ta có:

\(log^{\left(2a^2\right)}_2+\left(log_2^a\right)a^{log_a^{\left(log^a_1+1\right)}}+\frac{1}{2}log^2_2a^4=log_2^2+log_2^{a^2}+log_2^a\left(log^a_2+1\right)+\frac{1}{2}log^2_2a^4\)

\(=1+2log^a_2+log^a_2\left(1+log^a_2\right)+2log^2a_2\)

\(=3log^2_2a+3log^a_2+1\)

\(a,A=ln\left(\dfrac{x}{x-1}\right)+ln\left(\dfrac{x+1}{x}\right)-ln\left(x^2-1\right)\\ =ln\left(\dfrac{x}{x-1}\cdot\dfrac{x+1}{x}\right)-ln\left(x^2-1\right)\\ =ln\left(\dfrac{x+1}{x-1}\right)-ln\left(x^2-1\right)\\ =ln\left(\dfrac{x+1}{x-1}\cdot\dfrac{1}{x^2-1}\right)\\ =ln\left[\dfrac{1}{\left(x-1\right)^2}\right]\\ =2ln\left(\dfrac{1}{x-1}\right)\)

\(b,21log_3\sqrt[3]{x}+log_3\left(9x^2\right)-log_3\left(9\right)\\ =7log_3\left(x\right)+log_3x^2+log_39-log_39\\ =7log_3x+2log_3x\\ =9log_3x\)

a)

\(\begin{array}{c}A = {\log _{\frac{1}{3}}}5 + 2{\log _9}25 - {\log _{\sqrt 3 }}\frac{1}{5} = {\log _{{3^{ - 1}}}}5 + 2{\log _{{3^2}}}{5^2} - {\log _{{3^{\frac{1}{2}}}}}{5^{ - 1}}\\ =  - {\log _3}5 + 2{\log _3}5 + 2{\log _3}5 = 3{\log _3}5\end{array}\)                                     

b) \(B = {\log _a}{M^2} + {\log _{{a^2}}}{M^4} = 2{\log _a}M + \frac{1}{2}.4{\log _a}M = 4{\log _a}M\)

\(A=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)

\(A=\left(\frac{4\sqrt{a}}{a-1}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)

\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}=\frac{4a\left(a+1\right)}{a-1}\)

....... Tới đây được chưa bạn? 

a)

\(=\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{1}{2a\sqrt{a}}\)

\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{1}{2a\sqrt{a}}\)

\(=\frac{4a\sqrt{a}}{a-1}.\frac{1}{2a\sqrt{a}}=\frac{2}{a-1}\)

b) \(\frac{2}{a-1}=a\Rightarrow a^2-a-2=0\)

Ta có: 1+1+(-2)=0, nên pt có 2 nghiệm a₁=-1<0 (không thỏa mãn đk)=> loại

a₂=2(thỏa mãn đk)=> chọn

Vậy a=2 thì P=a

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

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

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

\(a.\sqrt{8}-2\sqrt{50}+\sqrt{18}=2\sqrt{2}-10\sqrt{2}+3\sqrt{2}=\sqrt{2}\left(2-10+3\right)=-5\sqrt{2}\)

\(b.\left(\dfrac{\sqrt{a}-a}{1-\sqrt{a}}+\sqrt{a}\right):\dfrac{2\sqrt{a}}{1+\sqrt{a}}\left(đk:a\ge0;a\ne1\right)\)

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

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

\(=1+\sqrt{a}\)

(Chỗ điều kiện bài b mik thấy a = 0 cũng có thể là nghiệm nên mik sửa lại nhé)

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

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

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

\(=1-a\)