a, \(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)ĐK : \(x\ge0;x\ne1\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}=\frac{x-2\sqrt{x}+1}{x-1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

b, \(B=\frac{3x-4}{x-2\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}}+\frac{\sqrt{x}-1}{2-\sqrt{x}}\)ĐK : \(x>0;x\ne4\)

\(=\frac{3x-4-\left(x-4\right)-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\frac{3x-4-x+4-x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)

c, \(Q=\frac{3}{\sqrt{a}-3}+\frac{2}{\sqrt{a}+3}+\frac{a-5\sqrt{a}-3}{a-9}\)ĐK : \(a\ge0;a\ne9\)

\(=\frac{3\sqrt{a}+9+2\sqrt{a}-6+a-5\sqrt{a}-3}{a-9}=\frac{a}{a-9}\)

d, \(B=\frac{x}{x-4}-\frac{1}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\)ĐK : \(x\ge0;x\ne4\)

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

\(ĐKXĐ:x\ge-\frac{1}{3}\)

\(x\sqrt{x^2-x+1}+2\sqrt{3x+1}=x^2+x+3\)

\(\left(x\sqrt{x^2-x+1}-1\right)+\left(2\sqrt{3x+1}-4\right)=x^2+x-2\)

\(\frac{x^2\left(x^2-x+1\right)-1}{x\sqrt{x^2-x+1}+1}+\frac{4\left(3x+1\right)-16}{2\sqrt{3x+1}+4}=\left(x-1\right)\left(x+2\right)\)

\(\frac{x^4-x^3+x^2-1}{x\sqrt{x^2-x+1}+1}+\frac{12x-12}{2\sqrt{3x+1}+4}-\left(x-1\right)\left(x+2\right)=0\)

\(\frac{\left(x-1\right)\left(x^3+x+1\right)}{x\sqrt{x^2-x+1}+1}+\frac{12\left(x-1\right)}{2\sqrt{3x+1}+4}-\left(x-1\right)\left(x+2\right)=0\)

\(\left(x-1\right)\left(\frac{x^3+x+1}{x\sqrt{x^2-x+1}+1}+\frac{12}{2\sqrt{3x+1}+4}-x-2\right)=0\)

\(\orbr{\begin{cases}x=1\left(TM\right)\\\frac{x^3+x+1}{x\sqrt{x^2-x+1}+1}+\frac{12}{2\sqrt{3x+1}+4}-x-2=0\end{cases}}\)

bạn cm \(\frac{x^3+x+1}{x\sqrt{x^2-x+1}+1}+\frac{12}{2\sqrt{3x+1}+4}-x-2\ne0\)

vậy pt có nghiệm duy nhất là x=1

\(A=\frac{3}{\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{x-1}\)

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

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

\(B=\left(1-\frac{\sqrt{2}}{x-\sqrt{2}}+\frac{\sqrt{2}}{x+\sqrt{2}}\right)\div\frac{x-\sqrt{6}}{x^2-2}\)

\(=\left[\frac{x^2-2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}-\frac{\sqrt{2}\left(x+\sqrt{2}\right)}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}+\frac{\sqrt{2}\left(x-\sqrt{2}\right)}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}\right]\cdot\frac{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}{x-\sqrt{6}}\)

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

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

Bài 1:

1) 3Mg + 4H2SO4 → 3MgSO4 + S + 4H2O

nMgSO4 = 3nS = 0,3 mol ⇒ mMgSO4 = 120.0,3 = 36 g

2) 

8Al + 30HNO3→8Al(NO3)3+3NH4NO3+9H2O

CK         Chất OXH

8×   Al → Al + 3e( sự oxh)

 3× N+8e→N(sự khử)

3) 

Quặng sắt tác dụng HNO3 không có khí thoát ra → quặng sắt chứa Fe2O3.

→ Quặng hematit

4)

Trả lời :

rảnh ak : )

Bài này làm sẵn rồi : )

~HT~

\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}-\frac{1}{2}\sqrt{60}\)

\(=\frac{\sqrt{3}}{\sqrt{5}}+\frac{\sqrt{5}}{\sqrt{3}}-\sqrt{\frac{1}{4}\cdot60}\)

\(=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}-\sqrt{15}\)

\(=\frac{3\sqrt{15}}{15}+\frac{5\sqrt{15}}{15}-\frac{15\sqrt{15}}{15}=\frac{-7\sqrt{15}}{15}\)

Tâm đường tròn ngoại tiếp tam giác ABC nằm trên trung điểm BC 

=> Tâm đường tròn là điểm M

tính bán kính nữa bạn ơi