Ko vì -5/2 khác -3

a: f(0)=1/2

f(1/2)=3/2

c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=5\)

\(\Leftrightarrow\sqrt{x-4}+2=5\)

\(\Leftrightarrow\sqrt{x-4}=3\)

\(\Leftrightarrow x-4=9\)

hay x=13

c: Ta có: 

hay x=13

\(a,ĐK:x\ge-1\\ PT\Leftrightarrow2x^2+2x-4=x^2+2x+1\\ \Leftrightarrow x^2=5\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}\left(tm\right)\\x=-\sqrt{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\sqrt{5}\\ b,ĐK:x\in R\\ PT\Leftrightarrow\left|x-2\right|=2x-5\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-5\left(x\ge2\right)\\x-2=5-2x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\dfrac{7}{3}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=3\)

\(c,ĐK:x\ge4\\ PT\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}=5\\ \Leftrightarrow\sqrt{x-4}=3\Leftrightarrow x-4=9\Leftrightarrow x=5\left(tm\right)\)

Bài 2: 

Ta có: \(3n^3+10n^2-5⋮3n+1\)

\(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Leftrightarrow3n\in\left\{0;-3;3\right\}\)

hay \(n\in\left\{0;-1;1\right\}\)

Phần BA là phần nào á b mình zoom ảnh thấy phần C

1: Khi x=3-2 căn 2 thì \(A=\dfrac{\sqrt{2}-1+2}{\sqrt{2}-1}=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)

2: \(B=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{x-4}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

3: \(P=A:B=\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{x-4}{x}\)

\(x\cdot P< =10\sqrt{x}-29-\sqrt{x-25}\)

=>\(x-4< =10\sqrt{x}-29-\sqrt{x-25}\)

\(\Leftrightarrow x-4-10\sqrt{x}+29< =-\sqrt{x-25}\)

=>\(x-10\sqrt{x}+25< =-\sqrt{x-25}\)

=>(căn x-5)^2<=-căn x-25

=>x-25=0

=>x=25

Giúp e mấy bài đó đi mn ạ