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

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

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

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

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

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

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

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

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

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 \(n\in\left\{2;0;6;-4\right\}\)

Mình ms bt lm đc câu c)

c) Để 4n+3⋮n+2

Thì (n+2)⋮(n+2)

=(4n+4)⋮(n+2)

⇒4n+3=4n+4-1⋮(n+2)

Vì n+2⋮n+2

Suy ra 1⋮(n+2) nên n+2∈ Ư(1)=

Mà (1,n+2)=1( nguyên tố cùng nhau)

Ta có bảng giá trị sau:

n+2 1 -1

n -1 -3

Mà n là số nguyên tố

Vậy n ∈ (-1;-3)

Tick cho mình!

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

=1/2023

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

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

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)