a: ĐKXĐ: x^2-1>=0 

=>x>=1 hoặc x<=-1

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

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

x>=căn 2

=>x^2>=2

=>x^2-1>=1

=>căn x^2-1>=1

=>căn(x^2-1)-1>=0

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

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

\(a,\) A có nghĩa \(\Leftrightarrow x^2-1\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)

\(b,x\ge\sqrt{2}\Leftrightarrow\sqrt{x^2-1}-1\ge\sqrt{\left(\sqrt{2}\right)^2-1}-1=0\\ \Rightarrow A=\sqrt{x^2-1}+1-\left(\sqrt{x^2-1}-1\right)=2\)

\(1,x^2=16\\ 2,\sqrt{x^2}=\left|x\right|=x\)

16

x

1.

\(\lim\dfrac{5\sqrt{3n^2+n}}{2\left(3n+2\right)}=\lim\dfrac{5\sqrt{3+\dfrac{1}{n}}}{2\left(3+\dfrac{2}{n}\right)}=\dfrac{5\sqrt{3}}{6}\Rightarrow a+b=11\)

2.

\(\lim\limits_{x\rightarrow2}\dfrac{x^2+ax+b}{x-2}=6\) khi \(x^2+ax+b=0\) có nghiệm \(x=2\)

\(\Rightarrow4+2a+b=0\Rightarrow b=-2a-4\)

\(\lim\limits_{x\rightarrow2}\dfrac{x^2+ax-2a-4}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)+a\left(x-2\right)}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+a+2\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\left(x+a+2\right)=a+4\Rightarrow a+4=6\Rightarrow a=2\Rightarrow b=-8\)

\(\Rightarrow a+b=-6\)

Đặt \(m=\sqrt[3]{x^2}\)và \(n=\sqrt[3]{y^2}\)

=> m³ = x² và n³ = y²

Ta có :\(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\)

=> \(\sqrt{m^3+\sqrt[3]{m^6n^3}}+\sqrt{n^3+\sqrt[3]{m^3n^6}}=a\)

=> \(\sqrt{m^3+m^2n}+\sqrt{n^3+mn^2}=a\)

=> \(\sqrt{m^2\left(m+n\right)}+\sqrt{n^2\left(m+n\right)}=a\)

=> \(\sqrt{m+n}\left(m+n\right)=a\)

=> \(\left(\sqrt{m+n}\right)^3=\left(\sqrt[3]{a}\right)^3\)

=>\(\sqrt{m+n}=\sqrt[3]{a}\)

=> \(m+n=\left(\sqrt[3]{a}\right)^2\)

=> \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)

Tính giải mà lười quá. Bạn cứ nhân vô là ra ah

@Doraemon2611.

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

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

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

a) A có nghĩa <=> \(x^2-1\ge0\Leftrightarrow x^2\ge1\Leftrightarrow\orbr{\begin{cases}x\ge1\\x\le-1\end{cases}}\)

b) Nếu \(x\ge\sqrt{2}\)khi đó \(\sqrt{x^2-1}-1\ge\sqrt{\left(\sqrt{2}\right)^2-1}-1=0\)

Ta có: \(A=\sqrt{x^2-1}+1-\left(\sqrt{x^2-1}-1\right)=2\)

VT=|căn(x-2)+1|+|căn (x-2)-1|

=|căn (x-2)+1|+|1-căn x-2|>=|căn(x-2)+1+1-căn(x-2)|=2

Xin người.