\(x^5+x^4+1\)

\(=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^3-x+1\right)\)

hack à anh nhanh thế hai cái cùng lúc

ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow x-2=x^2-2x+4\)

\(\Leftrightarrow x^2-3x+6=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{15}{4}=0\left(vl\right)\)

=> vô no

\(\left(x-2\right)\left(x^2-25\right)=0\)

\(\Rightarrow\) x-2=0      hoặc  \(x^2-25=0\)

     x=2                    \(x^2=25\)

                                \(\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

\(x^2=2x\)

Với x chưa khác 0 thì không thể chia (vì không có số nào chia đc cho 0)

a)Xét \(\Delta APD\) và \(\Delta AEB\) có:

\(\widehat{ADP}=\widehat{ABE}=90^o\)

AD = AB ( hvABCD)

\(\widehat{PAD}=\widehat{EAB}\) (cùng phụ \(\widehat{DAE}\))

=> \(\Delta APD\) = \(\Delta AEB\)  (gcg)

=>AP=AE

mà \(\widehat{PAE}=90^o\left(gt\right)\)

=>\(\Delta APE\) vuông cân tại A

b) Xét \(\Delta APF\) vuông tại A có:

\(\dfrac{1}{AP^2}+\dfrac{1}{AF^2}=\dfrac{1}{AD^2}\) ( hệ thức lượng trong tam giác vuông )

mà AP=AE ; AD=AB

=>\(\dfrac{1}{AE^2}+\dfrac{1}{AF^2}=\dfrac{1}{AB^2}\)

1) Với x=4 thì

\(A=\dfrac{2\sqrt{4}}{\sqrt{4}+3}=\dfrac{4}{2+3}=\dfrac{4}{5}\)

2) \(P=A+B\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x+3}\right)}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

3) Để P< 3 thì

\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}< 3\)

\(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}-3}-\dfrac{3\left(\sqrt{x}-3\right)}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\dfrac{9}{\sqrt{x}-3}< 0\)

\(\Rightarrow\sqrt{x}-3< 0\) ( vì 9>0)

<=> x<9

Vậy giá trị nguyên lớn nhất của x để P <3 là 8

Có:

\(x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}\ge0\)

\(x\left(\sqrt{x}-\sqrt{y}\right)-y\left(\sqrt{x}-\sqrt{y}\right)\ge0\)

\(\left(x-y\right)\left(\sqrt{x}-\sqrt{y}\right)\ge0\)

\(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\ge0\)

\(\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)\ge0\)  (luôn đúng)

Dấu = xảy ra khi x=y

\(\sqrt{x_1^2-1^2}+2\sqrt{x^2_2-2^2}+...+100\sqrt{x_{100}^2-100^2}=\dfrac{1}{2}\left(x_1^2+x^2_2+...+x_{100}^2\right)\)

\(\Leftrightarrow2\sqrt{x_1^2-1^2}+4\sqrt{x^2_2-2^2}+...+200\sqrt{x_{100}^2-100^2}=x_1^2+x^2_2+...+x_{100}^2\)

\(\Leftrightarrow x_1^2-1-2\sqrt{x_1^2-1}+1+x^2_2-4-4\sqrt{x^2_2-4}+4+...+x^2_{100}-10000-200\sqrt{x_{100}^2-10000}+10000=0\)

\(\Leftrightarrow\left(\sqrt{x^2_1-1}-1\right)^2+\left(\sqrt{x^2_2-4}-2\right)^2+....+\left(\sqrt{x^2_{100}-10000}-100\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2_1-1}-1=0\\\sqrt{x^2_2-4}-2=0\\....\\\sqrt{x^2_{100}-10000}-100=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\sqrt{1^2+1}=\sqrt{2}\\x_2=\sqrt{2^2+4}=2\sqrt{2}\\....\\x_{100}=\sqrt{100^2+10000}=100\sqrt{2}\end{matrix}\right.\)