Đề thiếu ko nhỉ? cộng b^2 nữa chứ 

\(\left(a-b\right)\left(a-2b\right)\left(a-3b\right)\left(a-4b\right)+b^2\)

\(=\left[\left(a-b\right)\left(a-4b\right)\right]\left[\left(a-2b\right)\left(a-3b\right)\right]+b^2\)

\(=\left(a^2-4ab-ab+4b^2\right)\left(a^2-3ab-2ab+6b^2\right)+b^2\)

\(=\left(a^2-5ab+4b^2\right)\left(a^2-5ab+6b^2\right)+b^2=\left(a^2-5ab+5b^2\right)^2-b^2+b^2\)

\(=\left(a^2-5ab+b^2\right)^2\rightarrowđpcm\)

\(a,\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\sqrt{2.3}+2}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

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

\(=3-2\)

\(=1\)

\(b,\sqrt{11+2\sqrt{6}}-3+\sqrt{2}\)

==>Đề sai???

Ta có:

$3^{2n}+3^n=9^n+3^n⋮12$ đồng dư 12 mod 13

${\mathrm{\Rightarrow 32n+3n+1⋮13}}^{}$

 học tốt

a có:

$3^{2n}+3^n=9^n+3^n⋮12$ đồng dư 12 mod 13

$\Rightarrow 3^{2n}+3^n+1⋮13$

Sai đâu có gì sửa giùm

\(a,\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{3+2\sqrt{2.3}+2}-\sqrt{3-2\sqrt{2.3}+2}\)

\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\)

\(=2\sqrt{2}\)

\(b,\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

\(=\sqrt{5-2\sqrt{2.5}+2}-\sqrt{5+2\sqrt{5.2}+2}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)

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

a) \(\sqrt{5+2\sqrt{6}}\) -\(\sqrt{5-2\sqrt{6}}\) 

=\(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\) 

=/\(\sqrt{3}+\sqrt{2}\)/  \(-\)/\(\sqrt{3}-\sqrt{2}\) /

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

=\(\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\) 

=\(2\sqrt{2}\) 

b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\) 

=\(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\) 

=/\(\sqrt{5}-\sqrt{2}\) / \(-\) /\(\sqrt{5}+\sqrt{2}\)/

=\(\sqrt{5}-\sqrt{2}-\left(\sqrt{5}+\sqrt{2}\right)\) 

=\(\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\) 

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

Theo Vi-ét cho 3 số (chứng minh bằng hệ số bất định)

\(\hept{\begin{cases}x_1+x_2+x_3=0\\x_1x_2+x_2x_3+x_1x_3=-3\\x_1x_2x_3=-1\end{cases}}\)

\(A=\frac{1+2x_1}{1+x_1}+\frac{1+2x_2}{1+x_2}+\frac{1+2x_3}{1+x_3}\)

   \(=3+\frac{x_1}{1+x_1}+\frac{x_2}{1+x_2}+\frac{x_3}{1+x_3}\)

   \(=3+\frac{x_1\left(1+x_2\right)\left(1+x_3\right)+x_2\left(1+x_1\right)\left(1+x_3\right)+x_3\left(1+x_1\right)\left(1+x_2\right)}{\left(1+x_1\right)\left(1+x_2\right)\left(1+x_3\right)}\)

    \(=3+\frac{x_1\left(1+x_2+x_3+x_2x_3\right)+x_2\left(1+x_1+x_3+x_1x_3\right)+x_3\left(1+x_1+x_2+x_1x_2\right)}{\left(1+x_1+x_2+x_1x_2\right)\left(1+x_3\right)}\)

    \(=3+\frac{\left(x_1+x_2+x_3\right)+2\left(x_1x_2+x_2x_3+x_3x_1\right)+3x_1x_2x_3}{1+x_1+x_2+x_3+x_1x_2+x_1x_3+x_2x_3+x_1.x_2.x_3}\)

   \(=3+\frac{0+2.\left(-3\right)+3.\left(-1\right)}{1+0-3-1}\)

   \(=6\)

Do x₁ là một nghiệm của đa thức f(x) nên ta có: \(x_1^3-3x_1+1=0\)

\(\Leftrightarrow\)\(\left(x_1+1\right)\left(x_1^2-x_1+1\right)=3x_1\)\(\Leftrightarrow\)\(x_1+1=\frac{3x_1}{x_1^2-x_1+1}\)

Có: \(A==\frac{1+2x_1}{1+x_1}+\frac{1+2x_2}{1+x_2}+\frac{1+2x_3}{1+x_3}=3+\left(\frac{x_1}{1+x_1}+\frac{x_2}{1+x_2}+\frac{x_3}{1+x_3}\right)\)

\(A=3+\left(\frac{x_1\left(x_1^2-x_1+1\right)}{3x_1}+\frac{x_2\left(x^2_2-x_2+1\right)}{3x_2}+\frac{x_3\left(x_3^2-x_3+1\right)}{3x_3}\right)\)

\(A=3+\frac{\left(x_1^2+x_2^2+x_3^2\right)-\left(x_1+x_2+x_3\right)+3}{3}\)

\(A=3+\frac{\left(x_1+x_2+x_3\right)^2-2\left(x_1x_2+x_2x_3+x_3x_1\right)-\left(x_1+x_2+x_3\right)+3}{3}\)

Đến đây theo Vi-et bậc 3 

\(\hept{\begin{cases}x_1+x_2+x_3=0\\x_1x_2+x_2x_3+x_3x_1=-3\end{cases}}\)

a4 = a + 1813

=> a + 1813 = a4

=> a + 1813 = a0 + 4

=> a + 1813 = 10 x a + 4

=> 10 x a - a = 1813 - 4

=> 9 x a        = 1809

=>      a         = 1809 : 9

=>      a         = 201

201 nha 

làm bạn mình mình chả lời hết cho bạn

3/2 . 5 + 3/2 . 95

= 3/2 . ( 5 + 95 )

= 3/2 . 10

= 30/2

= 15

3/2 x 5 + 3/2 x 95

= 3/2 x (5 + 95)

= 3/2 x 100

= 150