a,b dể tự làm nha

c)ta có:   \(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2-2ab\ge0\Leftrightarrow a^2+2ab+b^2-2ab-2ab\ge0\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)       mà a+b=1

\(\Rightarrow1\ge4ab\Leftrightarrow ab\le\frac{1}{4}\)

lại có: \(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2\ge2ab\) mà \(ab\le\frac{1}{4}\)

tahy vào có     \(a^2+b^2\ge2\times\frac{1}{4}\Leftrightarrow a^2+b^2\ge\frac{1}{2}\left(dpcm\right)\)

b mình tự làm, bạn làm phần a hộ mình với

\(a,\sqrt{68^2-32^2}\)

\(=\sqrt{\left(68-32\right)\left(68+32\right)}\)

\(=\sqrt{3600}=60\)

\(b,\sqrt{37^2-12^2}\)

\(=\sqrt{\left(37-12\right)\left(37+12\right)}\)

\(=\sqrt{25.49}=5.7=35\)

\(c,\sqrt{21,8^2-18,2^2}\)

\(=\sqrt{\left(21,8+18,2\right)\left(21,8-18,2\right)}\)

\(=\sqrt{40.3,6}=12\)

a) AM ứng với cạnh huyền BC nên AM = \(\frac{1}{2}\) x BC = \(\frac{4}{2}\) = 2 cm

AH = tan\(\widehat{ACH}\)x HM = tan 15⁰ x 2 = \(4-2\sqrt{3}\)cm

Sin \(\widehat{AMH}\)= \(\frac{AH}{AM}\)= \(\frac{4-2\sqrt{3}}{2}\)  = \(2-\sqrt{3}\)    cm

Định lí Pitago : AM² = AH² + HM²

HC = tan \(\widehat{ACH}\)x AH

\(\frac{x+5}{\sqrt{x}+1}=\frac{\left(x-1\right)+6}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+6}{\sqrt{x}+1}=\sqrt{x}-1+\frac{6}{\sqrt{x}+1}\)

\(=\sqrt{x}+1+\frac{6}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right)\frac{6}{\sqrt{x}+1}}-2=2\sqrt{6}-2\)(Cauchy)

\(=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\right]:\left(\frac{x+1}{x-1}-\frac{x-1}{x-1}\right)\)

\(=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\right]:\frac{2}{x-1}\)

\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{x-1}{2}\)

\(=\left[\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{x-1}{2}\)

\(=\frac{x-\sqrt{x}-2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)

\(=\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)

\(=\frac{1}{\sqrt{x}+1}\le1\)

Dấu "=" xảy ra khi x=0

thanks bạn