mk nhan nham

báo cáo nha

(ac+bd)2+(ad−bc)2=(a2+b2)(c2+d2)(ac+bd)2+(ad−bc)2=(a2+b2)(c2+d2) 

<=> a2c2+2abcd+b2d2+a2d2+b2c2−2abcd=a2c2+a2d2+b2c2+b2d2a2c2+2abcd+b2d2+a2d2+b2c2−2abcd=a2c2+a2d2+b2c2+b2d2

<=> a2b2+a2d2+b2c2+b2d2=a2c2+a2d2+b2c2+d2b2a2b2+a2d2+b2c2+b2d2=a2c2+a2d2+b2c2+d2b2 

1. a) (ac+bd)^2+(ad−bc)^2(ac+bd)^2+(ad−bc)^2

=a^2c^2+2abcd+b^2d^2+a^2d^2−2abcd+b^2c^2

=a^2.(c^2+d2)+b^2.(c^2+d^2)

=(c^2+d^2).(a^2+b^2)

b) Ta có (ac+bd)^2≤(a^2+b^2).(c^2+d^2)

⇔a^2c^2+2abcd+b^2d^2≤a^2c^2+b^2d^2+a^2d^2+b^2c^2

⇔a^2d^2−2abcd+b^2c^2≥0

⇔(ad−bc)^2≥0( Đúng )

Dấu "=" xảy ra ⇔ad=bc

Giả sử:

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

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

\(\Leftrightarrow2\sqrt{2}-2\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=2\left(\sqrt{2}-1\right)\)

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

\(\Leftrightarrow\sqrt{2-1}=1\)(đúng )

Vậy ta có điều phải chứng minh.

\(A=\sqrt{3-2\sqrt{2}}\)\(+\sqrt{6-4\sqrt{2}}\)

\(A=\sqrt{2-2\sqrt{2}+1}+\sqrt{4-2.2\sqrt{2}+2}\)

\(A=\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1}+\sqrt{2^2-2.2\sqrt{2}+\left(\sqrt{2}\right)^2}\)

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

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