là sao vậy bạn ???

KO CÓ J 

Biểu thức thứ 2 trong HPT chưa chính xác \(\left(y-1\right)\)xem lại đề bài nha bạn !

mik bt nè

a, Điều kiện : \(a\ge0,a\ne1\)

\(A=\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)

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

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

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

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

a, \(M=\frac{\sqrt{x}}{\sqrt{x}+6}+\frac{1}{\sqrt{x}-6}+\frac{17\sqrt{x}+30}{\left(\sqrt{x}+6\right)\left(\sqrt{x}-6\right)}\)

\(=\frac{x-6\sqrt{x}+\sqrt{x}+6+17\sqrt{x}+30}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\frac{12\sqrt{x}+x+36}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-6}\)

b, Ta có : \(L=N.M\Rightarrow L=\frac{\sqrt{x}+6}{\sqrt{x}-6}.\frac{24}{\sqrt{x}+6}=\frac{24}{\sqrt{x}+6}\)

Vì \(\sqrt{x}+6\ge6\)

\(\Rightarrow\frac{24}{\sqrt{x}+6}\le\frac{24}{6}=4\)

Dấu ''='' xảy ra khi \(\sqrt{x}+6=6\Leftrightarrow x=0\)

Vậy GTLN L là 4 khi x = 0

\(M=\frac{\left(a+1\right)^2+2a}{a\left(a+1\right)}+\frac{\left(b+1\right)^2+2b}{b\left(b+1\right)}+\frac{\left(c+1\right)^2+2c}{c\left(c+1\right)}\)

\(M=\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+1}{c}+2\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\)

\(M=3+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+2\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\)

\(M\ge3+\frac{9}{a+b+c}+2\left(\frac{9}{a+b+c+3}\right)\ge3+3+3=9\)

Dấu "=" xảy ra khi a=b=c=1