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

\(ĐKXĐ:1\le x\le9\)

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

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

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

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

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

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

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

\(9-x=4\)

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

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

\(ĐKXĐ:x\ge\frac{1}{2}\)

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

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

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

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

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

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

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

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

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

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

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

a, 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)

Ta có: \(\Delta=5^2-4\left(m-2\right)=25-4m+8=33-4m\)

Để pt có 2 nghiêm pb <=> \(\Delta>0\) <=> \(33-4m>0\) <=> \(m< \frac{33}{4}\)

Theo hệ thức vi-et, ta có: \(\hept{\begin{cases}x_1+x_2=-5\\x_1x_2=m-2\end{cases}}\)

Theo bài ra, ta có: \(\frac{1}{x_1-1}+\frac{1}{x_2-1}=2\) (Đk: (x₁ - 1)(x₂ - 1) \(\ne\)0 <=> x₁x₂ - (x₁ + x₂) + 1 = m - 2 + 5 + 1 = m + 4 \(\ne\)0 <=> m \(\ne\)-4

=> \(x_1+x_2-2=2\left(x_1-1\right)\left(x_2-1\right)\)

<=> \(-5-2=2\left(x_1x_2-x_1-x_2+1\right)\)

<=> \(2\left(m-2+5+1\right)=-7\)

<=> \(2m+8=-7\)

<=> \(m=-\frac{15}{2}\)(tm)

Vậy ...

\(x^2+5x+m-2=0\)

\(\Delta=5^2-4\left(m-2\right)=33-4m\)

Phương trình có hai nghiệm phân biệt nên \(\Delta>0\Rightarrow33-4m>0\Leftrightarrow m< \frac{33}{4}\).

Với \(m< \frac{33}{4}\)phương trình có hai nghiệm phân biệt \(x_1,x_2\).

Theo Viet: 

\(\hept{\begin{cases}x_1+x_2=-5\\x_1x_2=m-2\end{cases}}\)

\(\frac{1}{x_1-1}+\frac{1}{x_2-1}=\frac{x_2-1+x_1-1}{\left(x_1-1\right)\left(x_2-1\right)}=\frac{x_1+x_2-2}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\frac{-5-2}{m-2+5+1}=\frac{-7}{m+4}=2\)

\(\Leftrightarrow m=-7,5\)(thỏa mãn)

\(\frac{4}{2a+b+c}+\frac{4}{2b+c+a}+\frac{4}{2c+a+b}\ge\frac{\left(2+2+2\right)^2}{4\left(a+b+c\right)}=\frac{9}{a+b+c}\)

Dấu \(=\)khi \(\frac{2}{2a+b+c}=\frac{2}{2b+c+a}=\frac{2}{2c+a+b}\Leftrightarrow a=b=c>0\).

áp dụng tính chất j đẻ làm bài này vậy bạn