Đặt \(\left\{{}\begin{matrix}b+c-a=x\\c+a-b=y\\a+b-c=z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=2c\\y+z=2a\\z+x=2b\end{matrix}\right.\)

\(\Leftrightarrow P=\dfrac{2\left(y+z\right)}{x}+\dfrac{9\left(x+z\right)}{2y}+\dfrac{8\left(x+y\right)}{z}\\ \Leftrightarrow P=\dfrac{2y}{x}+\dfrac{2z}{x}+\dfrac{9x}{2y}+\dfrac{9z}{2y}+\dfrac{8x}{z}+\dfrac{8y}{z}\\ \Leftrightarrow P=\left(\dfrac{2y}{x}+\dfrac{9x}{2y}\right)+\left(\dfrac{2z}{x}+\dfrac{8x}{z}\right)+\left(\dfrac{9z}{2y}+\dfrac{8y}{z}\right)\\ \Leftrightarrow P\ge2\sqrt{\dfrac{18xy}{2xy}}+2\sqrt{\dfrac{16xz}{xz}}+2\sqrt{\dfrac{72yz}{2yz}}\\ \Leftrightarrow P\ge2\sqrt{9}+2\sqrt{16}+2\sqrt{36}=26\)

Nguyễn Hoàng Minh CTV, dấu "=" xảy ra khi nào vậy bn??

22,

1, Đặt √(3-√5) = A

=> √2A=√(6-2√5)

=> √2A=√(5-2√5+1)

=> √2A=|√5 -1|

=> A=\(\dfrac{\sqrt{5}-1}{\text{√2}}\)

=> A= \(\dfrac{\sqrt{10}-\sqrt{2}}{2}\)

2, Đặt √(7+3√5) = B

=> √2B=√(14+6√5)

 => √2B=√(9+2√45+5)

=> √2B=|3+√5|

=> B= \(\dfrac{3+\sqrt{5}}{\sqrt{2}}\)

=> B= \(\dfrac{3\sqrt{2}+\sqrt{10}}{2}\)

3, 

Đặt √(9+√17) - √(9-√17) -\(\sqrt{2}\)=C

=> √2C=√(18+2√17) - √(18-2√17) -\(2\)

=> √2C=√(17+2√17+1) - √(17-2√17+1) -\(2\)

=> √2C=√17+1- √17+1 -\(2\)

=> √2C=0

=> C=0

26,

|3-2x|=2\(\sqrt{5}\)

TH1: 3-2x ≥ 0 ⇔ x≤\(\dfrac{-3}{2}\)

3-2x=2\(\sqrt{5}\)

-2x=2\(\sqrt{5}\) -3

x=\(\dfrac{3-2\sqrt{5}}{2}\) (KTMĐK)

TH2: 3-2x < 0 ⇔ x>\(\dfrac{-3}{2}\)

3-2x=-2\(\sqrt{5}\)

-2x=-2√5 -3

x=\(\dfrac{3+2\sqrt{5}}{2}\) (TMĐK)

Vậy x=\(\dfrac{3+2\sqrt{5}}{2}\)

2, \(\sqrt{x^2}\)=12 ⇔ |x|=12 ⇔ x=12, -12

3, \(\sqrt{x^2-2x+1}\)=7

⇔ |x-1|=7 

TH1: x-1≥0 ⇔ x≥1

x-1=7 ⇔ x=8 (TMĐK)

TH2: x-1<0 ⇔ x<1

x-1=-7 ⇔ x=-6 (TMĐK)

Vậy x=8, -6

4, \(\sqrt{\left(x-1\right)^2}\)=x+3

⇔ |x-1|=x+3

TH1: x-1≥0 ⇔ x≥1

x-1=x+3 ⇔ 0x=4 (KTM)

TH2: x-1<0 ⇔ x<1

x-1=-x-3 ⇔ 2x=-2 ⇔x=-1 (TMĐK)

Vậy x=-1

đề như thế này à \(\dfrac{\sqrt{27-3\sqrt{2}+2\sqrt{6}}}{3\sqrt{3}}\)

và bài này luôn quên không viết 
√(√3 +1)^2 + √(1- √3)^2

\(M=\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\dfrac{\sqrt{x}-2}{2}\)

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

2. Ta có: 

\(\sqrt{x}>0\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>0\) hay \(M>0\)

Lại có: \(M=\dfrac{\sqrt{x}+2-1}{\sqrt{x}+2}=1-\dfrac{1}{\sqrt{x}+2}< 1\)

\(\Rightarrow0< M< 1\Rightarrow M>M^2\)

1) Ta có: \(M=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-2}{2}\)

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

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

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

\(P=\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\) (đk:\(a\ge0;a\ne1\))

\(=\left[\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]:\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

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

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

2) \(\dfrac{1}{P}\ge\dfrac{\sqrt{a}+9}{8}\)

\(\Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}+1}\ge\dfrac{\sqrt{a}+9}{8}\)

\(\Leftrightarrow16\sqrt{a}\ge\left(\sqrt{a}+9\right)\left(\sqrt{a}+1\right)\)

\(\Leftrightarrow a-6\sqrt{a}+9\le0\)

\(\Leftrightarrow\left(\sqrt{a}-3\right)^2\le0\)

Dấu "=" xảy ra khi \(\sqrt{a}-3=0\Leftrightarrow a=9\) (tm)

Vậy...

1) ĐKXĐ: \(a\ge0;a\ne1\)

\(P=\left[\dfrac{a+\sqrt{a}+2\sqrt{a}+2}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}.\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\right]\)\(:\left[\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\right]\)

\(\Leftrightarrow P=\left[\dfrac{\sqrt{a}.\left(\sqrt{a}+1\right)+2.\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right]\)\(:\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\)

\(\Leftrightarrow P=\left[\dfrac{\left(\sqrt{a}+2\right).\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right].\dfrac{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}{2\sqrt{a}}\)

\(\Leftrightarrow P=\dfrac{\sqrt{a}+1-\sqrt{a}}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}{2\sqrt{a}}\)

\(\Leftrightarrow P=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)

2) Có : \(\dfrac{1}{P}\ge\dfrac{\sqrt{a}+9}{8}\)

\(\Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}+1}\ge\dfrac{\sqrt{a}+9}{8}\)

\(\Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}+1}-\dfrac{\sqrt{a}+9}{8}\ge0\)

\(\Leftrightarrow\dfrac{16\sqrt{a}-\left(\sqrt{a}+9\right).\left(\sqrt{a}+1\right)}{8.\left(\sqrt{a}+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{16\sqrt{a}-a-10\sqrt{a}-9}{8.\left(\sqrt{a}+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{-\left(a-6\sqrt{a}+9\right)}{8.\left(\sqrt{a}+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{a}-3\right)^2}{8.\left(\sqrt{a}+1\right)}\le0\)

Vì \(\sqrt{a}\ge0\Rightarrow8.\left(\sqrt{a}+1\right)>0\)  mà \(\left(\sqrt{a}-3\right)^2\) \(\ge0\) 

\(\Rightarrow\) \(\dfrac{\left(\sqrt{a}-3\right)^2}{8.\left(\sqrt{a}+1\right)}=0\) \(\Rightarrow\left(\sqrt{a}-3\right)^2=0\) \(\Leftrightarrow\sqrt{a}-3=0\Leftrightarrow\sqrt{a}=3\Leftrightarrow a=9\)

Vậy để\(\dfrac{1}{P}\ge\dfrac{\sqrt{a}+9}{8}\) thì \(a=9\)