}\right)\\ =\dfrac{1}{8}\cdot\dfrac{54}{5}\\ =\dfrac{27}{20}\)
Bài 1 :
a, \(\frac{3}{4}:x=\frac{5}{12}\)
\(x=\frac{3}{4}:\frac{5}{12}\)
\(x=\frac{9}{5}\)
b, \(x-\frac{1}{2}=\frac{3}{4}:\frac{3}{2}\)
\(x-\frac{1}{2}=\frac{1}{2}\)
\(x=\frac{1}{2}+\frac{1}{2}\)
\(x=1\)
c, \(1\frac{1}{2}x-\frac{1}{2}=\frac{3}{4}\)
\(\frac{3}{2}x-\frac{1}{2}=\frac{3}{4}\)
\(\frac{3}{2}x=\frac{3}{4}+\frac{1}{2}\)
\(\frac{3}{2}x=\frac{5}{4}\)
\(x=\frac{5}{4}:\frac{3}{2}\)
\(x=\frac{5}{6}\)
Bài 2 :
\(A=\frac{-3}{5}+\left(\frac{-2}{5}-99\right)\)
\(A=\frac{-3}{5}+\frac{-2}{5}-99\)
\(A=\left(-1\right)-99\)
\(A=-100\)
\(B=\left(7\frac{2}{3}+2\frac{3}{5}\right)-6\frac{2}{3}\)
\(B=\left(\frac{23}{3}+\frac{13}{5}\right)-\frac{20}{3}\)
\(B=\frac{23}{3}+\frac{13}{5}-\frac{20}{3}\)
\(B=\left(\frac{23}{3}-\frac{20}{3}\right)+\frac{13}{5}\)
\(B=1+\frac{13}{5}\)
\(B=\frac{18}{5}\)
a, \(\left(5x-10\right)\left(6x-\frac{1}{3}\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x-10=0\\6x-\frac{1}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=10\\6x=\frac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{18}\end{matrix}\right.\)
Vậy \(x\in\left\{2;\frac{1}{18}\right\}\)
b, \(\frac{-3}{4}-\left|\frac{4}{5}-x\right|=-10\\ \frac{-3}{4}+10=\left|\frac{4}{5}-x\right|\\ \left|\frac{4}{5}-x\right|=\frac{37}{4}\\ \Rightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{37}{4}\\\frac{4}{5}-x=\frac{-37}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{4}{5}-\frac{37}{4}\\x=\frac{4}{5}-\frac{-37}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-169}{20}\\x=\frac{201}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{-169}{20};\frac{201}{20}\right\}\)
c, \(\left|5+x\right|-\frac{-2}{3}=3\\ \left|5+x\right|=3+\frac{-2}{3}\\ \left|5+x\right|=\frac{7}{3}\\ \Rightarrow\left[{}\begin{matrix}5+x=\frac{7}{3}\\5+x=\frac{-7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{3}-5\\x=\frac{-7}{3}-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-8}{3}\\x=\frac{-22}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{-8}{3};\frac{-22}{3}\right\}\)
d, Xem lại đề nhé vì không xuất hiện x thì đẳng thức sai.
a, [ (- 1) . 3 ] + (- 4) - (7.0) + 1
= [- 3 + ( - 4 )] - 0 + 1
= - 7 - 0 + 1
= - 6
b, (-1).(-2)+(-3).(-4)-(-2).(-3)
= [(-1).(-2)] + [(-3).(-4)] - [(-2).(-3)]
= 2 + 12 - 6
= 8
c, (-1).(-2).(-3).(-4).(-5):[(-3)-(-5)]
= (-1).(-2).(-3).(-4).(-5) : 2
= - 120 : 2
= - 60
Chắc chắn đấy!
a, (-1).3+(-4)-7.0+1=-3-4-0+1=-6
b, (-1).(-2)+(-3).(-4)-(-2).(-3)=2+12-6=8
c, (-1).(-2).(-3).(-4).(-5):[(-3)-(-5)]=(-120):2=-60
a) (-2).3 + (-4) - 7.0 + 1
=-6+(-4)-0+1
=-10-0+1
=-10+1
=-9
b) (-1).(-2) + (-3).(-4) - (-2).(-3)
=2+12-6
=14-6
=8
c)(-1).(-2).(-3).(-4).(-5) : [(-3) – (-5)]
=2.(-3).(-4).(-5):[(-3)+5]
=-6.(-4).(-5):2
=24.(-5):2
=-120:2
=-60
nhớ k giùm mình nhaa~