\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)

\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)

\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)

\(B=3-\sqrt{x}-\sqrt{x}+3-6\)

\(B=-2\sqrt{x}\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)

\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)

\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3}{\sqrt{x}-6}\)

ĐKXĐ : a;b;c>0;a≠−(b+c);b≠−(c+a);c≠−(a+b)a;b;c≠0;a≠−(b+c);b≠−(c+a);c≠−(a+b)

a+b−xc+b+c−xa+c+a−xb+4xa+b+c=1a+b−xc+b+c−xa+c+a−xb+4xa+b+c=1

⇔(a+b−xc+1)+(b+c−xa+1)+(c+a−xb+1)+4xa+b+c−3−1=0⇔(a+b−xc+1)+(b+c−xa+1)+(c+a−xb+1)+4xa+b+c−3−1=0

⇔a+b+c−xc+a+b+c−xa+a+b+c−xb+4xa+b+c−4=0⇔a+b+c−xc+a+b+c−xa+a+b+c−xb+4xa+b+c−4=0

⇔(a+b+c−x)(1a+1b+1c)+4(x−a−b−c)a+b+c=0⇔(a+b+c−x)(1a+1b+1c)+4(x−a−b−c)a+b+c=0

⇔(a+b+c−x)(1a+1b+1c−4a+b+c)=0⇔(a+b+c−x)(1a+1b+1c−4a+b+c)=0

Do 1a+1b+1c−4a+b+c≠01a+1b+1c−4a+b+c≠0

⇒a+b+c−x=0⇔x=a+b+c⇒a+b+c−x=0⇔x=a+b+c

Vậy ...

Ta có pt : \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\) (1)

( ĐK: Do bài cho a,b,c > 0 rồi nên không cần nhé bạn )

Pt (1) \(\Leftrightarrow\left(\frac{a+b-x}{c}+1\right)+\left(\frac{b+c-x}{a}+1\right)+\left(\frac{c+a-x}{b}+1\right)+\left(\frac{4x}{a+b+c}-4\right)=0\)

\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}-\frac{4\left(a+b+c-x\right)}{a+b+c}=0\)

\(\Leftrightarrow\left(a+b+c-x\right)\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)

Do : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}\ne0\forall a,b,c>0\)

Nên : \(a+b+c-x=0\)

\(\Leftrightarrow a+b+c=x\)

Vậy : pt (1) có tập nghiệm \(S=\left\{a+b+c\right\}\)

\(\Rightarrow\)\(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}=1-\frac{4x}{a+b+c}\)

\(\Leftrightarrow\)\(\frac{a+b+c-x}{c}+\frac{b+c+a-x}{a}+\frac{c+a+b-x}{b}=4-\frac{4x}{a+b+c}\)(Vế trái cộng mỗi phân số với 1 thì vế phải +3)

\(\Leftrightarrow\)\(\left(a+b+c-x\right)\left(\frac{1}{c}+\frac{1}{b}+\frac{1}{a}\right)=4\left(a+b+c-x\right).\frac{1}{a+b+c}\)

+ Xét \(a+b+c-x=0\Rightarrow x=a+b+c\)

+ Xét \(a+b+c-x\)khác 0 \(\Rightarrow\)\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=4\left(\frac{1}{a+b+c}\right)\)

Ta có : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{a+b+c}>4\left(\frac{1}{a+b+c}\right)\)(bất đẳng thức COSY đó bạn)

như vậy là phương trình vô nghiệm

Sai rồi nha bạn Nguyễn Thuỳ Trang.

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{4}{a+b+c}\) vẫn được mà.

Đề có cho \(a,b,c\) dương đầu mà dùng Cauchy như đúng rồi vậy! Cẩn thận một chút.

\(\Leftrightarrow\frac{\left(x+1\right)+a\left(b+1\right)}{\left(a+1\right)}+\frac{\left(x+1\right)+c\left(b+1\right)}{\left(c+1\right)}+\frac{\left(x+1\right)+b\left(b+1\right)}{\left(b+1\right)}=3\left(b+1\right)\)

\(\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\left(x+1\right)=\left(b+1\right)\left(3-\frac{a}{a+1}-\frac{b}{b+1}-\frac{c}{c+1}\right)\)

\(\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\left(x+1\right)=\left(b+1\right)\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\)

\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=A=0\) pt N₀ đúng mọi x. thuộc R

Nếu A khác 0 pt có nghiệm duy nhất x=b

\(\frac{x-ab}{a+b}+\frac{x-ac}{a+c}+\frac{x-bc}{b+c}=a+b+c\)

\(\frac{x-ab}{a+b}-c+\frac{x-ac}{a+c}-b+\frac{x-bc}{b+c}-a=0\)

\(\frac{x-ab-ac-bc}{a+b}+\frac{x-ac-ba-bc}{a+c}+\frac{x-bc-ab-ac}{b+c}=0\)

\(\left(x-ab-ac-bc\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=0\)

\(x-ab-ac-bc=0\)

\(x=ab+ac+bc\)

<=> \(\left(\frac{x-ab}{a+b}-c\right)+\left(\frac{x-ac}{a+c}-b\right)+\left(\frac{x-bc}{b+c}-a\right)=0\)

<=>\(\frac{x-ab-ac-bc}{a+b}+\frac{x-ab-ac-bc}{a+c}+\frac{x-ab-ac-bc}{b+c}=0\)

<=>\(\left(x-ab-ac-bc\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=0\)

Vì \(a\ne-b;b\ne-c;c\ne-a\) nên tổng 3 phân số kia khác 0

=> (x-ab-ac-ca)=0

=>x=ab+ac+ca