\(ab=\dfrac{\left(a+b\right)^2-a^2-b^2}{2}=\dfrac{13^2-89}{2}=\dfrac{80}{2}=40\)

\(P=\left(a+b\right)^3-3ab\left(a+b\right)=13^3-3\cdot40\cdot13=637\)

\(\left\{{}\begin{matrix}a+b=13\\a^2+b^2=89\end{matrix}\right.\)

\(\left(a+b\right)^2=169\)

\(a^2+2ab+b^2=169\)

\(ab=40\)

\(P=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=13^3-3.40.13=637\)

Ta có: \(A=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}-\dfrac{b^2}{\left(b-a\right)\left(c-b\right)}-\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\)

\(=\dfrac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\dfrac{b^2\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\dfrac{c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{\left(a-b\right)\left(ab+c^2\right)-c\left(a-b\right)\left(a+b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{\left(a-b\right)\left(ab+c^2-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{c^2+ab-c}{\left(a-c\right)\left(b-c\right)}\)

a: \(B=\frac{2\sqrt{x}+13}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

\(=\frac{2\sqrt{x}+13+\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2\sqrt{x}+13+x+\sqrt{x}-6}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}+3\right)}=\frac{x+3\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

b: C=A-B

\(=\frac{2\sqrt{x}-1}{\sqrt{x}+3}-\frac{x+3\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(x+3\sqrt{x}+7\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}=\frac{2x+3\sqrt{x}-2-x-3\sqrt{x}-7}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-3}{\sqrt{x}+2}\)

C nguyên khi \(\sqrt{x}-3\) ⋮\(\sqrt{x}+2\)

=>\(\sqrt{x}+2-5\) ⋮\(\sqrt{x}+2\)

=>-5⋮\(\sqrt{x}+2\)

=>\(\sqrt{x}+2=5\) (Do \(\sqrt{x}+2\ge2\forall x\) thỏa mãn ĐKXĐ)

=>\(\sqrt{x}=3\)

=>x=9(nhận)

Toán lớp 9 như này á hả

Chắc nổ não sớm quá

a: Ta có: \(x^2=3-2\sqrt{2}\)

nên \(x=\sqrt{2}-1\)

Thay \(x=\sqrt{2}-1\) vào A, ta được:

\(A=\dfrac{\left(\sqrt{2}+1\right)^2}{\sqrt{2}-1}=\dfrac{3+2\sqrt{2}}{\sqrt{2}-1}=7+5\sqrt{2}\)

\(1,M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

Thay \(a+b=1\) vào ta được:

\(1\left(1-3ab\right)+3ab\left(1-2ab\right)+6a^2b^2\)

\(=1-3ab+3ab-6a^2b^2+6a^2b^2\)

\(=1\)

Vậy ......................

*) \(MinA\) :

Ta thấy: a,b,c đều là các số thực không âm.

Do đó : \(A\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=0,c=1\) và các hoán vị.

\(*)MaxA\) :

Giả sử \(a\ge b\ge c\) \(\Rightarrow3a\ge a+b+c=1\) 

\(\Rightarrow1-3a\le0\)

Ta có : \(A=a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)\)

\(=a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)+3abc-3abc\)

\(=\left(a+b+c\right)\left(ab+bc+ca\right)-3abc\)

\(=ab+bc+ca-3abc\)

\(=a\left(b+c\right)+bc\left(1-3a\right)\) \(\le\frac{\left(a+b+c\right)^2}{4}+0\) ( do \(1-3a\le0\) )    \(=\frac{1}{4}\)

hay \(A\le\frac{1}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=\frac{1}{2},c=0\) và các hoán vị.

\(\)