sửa lại đề\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)

\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)

\(\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\)

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

đề là dấu trừ mà

Đặt $P=a^2+b^2+c^2+ab+bc+ca$

$P=\frac{1}{2}{\left(a+b+c\right)}^2+\frac{1}{2}\left(a^2+b^2+c^2\right)$

$P\ge \frac{1}{2}{\left(a+b+c\right)}^2+\frac{1}{6}{\left(a+b+c\right)}^2=6$

Dấu "=" xảy ra khi $a=b=c=1$

Ta có: \(\frac{a^2b^2+7}{\left(a+b\right)^2}=\frac{a^2b^2+1+6}{\left(a+b\right)^2}\ge\frac{2ab+2\left(a^2+b^2+c^2\right)}{\left(a+b\right)^2}\)( cô-si )

\(=\frac{\left(a+b\right)^2+a^2+b^2+2c^2}{\left(a+b\right)^2}=1+\frac{a^2+b^2+2c^2}{\left(a+b\right)^2}\)\(\ge1+\frac{a^2+b^2+2c^2}{2\left(a^2+b^2\right)}=1+\frac{1}{2}+\frac{c^2}{a^2+b^2}=\frac{3}{2}+\frac{c^2}{a^2+b^2}\)

CMTT \(\Rightarrow\)\(VT\ge\frac{9}{2}+\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}\)

\(P=\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}\)

Đặt \(\hept{\begin{cases}b^2+c^2=x>0\\a^2+c^2=y>0\\a^2+b^2=z>0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a^2=\frac{y+z-x}{2}\\b^2=\frac{z+x-y}{2}\\c^2=\frac{x+y-z}{2}\end{cases}}\)

\(\Rightarrow P=\frac{y+z-x}{2x}+\frac{z+x-y}{2y}+\frac{x+y-z}{2z}\)

\(=\frac{y}{2x}+\frac{z}{2x}-\frac{1}{2}+\frac{z}{2y}+\frac{x}{2y}-\frac{1}{2}+\frac{x}{2z}+\frac{y}{2z}-\frac{1}{2}\)

\(=\left(\frac{y}{2x}+\frac{x}{2y}\right)+\left(\frac{z}{2x}+\frac{x}{2z}\right)+\left(\frac{z}{2y}+\frac{y}{2z}\right)-\frac{3}{2}\)

\(\ge1+1+1-\frac{3}{2}=\frac{3}{2}\)( bđt cô si )

\(\Rightarrow VT\ge\frac{9}{2}+\frac{3}{2}=6\) ( đpcm)

Dấu "=" xảy ra <=> a=b=c=1

Do đa thức chia là \(x^2-4x+3\)là đa thức bậc hai nên đa thức dư là đa thức bậc nhất, có dạng \(ax+b\).

Đặt \(P\left(x\right)=Q\left(x\right)\left(x^2-4x+3\right)+ax+b\)

\(P\left(1\right)=Q\left(1\right)\left(1-4+3\right)+a+b\Leftrightarrow a+b=3\)

\(P\left(3\right)=Q\left(3\right)\left(9-12+3\right)+3a+b\Leftrightarrow3a+b=7\)

Ta có hệ: 

\(\hept{\begin{cases}a+b=3\\3a+b=7\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=1\end{cases}}\).

Vậy đa thức dư là: \(2x+1\).

a) ĐKXĐ: 

\(3-2x-x^2\ge0\Leftrightarrow\left(1-x\right)\left(x+3\right)\ge0\Leftrightarrow-3\le x\le1\).

b) Đặt \(\sqrt{3-2x-x^2}=t\ge0\)

\(3-2x-x^2=4-\left(x+1\right)^2\le4\Rightarrow t\le2\).

\(P=t+t^2-3\le2+2^2-3=3\)

Dấu \(=\)khi \(t=2\Rightarrow x=-1\).

Vậy \(maxP=3\).

\(Q=x+\frac{3}{\sqrt{x}}=x+\frac{3}{2\sqrt{x}}+\frac{3}{2\sqrt{x}}\ge3\sqrt[3]{x.\left(\frac{3}{2\sqrt{x}}\right)^2}=3\sqrt[3]{\frac{9}{4}}\)

Dấu \(=\)khi \(x=\frac{3}{2\sqrt{x}}\Leftrightarrow x=\sqrt[3]{\frac{9}{4}}\).

Tổng quát: 

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

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

Suy ra: \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)

\(A=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}< 1+2\left(\sqrt{2}-\sqrt{1}+...+\sqrt{100}-\sqrt{99}\right)\)

\(=1+2\left(\sqrt{100}-\sqrt{1}\right)=19\)

\(A=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)

\(=2\left(\sqrt{100}-\sqrt{1}\right)=18\)

Do đó ta có đpcm. 

Dựng hình bình hành \(ABEC\).

Khi đó \(E\in DC\).

Vì \(BD\perp AC\)mà \(AC//BE\)nên \(BE\perp BD\).

Kẻ \(BH\perp DE\). 

Xét tam giác \(BED\)vuông tại \(B\)đường cao \(BH\): 

\(\frac{1}{BH^2}=\frac{1}{BD^2}+\frac{1}{BE^2}\Leftrightarrow\frac{1}{4^2}=\frac{1}{5^2}+\frac{1}{BE^2}\Leftrightarrow BE=\frac{20}{3}\left(cm\right)\)

\(S_{ABCD}=\frac{1}{2}.AC.BD=\frac{1}{2}.BD.BE=\frac{1}{2}.5.\frac{20}{3}=\frac{50}{3}\left(cm^2\right)\)

Có ai biết đổi tên cho mình hông?

\(\frac{3+\sqrt{3}}{3}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=1+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}.\sqrt{2}-\sqrt{3}}{1-\sqrt{2}}=1+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}\)

\(=1+\frac{\sqrt{3}}{3}-\sqrt{3}\)

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

\(\frac{\sqrt{3}+1}{\sqrt{3}}-\sqrt{3}\)

\(\frac{\sqrt{3}-2}{\sqrt{3}}\)

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

\(\sqrt{193-132\sqrt{2}}+\sqrt{193+132\sqrt{2}}=\sqrt{121-2.11.6\sqrt{2}+72}+\sqrt{121+2.11.6\sqrt{2}+72}\)

\(=\sqrt{11^2-2.11.6\sqrt{2}+\left(6\sqrt{2}\right)^2}+\sqrt{11^2+2.11.6\sqrt{2}+\left(6\sqrt{2}\right)^2}\)

\(=\sqrt{\left(11-6\sqrt{2}\right)^2}+\sqrt{\left(11+6\sqrt{2}\right)^2}=\left|11-6\sqrt{2}\right|+\left|11+6\sqrt{2}\right|\)

\(=11-6\sqrt{2}+11+6\sqrt{2}=22\)

p/s : cách khác 

Đặt \(Nghia=\sqrt{193-132\sqrt{2}}+\sqrt{193+132\sqrt{2}}\)

\(\Rightarrow Nghia^2=193-132\sqrt{2}+193+132\sqrt{2}+2\sqrt{\left(193-132\sqrt{2}\right)\left(193+132\sqrt{2}\right)}\)

\(=386+2\sqrt{2401}=386+2.49=484\)

\(\Rightarrow Nghia=\sqrt{484}=22\)