b) Ta có: \(\left(-\dfrac{1}{2}\right)^2\cdot2\dfrac{6}{7}-\dfrac{14}{15}:2\dfrac{1}{3}+\left(-1.21\right)^0\)

\(=\dfrac{1}{4}\cdot\dfrac{20}{7}-\dfrac{14}{15}:\dfrac{7}{3}+1\)

\(=\dfrac{5}{7}-\dfrac{14}{15}\cdot\dfrac{3}{7}+1\)

\(=\dfrac{5}{7}-\dfrac{2}{5}+1\)

\(=\dfrac{25-14-35}{35}=\dfrac{-24}{35}\)

Du la fan cua ai di chang nua,Online Math nay cung khong phai la cai khu cua cac cau thich lam gi thi lam.Day la khu hoc!!!!

Yes,sir

BẠN ĐỪNG ĐĂNG NHỮNG THỨ LINH TINH LÊN ONLINE MATH, BẠN SẼ LÀM NHIỄU DIỄN ĐÀN ĐẤY BẠN AH.?

\(B=\dfrac{2x+3\sqrt{x}+9}{x-9}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(\Rightarrow B=\dfrac{2x+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow B=\dfrac{2x+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow B=\dfrac{2x+3\sqrt{x}+9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow B=\dfrac{x+6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow B=\dfrac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow B=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(\dfrac{2x+3\sqrt{x}+9-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(B=\dfrac{2x+3\sqrt{x}+9-x+3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)

bạn ghi rõ nội dung, yêu cầu đề của bài đó ra cho mình nha

b: \(\Leftrightarrow n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{0;-2;1;-3;3;-5\right\}\)

c: \(\Leftrightarrow n+2\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{-1;-3;3;-7\right\}\)

d: \(\Leftrightarrow n+2\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{-1;-3;0;-4;2;-6\right\}\)

a: \(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{2;0;6;-4\right\}\)

Bài 7:

a: a>b

nên -2a<-2b

=>-2a-6<-2b-6<-2b

b: a>b

nên 3a>3b

=>3a-9>3b-9

=>3(a-3)>3(b-3)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2021}{2022}\cdot\dfrac{2022}{2023}\)

=1/2023

\(B=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2021}{2022}.\dfrac{2022}{2023}\)

\(=\dfrac{1.2.3...2022}{2.3.4...2023}=\dfrac{1}{2023}\)

a)

const fo='bt4_20.pas';
var f:text;
a,b,c,s,xx:longint;
begin
assign(f,fo); rewrite(f);
a:=30000;
b:=a*2;
c:=b-10000;
s:=a+b+c;
xx:=9500;
writeln(f,s div xx);
close(f);
end.

b)

const fo='bt4_20c2.pas';
var f:text;
a,b,c,s,xx:longint;
begin
assign(f,fo); rewrite(f);
a:=30000;
b:=a*2;
c:=b-10000;
s:=a+b+c;
xx:=9500;
writeln(f,s/xx:4:0);
close(f);
end.