Câu hỏi của Hoan Mạnh

Question

Cho a =  $\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}$. Tính P =  $\left(a^3-4a+1\right)^{2011}$

Trần Thanh Phương · Accepted Answer

$a=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}$

$a^2=4+2\sqrt{3}-2\sqrt{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}+4-2\sqrt{3}$

$a^2=8-2\sqrt{16-12}$

$a^2=8-4$

$a^2=4$

Thay vào P ta được :

$P=\left(a^3-4a+1\right)^{2011}$

$P=\left[a\left(a^2-4\right)+1\right]^{2011}$

$P=\left[a\left(4-4\right)+1\right]^{2011}$

$P=\left(0+1\right)^{2011}$

$P=1^{2011}=1$

Vậy...Câu trả lời: $a=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}$

$a^2=4+2\sqrt{3}-2\sqrt{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}+4-2\sqrt{3}$

$a^2=8-2\sqrt{16-12}$

$a^2=8-4$

$a^2=4$

Thay vào P ta được :

$P=\left(a^3-4a+1\right)^{2011}$

$P=\left[a\left(a^2-4\right)+1\right]^{2011}$

$P=\left[a\left(4-4\right)+1\right]^{2011}$

$P=\left(0+1\right)^{2011}$

$P=1^{2011}=1$

Vậy...

nguyễn hà linh · Answer

a= $\sqrt{\left(\sqrt{3}+1\right)}^2$-$\sqrt{\left(\sqrt{3}-1\right)}^2$

= $\left(\sqrt{3}+1\right)$-$\left(\sqrt{3}-1\right)$

= 2

Có P =$\left(2^3-4.2+1\right)^{2011}$

= 1Câu trả lời: a= $\sqrt{\left(\sqrt{3}+1\right)}^2$-$\sqrt{\left(\sqrt{3}-1\right)}^2$

= $\left(\sqrt{3}+1\right)$-$\left(\sqrt{3}-1\right)$

= 2

Có P =$\left(2^3-4.2+1\right)^{2011}$

= 1

Bài 3: Liên hệ giữa phép nhân và phép khai phương