Ta có: \(x^4-x^2-2mx-m^2=0\)

<=> \(x^4-\left(x+m\right)^2=0\)

<=> \(\left(x^2-x-m\right)\left(x^2+x+m\right)=0\)

<=> \(\orbr{\begin{cases}x^2-x-m=0\left(1\right)\\x^2+x+m=0\left(2\right)\end{cases}}\)

<=> \(\Delta_1=\left(-1\right)^2+4m=4m+1\)

 \(\Delta_2=1^2-4m=1-4m\)

Để pt có 4 nghiệm phân biệt <=> pt (1) và pt (2) cùng có 2 nghiệm pb

<=> \(\hept{\begin{cases}\Delta_1>0\\\Delta_2>0\end{cases}}\) <=> \(\hept{\begin{cases}4m+1>0\\1-4m>0\end{cases}}\) <=> \(\hept{\begin{cases}m>-\frac{1}{4}\\m< \frac{1}{4}\end{cases}}\) <=> \(-\frac{1}{4}< m< \frac{1}{4}\)

Vậy ...

\(\sqrt{18+6\sqrt{5}}+\sqrt{18-6\sqrt{5}}=\sqrt{\sqrt{15}^2+2\sqrt{45}+\sqrt{3}^2}+\sqrt{\sqrt{15}^2-2\sqrt{45}+\sqrt{3}^2}\)

\(=\sqrt{15}+\sqrt{3}+\sqrt{15}-\sqrt{3}\)

\(=2\sqrt{15}\)

2v15 nha

bạn đăng tách câu hỏi ra cho mn cùng giúp nhé 

Bài 4 : 

\(A=3x+\sqrt{16-24x+9x^2}=3x+\sqrt{\left(4-3x\right)^2}=3x+\left|3x-4\right|\)

Thay x = -3 vào A ta được : \(=-9+\left|-13\right|=-9+13=4\)

\(B=5x-\sqrt{4x^2+12x+9}=5x-\sqrt{\left(2x+3\right)^2}=5x-\left|2x+3\right|\)

Thay x = -\(\sqrt{5}\)vào B ta được : \(=-5\sqrt{5}-\left|-2\sqrt{5}+3\right|\)

\(=-5\sqrt{5}+2\sqrt{5}-3=-3\sqrt{5}-3\)

Bài 5 

a, \(\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=3\)

\(VT=9-5-\left(4-3\right)==4-1=3=VP\)

Vậy ta có đpcm 

b, \(2\sqrt{3}\left(\sqrt{3}-1\right)+\left(2-\sqrt{3}\right)^2+6\sqrt{3}=13\)

\(VT=2.3-2\sqrt{3}+4-4\sqrt{3}+3+6\sqrt{3}=6-2\sqrt{3}+7-4\sqrt{3}+6\sqrt{3}\)

\(=13=VP\)Vậy ta có đpcm 

Áp dụng BĐT Cauchy - Shwarz ta có : 

\(B=\frac{1}{a}+\frac{1}{b}\ge\frac{\left(1+1\right)^2}{a+b}=\frac{4}{3}\)

Dấu ''='' xảy ra khi \(a=b=\frac{3}{2}\)

Cauchy Schwarz nhé 

a, \(\sqrt{x^2-2x+1}=5\Leftrightarrow\left|x-1\right|=5\)

TH1 : \(x-1=5\Leftrightarrow x=6\)

TH2 : \(x-1=-5\Leftrightarrow x=-4\)

b, \(\sqrt{9x^2-6x+1}=2x-3\Leftrightarrow\left|3x-1\right|=2x-3\)

ĐK : \(x\ge\frac{3}{2}\)

TH1 : \(3x-1=2x-3\Leftrightarrow x=-2\)( ktm )

TH2 : \(3x-1=3x-2\Leftrightarrow0x=-1\)( ktm )

Vậy pt vô nghiệm 

c, \(\sqrt{9x^2}=2x+1\Leftrightarrow\left|3x\right|=2x+1\)

ĐK : \(x\ge-\frac{1}{2}\)

TH1 : \(3x=2x+1\Leftrightarrow x=1\)

TH2 : \(3x=-2x-1\Leftrightarrow5x=-1\Leftrightarrow x=-\frac{1}{5}\)

\(a,\sqrt{x^2-2x+1}=5\)

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

\(\left|x-1\right|=5\)

\(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}\orbr{\begin{cases}x=6\left(TM\right)\\x=-4\left(TM\right)\end{cases}}}\)

\(b,\sqrt{9x^2-6x+1}=2x-3\)

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

\(\left|3x-1\right|=2x-3\)

\(\orbr{\begin{cases}3x-1=2x-3\\3x-1=3-2x\end{cases}\orbr{\begin{cases}x=-2\left(TM\right)\\x=\frac{4}{5}\left(TM\right)\end{cases}}}\)

\(c,\sqrt{9x^2}=2x+1\)

\(\left|3x\right|=2x+1\)

\(\orbr{\begin{cases}3x=2x+1\\3x=-2x-1\end{cases}\orbr{\begin{cases}x=1\left(TM\right)\\x=-\frac{1}{5}\left(TM\right)\end{cases}}}\)

90

ho tốt

=90 nha

k cho mk vs

hok tốt

\(\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\)

\(=4-\sqrt{7}-2\sqrt{16-7}+4+\sqrt{7}=8-2.3=2\)