(a+b)*(a^2-ab+b^2)+(a-b)*(a^2+ab+b^2)

=a^3+b^3+a^3-b^3

=2a^3

tks cho mk nhe

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

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

=\(\frac{2\sqrt{a^2-b^2}}{2b}\):​\(\frac{\sqrt{a^2-b^2}}{a^2+b^2}\)​

=\(\frac{\sqrt{a^2-b^2}}{b}\)*\(\frac{a^2+b^2}{\sqrt{a^2-b^2}}\)

=\(\frac{a^2+b^2}{b}\)

b/ Thế \(b=a-1\)thì ta có

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

\(\Leftrightarrow2a^2-\left(2+P\right)a+1+P=0\)

\(\Rightarrow\Delta_a=\left(2+P\right)^2-4.2.\left(1+P\right)\ge0\)

\(\Leftrightarrow P\ge2+2\sqrt{2}\)

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

\(\left(a-b\right)\left(a^2+ab+b^2\right)=a\left(a^2+ab+b^2\right)-b\left(a^2+ab+b^2\right)\)

                                                \(=a^3+a^2b+ab^2-a^2b-ab^2-b^3\)

                                                  \(=a^3-b^3\)

\(\left(a+b\right)\left(a^2-ab+b^2\right)=a\left(a^2-ab+b^2\right)+b\left(a^2-ab+b^2\right)\)

                                                \(=a^3-a^2b+ab^2+a^2b-ab^2+b^3\)

                                                 \(=a^3+b^3\)

(a-b)(a^2+ab+b^2)=a^3-b^3

(a+b)(a^2-ab+b^2)=a^3+b^3

a: ĐKXĐ: a>=0; b>=0; ab<>1

Ta có: \(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\)

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

\(=\frac{\sqrt{a}+a\cdot\sqrt{b}+\sqrt{b}+b\cdot\sqrt{a}+\sqrt{a}-a\cdot\sqrt{b}-\sqrt{b}+b\cdot\sqrt{a}}{1-ab}=\frac{2\cdot\sqrt{a}+2b\cdot\sqrt{a}}{1-ab}\)

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

Ta có: \(D=\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)

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

\(=\frac{2\sqrt{a}\left(b+1\right)}{ab+a+b+1}=\frac{2\sqrt{a}\left(b+1\right)}{\left(b+1\right)\left(a+1\right)}=\frac{2\sqrt{a}}{a+1}\)

b: \(a=\frac{2}{2+\sqrt3}=\frac{2\left(2-\sqrt3\right)}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}\)

\(=\frac{4-2\sqrt3}{4-3}=4-2\sqrt3=\left(\sqrt3-1\right)^2\)

Thay \(a=\left(\sqrt3-1\right)^2\) vào D, ta được:

\(D=\frac{2\cdot\sqrt{\left(\sqrt3-1\right)^2}}{\left(\sqrt3-1\right)^2+1}\)

\(=\frac{2\left(\sqrt3-1\right)}{4-2\sqrt3+1}=\frac{2\sqrt3-2}{5-2\sqrt3}=\frac{\left(2\sqrt3-2\right)\left(5+2\sqrt3\right)}{\left(5-2\sqrt3\right)\left(5+2\sqrt3\right)}\)

\(=\frac{10\sqrt3+12-10-4\sqrt3}{25-12}=\frac{6\sqrt3+2}{13}\)

c: \(\frac{1}{D}=\frac{a+1}{2\sqrt{a}}\)

=>\(\frac{1}{D}-1=\frac{a+1-2\sqrt{a}}{2\sqrt{a}}=\frac{\left(\sqrt{a}-1\right)^2}{2\sqrt{a}}\ge0\forall a\) thỏa mãn ĐKXĐ

=>\(\frac{1}{D}\ge1\forall a\) thỏa mãn ĐKXĐ

=>D<=1∀a thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\sqrt{a}-1=0\)

=>a=1(nhận)

a)  ĐK:  a > 0;  b > 0

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

\(=\frac{\sqrt{a}+\sqrt{b}+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}-b\)

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

\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}-b\)

\(=2\sqrt{b}-b\)

b)  \(A=1\)\(\Rightarrow\)\(2\sqrt{b}-b=1\)

                    \(\Leftrightarrow\)\(b-2\sqrt{b}+1=0\)

                   \(\Leftrightarrow\) \(\left(\sqrt{b}-1\right)^2=0\)

                   \(\Leftrightarrow\)\(\sqrt{b}-1=0\)

                   \(\Leftrightarrow\)\(\sqrt{b}=1\)

                   \(\Leftrightarrow\)\(b=1\)   (t/m ĐKXĐ)

Vậy  b=1