Đặt A = .....

A^2 = \(-x^2+4x+12+-x^2+2x+3+2\sqrt{\left(-x^2+4x+12\right)\left(-x^2+2x+3\right)}\)

    = \(-2x^2+6x+15+2\sqrt{\left(x+2\right)\left(6-x\right)\left(x+1\right)\left(3-x\right)}\)

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

  = \(l\sqrt{\left(x+2\right)\left(3-x\right)}+\sqrt{\left(x+1\right)\left(6-x\right)}l+3\ge3\)

=> P \(\ge\sqrt{3}\)

Vậy GTNN là .... tại x = 0 

Tick cho mình trước khi đọc nha thể nào cũng đúng

Ta có \(x^2+6x^2+6+\left(\frac{x+3}{x+4}\right)^2=0\)

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

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

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

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

\(\Leftrightarrow\left(\frac{x^2+7x+12-\left(x+3\right)}{x+4}\right)^2+2\frac{\left(x+3\right)^2}{\left(x+4\right)}-3=0\)

\(\Leftrightarrow\left(\frac{x^2+6x+9}{x+4}\right)^2+2\frac{\left(x+3\right)^2}{\left(x+4\right)}-3=0\)

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

Đặt \(\frac{\left(x+3\right)^2}{x+4}=a\) pt <=> \(a^2+2a-3=0\Leftrightarrow\left(a+3\right)\left(a-1\right)=0\)

nên a=-3 hoặc a=1

Với a=-3 thì \(\frac{\left(x+3\right)^2}{x+4}=-3\Leftrightarrow x^2+6x+9=-3\left(x+4\right)\Leftrightarrow x^2+9x+21=0\)

nên pt này vô nghiệm

Với a=1 thì \(\frac{\left(x+3\right)^2}{x+4}=1\Leftrightarrow x^2+6x+9=\left(x+4\right)\Leftrightarrow x^2+5x+5=0\)

Giải ra được 2 nghiệm

Vậy....

\(B=l7x-3l+l7x+3l\)

     = \(l3-7xl+l7x+3l\) \(\ge l3-7x+7x+3l=6\)

Vậy GTNN là 6 khi -7/3 <= x <= 7/3 

Ngu Người **** cho tui đi 

BÌNH PHƯƠNG 2 VẾ HAI LẦN => \( \sqrt{(x+1)(x+10)} =-(x+1)\)

vì x> hoặc= -1 thì PT mới có nghĩa nên nghiệm là -1 khi 2 vế bằng 0
 

ĐK : x > = - 1

PT <=> \(\left(x+1\right)+\left(x+10\right)+2\sqrt{x+1}.\sqrt{x+10}=\left(x+2\right)+\left(x+5\right)+2\sqrt{x+2}.\sqrt{x+5}\)

<=> \(2+\sqrt{x+1}.\sqrt{x+10}=\sqrt{x+2}.\sqrt{x+5}\)

<=> \(4+4\sqrt{x+1}.\sqrt{x+10}+\left(x+1\right)\left(x+10\right)=\left(x+2\right)\left(x+5\right)\)

<=> \(4+4\sqrt{x+1}.\sqrt{x+10}+4x=0\)

<=> \(\left(\sqrt{x+1}\right)^2+\sqrt{x+1}.\sqrt{x+10}=0\)

<=> \(\sqrt{x+1}.\left(\sqrt{x+1}+\sqrt{x+10}\right)=0\)

<=> x+ 1 = 0 ( vì \(\sqrt{x+1}+\sqrt{x+10}>0\) với mọi x >=-1)

<=> x = - 1 (t/m)

Vậy...

cx phân tích thành nhân tử nha bạn

dùng hđt đó, sau đó lí luận

nhầm 

\(=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)\)

\(=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(=8\)

ĐK: 4x² - 1 > = 0 ; 2x + 1 > = 0  <=> x >=  -1/2 và x >= 1/2

PT <=> \(2+2x-4x^2=2\left(\sqrt{4x^2-1}-\sqrt{2x+1}\right)\)

<=> \(\left(\sqrt{2x+1}\right)^2-\left(\sqrt{4x^2-1}\right)^2=2\left(\sqrt{4x^2-1}-\sqrt{2x+1}\right)\)

<=> \(\left(\sqrt{2x+1}-\sqrt{4x^2-1}\right).\left(\sqrt{4x^2-1}+\sqrt{2x+1}\right)=2\left(\sqrt{4x^2-1}-\sqrt{2x+1}\right)\)

<=> \(\left(\sqrt{2x+1}-\sqrt{4x^2-1}\right).\left(\sqrt{4x^2-1}+\sqrt{2x+1}+2\right)=0\)

<=> \(\sqrt{2x+1}-\sqrt{4x^2-1}=0\left(Vì\sqrt{4x^2-1}+\sqrt{2x+1}+2>0\right)\)

<=> 2x + 1 = 4x² - 1

<=> (2x + 1).(1 - 2x + 1) = 0 <=> x = -1/2 hoặc x = 1 (chọn)

Vậy...