Bài 1
a) -115.137 phần 115.138
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
18:
a: \(S=3\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)
=3*(1/2-1/4+1/4-1/6+...+1/98-1/100)
=3*49/100=147/100
b: Để A là số nguyên thì n-1 thuộc Ư(2)
=>n-1 thuộc {1;-1;2;-2}
=>n thuộc {2;0;3;-1}
\(a.32,5-3\cdot0,87=32,5-2,61=29,89\)
\(8,5\cdot\left(1\dfrac{1}{2}+\dfrac{4}{4}\right):5=8,5\cdot\left(\dfrac{3}{2}+\dfrac{4}{4}\right):5\\ =8,5\cdot\left(\dfrac{6}{4}+\dfrac{4}{4}\right):5\\ =8,5\cdot\dfrac{10}{4}:5\\ =\dfrac{85}{4}:5\\ =\dfrac{17}{4}\)
\(b.30,96-6,45+14,4:3=30,96-6,45+4,8\\ =29,31\)
\(\dfrac{2}{5}\cdot\left(\dfrac{4}{5}-\dfrac{1}{2}\right)=\dfrac{2}{5}\cdot\left(\dfrac{8}{10}-\dfrac{5}{10}\right)\\ =\dfrac{2}{5}\cdot\dfrac{3}{10}=\dfrac{3}{25}\)
bài 2
\(a.2,5\cdot12,5\cdot8\cdot0,4=\left(2,5\cdot0,4\right)\left(12,5\cdot8\right)\\ =1\cdot100=100\)
b,\(\dfrac{12}{15}\cdot\dfrac{5}{6}\cdot\dfrac{3}{20}\cdot\dfrac{32}{5}=\dfrac{12\cdot5\cdot3\cdot32}{15\cdot6\cdot20\cdot5}\\ =\dfrac{3\cdot4\cdot5\cdot3\cdot4\cdot8}{3\cdot5\cdot2\cdot3\cdot5\cdot4\cdot5}=\dfrac{16}{25}\)
Bài 1:
a) \(32.5-3\cdot0.87=32.5-2.61=29.89\)
\(8.5\cdot\left(1\dfrac{1}{2}+\dfrac{4}{4}\right):5=8.5\cdot\dfrac{5}{2}:5=\dfrac{17}{2}\cdot\dfrac{5}{2}:5=\dfrac{85}{4}\cdot\dfrac{1}{5}=\dfrac{17}{4}\)
#)Giải :
Bài 1 :
\(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+...+\frac{1}{1280}\)
\(\Rightarrow A\times2=\frac{2}{5}-\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+...+\frac{1}{1280}\right)-\frac{1}{1280}\)
\(\Rightarrow A\times2=\frac{2}{5}-A-\frac{1}{1280}\)
\(\Rightarrow A\times2+A=\frac{2}{5}-\frac{1}{1280}\)
\(\Rightarrow A=\frac{2}{5}-\frac{1}{1280}\)
\(\Rightarrow A=\frac{511}{1280}\)
#)Giải :
Bài 2 :
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{59049}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{10}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^9}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}...+\frac{1}{3^9}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{10}}\right)\)
\(2B=1-\frac{1}{3^{10}}\)
\(B=\frac{1-\frac{1}{3^{10}}}{2}\)
\(a,MSC:3\)
Vì \(2>1\)
\(\Rightarrow\dfrac{2}{3}>\dfrac{1}{3}\)
\(b,MSC:42\\ \dfrac{4}{6}=\dfrac{4.7}{6.7}=\dfrac{28}{42}\\ \dfrac{4}{7}=\dfrac{4.6}{7.6}=\dfrac{24}{42}\)
Vì \(28>24\\ \Rightarrow\dfrac{4}{6}>\dfrac{4}{7}\)
\(c,MSC:22\)
\(\dfrac{6}{11}=\dfrac{6.2}{11.2}=\dfrac{12}{22}=\dfrac{12}{22}\\ \Rightarrow\dfrac{6}{11}=\dfrac{12}{22}\)
\(d,MSC:2\\ 1=\dfrac{2}{2}\)
Vì \(3>2\\ \Rightarrow\dfrac{3}{2}>1\)
Bài 2:
Uses crt;
Var a:array[1..200]of integer;
i,n,t:integer;
Begin
Clrscr;
Write('Nhap n='); readln(n);
For i:=1 to n do
Begin
Write('A[',i,']='); readln(a[i]);
End;
t:=0;
for i:=1 to n do
if a[i] mod 2=0 then t:=t+a[i];
writeln(t);
readln;
end.
Bài 1. Nhập mảng A gồm n phần tử, rồi in mảng đó ra màn hình.
program BaiTap;
var
A: array[1..150] of integer;
N, i: integer;
begin
write('Nhap so phan tu cua mang A (N <= 150): ');
readln(N);
for i := 1 to N do
begin
write('Nhap phan tu thu ', i, ': ');
readln(A[i]);
end;
writeln('Mang A vua nhap la:');
for i := 1 to N do
write(A[i], ' ');
readln;
end.
Bài 2. Nhập mảng A gồm n phần tư, rồi in tổng các phần tử mảng đó ra màn hình.
program BaiTap;
var
A: array[1..150] of integer;
N, i, sum: integer;
begin
write('Nhap so phan tu cua mang A (N <= 150): ');
readln(N);
for i := 1 to N do
begin
write('Nhap phan tu thu ', i, ': ');
readln(A[i]);
end;
sum := 0;
for i := 1 to N do
sum := sum + A[i];
writeln('Tong cac phan tu trong mang la: ', sum);
readln;
end.
Bài 3. Nhập mảng A gồm n phần tử, rồi in tổng các phần tử dầu và phần tủ cuối của mảng đó ra màn hình.
program BaiTap;
var
A: array[1..150] of integer;
N, i, sum: integer;
begin
write('Nhap so phan tu cua mang A (N <= 150): ');
readln(N);
for i := 1 to N do
begin
write('Nhap phan tu thu ', i, ': ');
readln(A[i]);
end;
sum := A[1] + A[N];
writeln('Tong cua phan tu dau va cuoi mang la: ', sum);
readln;
end.
\(\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow-x\cdot x=-9\cdot4\)
\(\Rightarrow-x^2=-36\)
\(\Rightarrow-x^2=-6^2\)
\(\Rightarrow-x=-6\)
\(\Rightarrow\) \(x=6\)
\(\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow\left(x-1\right)\cdot3=9\cdot8\)
\(\Rightarrow\) \(3x-3=72\)
\(\Rightarrow\) \(3x=72+3\)
\(\Rightarrow\) \(3x=75\)
\(\Rightarrow\) \(x=75\div3\)
\(\Rightarrow\) \(x=25\)
BÀI 1: \(a,1-\frac{3}{7}=\frac{4}{7}\)
BÀI 2:TÍNH
\(a,\frac{3}{5}+\frac{5}{7}-\frac{1}{2}=\frac{46}{35}-\frac{1}{2}=\frac{57}{70}\)
\(b,\frac{1}{6}+\frac{3}{5}+\frac{2}{7}=\frac{23}{30}+\frac{2}{7}=\frac{221}{210}\)
\(c,1-\left(\frac{3}{15}+\frac{4}{5}\right)=1-\left(\frac{1}{5}+\frac{4}{5}\right)=1-1=0\)
\(d,\frac{4}{5}-\frac{3}{7}+3=\frac{13}{35}+3=\frac{118}{35}\)
HỌC TỐT ~~~
a: -3/4x12=-36/4=-9
b: \(=\dfrac{7}{15}\cdot\dfrac{10}{21}=\dfrac{7}{21}\cdot\dfrac{10}{15}=\dfrac{1}{3}\cdot\dfrac{2}{3}=\dfrac{2}{9}\)
c: \(=\dfrac{7}{18}\cdot\dfrac{12}{21}=\dfrac{12}{18}\cdot\dfrac{7}{21}=\dfrac{2}{3}\cdot\dfrac{1}{3}=\dfrac{2}{9}\)
d: \(=\dfrac{11}{15}\cdot\dfrac{5}{22}=\dfrac{11}{22}\cdot\dfrac{5}{15}=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}\)
e: \(=-\dfrac{8}{15}\cdot\dfrac{5}{4}=\dfrac{-40}{60}=\dfrac{-2}{3}\)
f: \(=-15\cdot\dfrac{3}{5}=-9\)
\(\dfrac{-11^5\cdot13^7}{11^5\cdot13^8}=-\dfrac{11^5\cdot13^7}{11^5\cdot13^7\cdot13}\)
\(=\dfrac{-1}{13}\)
Bài 1:
a) \(\dfrac{-11^5.13^7}{11^5.13^8}\)=\(\dfrac{-11^5}{11^5.13}=\dfrac{-11^5}{-11^5.-13}=\dfrac{1}{-13}\)