\(x=6-2\sqrt{5}\)

\(x=\sqrt{5}^2-2\sqrt{5}+1\)

\(x= \left(\sqrt{5}-1\right)^2\)

thay vào P ta đc:

\(P=\frac{2\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}\)

\(P=\frac{2\left(\sqrt{5}-1\right)-1}{\sqrt{5}-1+1}\)

\(P=\frac{2\sqrt{5}-2-1}{\sqrt{5}}\)

\(P=\frac{2\sqrt{5}-3}{\sqrt{5}}\)

\(P=\frac{10-3\sqrt{5}}{5}\)

\(ĐKXĐ:x\ge0;x\ne1\)

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

\(2x-\sqrt{x}=\sqrt{x}-1\)

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

\(\Delta=\left(-2\right)^2-\left(4.2.-1\right)=4+8=12\)

\(\sqrt{\Delta}=2\sqrt{3}\)

\(x_1=\frac{2+2\sqrt{3}}{4}=\frac{1+\sqrt{3}}{2}\left(TM\right)\)

\(x_2=\frac{2-2\sqrt{3}}{4}=\frac{1-\sqrt{3}}{2}\left(KTM\right)\)

\(a,\sqrt{x-1}+\sqrt{9-x}=4\)

\(ĐKXĐ:1\le x\le9\)

\(\sqrt{x-1}=4-\sqrt{9-x}\)

\(x-1=16-8\sqrt{9-x}+9-x\)

\(26-8\sqrt{9-x}-2x=0\)

\(13-4\sqrt{9-x}-x=0\)

\(9-x-4\sqrt{9-x}+4=0\)

\(\left(\sqrt{9-x}-2\right)^2=0\)

\(\sqrt{9-x}=2\)

\(9-x=4\)

\(x=5\left(TM\right)\)

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

\(ĐKXĐ:x\ge\frac{1}{2}\)

\(x+4=36-12\sqrt{2x-1}+2x-1\)

\(x+4=35-12\sqrt{2x-1}+2x\)

\(31-12\sqrt{2x-1}+x=0\)

\(\left(31+x\right)^2=\left(12\sqrt{2x-1}\right)^2\)

\(961+62x+x^2=144\left(2x-1\right)\)

\(961+62x+x^2=288x-144\)

\(x^2-226x+1105=0\)

\(\sqrt{\Delta}=216\)

\(x_1=\frac{226+216}{2}=221\left(TM\right)\)

\(x_2=\frac{226-216}{2}=5\left(TM\right)\)

................................................. tui ko bít

A B C H 29 20 D

a, Xét tam giác ABC vuông tại A, đường cao AH 

Áp dụng định lí Pytago cho tam giác ABC vuông tại A

\(AB^2+AC^2=BC^2\Rightarrow AB^2=BC^2-AC^2=29^2-20^2=441\)

\(\Rightarrow AB=\sqrt{441}=21\)cm 

* Áp dụng hệ thức :

 \(AH.BC=AB.AC\Rightarrow AH=\frac{AB.AC}{BC}=\frac{21.20}{29}=\frac{420}{29}\)cm

b, Vì AD là tia phân giác nên : \(\frac{BD}{DC}=\frac{AB}{AC}\Rightarrow\frac{BD}{AB}=\frac{DC}{AC}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{BD}{AB}=\frac{DC}{AC}=\frac{BD+DC}{AB+AC}=\frac{29}{41}\)

\(\Rightarrow DC=\frac{29}{41}.20=\frac{580}{41}\)cm 

Diện tích tam giác ADC là :

 \(S_{ADC}=\frac{1}{2}.AH.DC=\frac{1}{2}.\frac{420}{29}.\frac{580}{41}=\frac{4200}{41}\)cm²

\(A=\frac{1-\sqrt{x-1}}{\sqrt{x-2\sqrt{x-1}}}\)

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

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

\(A=\frac{1-\sqrt{x-1}}{\left|\sqrt{x-1}-1\right|}\)

\(TH1:1\le x\le2\)

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

\(TH2:2< x\)

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

\(Q=\frac{x+2\sqrt{x}-10}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{\sqrt{x}-3}-\frac{1}{\sqrt{x}-2}\)

\(Q=\frac{\left(x+2\sqrt{x}-10\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)\left(x-4\right)-x+\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(Q=\frac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}-6\right)-x+\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(x-4\right)}\)

\(Q=\frac{x-9\sqrt{x}+18}{\left(\sqrt{x}-3\right)\left(x-4\right)}\)

\(Q=\frac{\left(\sqrt{x}-6\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(x-4\right)}=\frac{\sqrt{x}-6}{x-4}\)

\(Q=\frac{1}{3}\Leftrightarrow\frac{\sqrt{x}-6}{x-4}=\frac{1}{3}\)

\(3\sqrt{x}-18=x-4\)

\(x-3\sqrt{x}+14=0\)(vo nghiem)