ĐK : \(x\ne-\frac{1}{2}\);\(x\ge0\)

\(\frac{\left(2x^2+8x+1\right)^2}{\left(2x+1\right)^2}=25x\)

\(25\left(4x^2+4x+1\right)=\left(4x^4+64x^2+1+32x^3+4x^2+16x\right)\)

\(4x^4+32x^3-32x^2-84x-24=0\)

giải tiếp đc nghiệm

bạn sửa lại ở dưới x là P nhé

\(P=\frac{3x\left(x-2\right)+x^2-x+3}{x-2}\)

\(P=\frac{3x^2-6x+x^2-x+3}{x-2}\)

\(P=\frac{4x^2-7x+3}{x-2}\)

\(P\left(x-2\right)=4x^2-7x+3\)

\(Px-2P=4x^2-7x+3\)

\(4x^2-x\left(7+P\right)+3+2P=0\)

Ta có : \(\Delta=\left(7+P\right)^2-4.4\left(3+2P\right)\)

\(\Delta=7^2+14P+P^2-48-32P\)

\(\Delta=P^2-18P+1\ge0\)

\(\left\{{}\begin{matrix}x\le9-4\sqrt{2}\\x\ge9+4\sqrt{2}\end{matrix}\right.\)

Vậy \(P_{min}=9+4\sqrt{2}\)

thấy đề hơi sai sai nha bạn :

\(\overrightarrow{AI}=\overrightarrow{CI}-\overrightarrow{CA}=\frac{1}{2}\overrightarrow{BC}+\overrightarrow{AC}\)

\(\frac{1}{2}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)+\overrightarrow{AC}=\frac{3}{2}\overrightarrow{AC}-\frac{1}{2}\overrightarrow{AB}\)

g/ \(\left(x-1\right)^3-x\left(x-2\right)^2-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^3-3x^2+3x-1\right)-x\left(x^2-4x+4\right)-\left(x-2\right)=0\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3+4x^2-4x-x+2=0\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2\)

\(\Leftrightarrow x=1\)

h) \(x^2-5x+6=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

i) \(2x^3+5x^2-3x=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x+3\right)x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-3\\x=0\end{matrix}\right.\)

=112 nha

2/ (6+x)^2 = 49

=> 6+x = 7

=> x =1

ta có \(P=\frac{2\sqrt{x}-1}{\sqrt{x}+1}=\frac{2\left(\sqrt{x}+1\right)-3}{\left(\sqrt{x}+1\right)}=2-\frac{3}{\sqrt{x}+1}\ge2\)

Vậy \(P_{min}=2\)

Neu tang so lon them 8 don vi và giam so bé 3 don vi thì ta có hieu moi la : 
161 + 8. + 3 = 172 
Khi do so lon gap 3 lan so bé 
Ta co so do : 
So be khi giam 3 don vi : l----l 
So lon tang tám don vi. : l----l----l----l 
Hieu so phan bang nhau la 3 - 1 = 2 phan 
So bé khi giam 3 don vi la : 
( 172 : 2 ) x 1 = 86 
So be can tim la 86 + 3 = 89 
So lon can tim la 89 + 161 = 250

CHÚC BẠN HỌC TỐT

Đặt : \(\sqrt{a}=x\left(x\ge0\right);\sqrt{b}=y\left(y\ge0\right)\)

BPT \(\Leftrightarrow\frac{x^2}{y}+\frac{y^2}{x}-x-y\ge0\)

BĐT dĩ nhiên đúng vì theo BĐT caushy-schwars,ta có:

\(\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{\left(x+y\right)^2}{x+y}=x+y\left(dpcm\right)\)

Dấu "=" xảy ra khi x = y <=> a = b