1) Vì x=25 thỏa mãn ĐKXĐ nên Thay x=25 vào biểu thức \(A=\dfrac{\sqrt{x}-2}{x+1}\), ta được:

\(A=\dfrac{\sqrt{25}-2}{25+1}=\dfrac{5-2}{25+1}=\dfrac{3}{26}\)

Vậy: Khi x=25 thì \(A=\dfrac{3}{26}\)

2) Ta có: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}+\dfrac{2x+8\sqrt{x}-6}{x-\sqrt{x}-2}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-5\sqrt{x}+6+2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3x+3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

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

câu 3 chứ

Để hệ có nghiệm duy nhất thì \(\frac{2}{m}<>-\frac{m}{1}\)

=>\(m^2<>-2\) (luôn đúng)

=>Hệ luôn có nghiệm duy nhất

Ta có: \(\begin{cases}2x-my=3\\ mx+y=2\end{cases}\Rightarrow\begin{cases}2x-my=3\\ m^2x+my=2m\end{cases}\)

=>\(\begin{cases}2x-my+m^2x+my=3+2m\\ mx+y=2\end{cases}\Rightarrow\begin{cases}x\left(m^2+2\right)=3m+2\\ y=2-mx\end{cases}\)

=>\(\begin{cases}x=\frac{3m+2}{m^2+2}\\ y=2-mx=2-\frac{m\left(3m+2\right)}{m^2+2}=\frac{2\left(m^2+2\right)-3m^2-2m}{m^2+2}=\frac{-m^2-2m+4}{m^2+2}\end{cases}\)

A=3x+2y

\(=\frac{3\left(3m+2\right)}{m^2+2}+2\cdot\frac{-m^2-2m+4}{m^2+2}=\frac{9m+6-2m^2-4m+8}{m^2+2}=\frac{-2m^2+5m+14}{m^2+2}\)

Để A nguyên thì \(-2m^2+5m+14\) ⋮\(m^2+2\)

=>\(-2m^2-4+5m+18\) ⋮\(m^2+2\)

=>5m+18⋮\(m^2+2\)

=>(5m+18)(5m-18)⋮\(m^2+2\)

=>\(25m^2-324\) ⋮\(m^2+2\)

=>\(25m^2+50-374\) ⋮\(m^2+2\)

=>-374⋮\(m^2+2\)

=>\(m^2+2\in\left\lbrace2;11;17;22;34;187;374\right\rbrace\)

=>\(m^2\) ∈{0;9;15;20;32;185;372}

mà m nguyên

nên m∈{0;3;-3}

10. Câu này chứng minh BĐT BSC:

\(\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge\sqrt{\left(ab+bc\right)^2}=b\left(a+c\right)\)

11.

Ta có: \(\dfrac{1}{1+a}+\dfrac{1}{1+b}-\dfrac{2}{1+\sqrt{ab}}\)

\(=\dfrac{\left(1+b\right)\left(1+\sqrt{ab}\right)}{\left(1+a\right)\left(1+b\right)\left(1+\sqrt{ab}\right)}+\dfrac{\left(1+a\right)\left(1+\sqrt{ab}\right)}{\left(1+a\right)\left(1+b\right)\left(1+\sqrt{ab}\right)}-\dfrac{2\left(1+a\right)\left(1+b\right)}{\left(1+a\right)\left(1+b\right)\left(1+\sqrt{ab}\right)}\)

\(=\dfrac{1+b+\sqrt{ab}+b\sqrt{ab}}{\left(1+a\right)\left(1+b\right)\left(1+\sqrt{ab}\right)}+\dfrac{1+a+\sqrt{ab}+a\sqrt{ab}}{\left(1+a\right)\left(1+b\right)\left(1+\sqrt{ab}\right)}-\dfrac{2+2a+2b+2ab}{\left(1+a\right)\left(1+b\right)\left(1+\sqrt{ab}\right)}\)

\(=\dfrac{-a-b+2\sqrt{ab}+a\sqrt{ab}+b\sqrt{ab}-2ab}{\left(1+a\right)\left(1+b\right)\left(1+\sqrt{ab}\right)}\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{ab}-1\right)}{\left(1+a\right)\left(1+b\right)\left(1+\sqrt{ab}\right)}\ge0\forall x,y\ge1\)

Đẳng thức xảy ra khi \(a=b=1\)

dùng phương pháp hình học cm câu a 

đặt BH =a , HC =c kẻ HA =b 

theo định lí py ta go ta có 

AB=a²+b²;AC=b²+c²;BC=a+b

dễ thấy AB.AC\(\ge\) 2SABC=BC.AH

(a²+b²).(b²+c²)\(\ge\)b.(a+c)

a) Phương trình hoành độ giao điểm là: 

\(x^2=\left(m+2\right)x-2m\)

\(\Leftrightarrow x^2-\left(m+2\right)x+2m=0\)

\(\Delta=\left(m+2\right)^2-8m=m^2+4m+4-8m=m^2-4m+4=\left(m-2\right)^2\)

Để (d) và (P) cắt nhau tại hai điểm phân biệt thì \(\Delta>0\)

\(\Leftrightarrow\left(m-2\right)^2>0\)

mà \(\left(m-2\right)^2\ge0\)

nên \(m-2\ne0\)

hay \(m\ne2\)

Vậy: Để (d) và (P) cắt nhau tại hai điểm phân biệt thì \(m\ne2\)

Bài 1:ĐKXĐ: x>0; x<>1; x<>1/4

a: Ta có: \(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\cdot\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\)

\(=\left(2x+\sqrt{x}-1\right)\left(\frac{-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)

\(=\left(2x+\sqrt{x}-1\right)\cdot\frac{-x+\sqrt{x}-1+\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\frac{-x+\sqrt{x}-1+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\left(2\sqrt{x}-1\right)\cdot\frac{-1}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\frac{-\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)

TA có: \(A=\left(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\cdot\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right):\frac{2\sqrt{x}-1}{-x+\sqrt{x}}\)

\(=\frac{-\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{-\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)

b: \(A=\frac23\)

=>\(\frac{\sqrt{x}}{x-\sqrt{x}+1}=\frac23\)

=>\(2x-2\sqrt{x}+2=3\sqrt{x}\)

=>\(2x-5\sqrt{x}+2=0\)

=>\(2x-4\sqrt{x}-\sqrt{x}+2=0\)

=>\(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)

=>x=4(nhận) hoặc x=1/4(loại)