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

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

c, \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

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

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

\(=\sqrt{9-\left(5+\sqrt{2}\right)}=\sqrt{9-5-\sqrt{2}}=\sqrt{4-\sqrt{2}}\)

\(\sqrt{2x-2+2\sqrt{2x-3}+}+\sqrt{2x+13+8\sqrt{2x-3}}=7\)          đkxđ \(x\ge\frac{3}{2}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=7\)

\(\Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=7\)

\(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-3}+4=7\)

\(\Leftrightarrow2\sqrt{2x-3}=2\)

\(\Leftrightarrow\sqrt{2x-3}=1\)

\(\Leftrightarrow2x-3=1\)

\(\Leftrightarrow x=2\)(chọn)

KL vậy x=2 là ngiệm của phương trình

Đặt \(y=\sqrt{2x-3}\left(y\ge0\right)\Rightarrow x=\frac{y^2+3}{2}\)

\(pt\Leftrightarrow\sqrt{y^2+2y+1}+\sqrt{y^2+8y+16}=7\)

\(\Leftrightarrow\sqrt{\left(y+1\right)^2}+\sqrt{\left(y+4\right)^2}=7\Leftrightarrow\left|y+1\right|+\left|y+4\right|=7\)

=> y=1 hay 2x-3 =1 => x=2

Vậy pt có nghiệm x=2

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

\(ĐKXĐ:\orbr{\begin{cases}x\ge1\\x\le-2\end{cases}}\)hoặc x=0

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

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

\(x^2-x+x^2+2x+2x\sqrt{x^2+x-2}=4x^2\)

\(2x\sqrt{x^2+x-2}=2x^2-x\)

\(2x\sqrt{x^2+x-2}=2x^2-x\)

\(ĐXKĐ:x\ge\frac{1}{2}\)

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

\(4x^4+4x^3-8x^2=4x^4-4x^3+x^2\)

\(8x^3-9x^2=0\)

\(x^2\left(8x-9\right)=0\)

\(\orbr{\begin{cases}x^2=0\\8x-9=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=\frac{9}{8}\left(TM\right)\end{cases}}}\)

\(\)

Mình chỉ lo bạn không dịch được chữ =)

Cho \(f\left(x\right)=mx-2\) và \(g\left(x\right)=\left(m^2+1\right)x+5\)

\(f\left(x\right)+g\left(x\right)=mx-2+m^2x+x+5\)

\(=mx+m^2x+x+3\)

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

Ta có: \(m^2+m+1=\left(m+1\right)^2+\frac{3}{4}>0\forall m\)

\(\Rightarrow f\left(x\right)+g\left(x\right)\) là hàm số bậc nhất đồng biến

\(f\left(x\right)-g\left(x\right)=mx-2-m^2x-x-5\)

\(=x\left(m-m^2-1\right)-5\)

Ta có: \(-m^2+m-1=-m^2+2.\frac{1}{2}m-\frac{1}{4}-\frac{3}{4}=-\left(m-\frac{1}{2}\right)^2-\frac{3}{4}< 0\forall m\)

\(\Rightarrow f\left(x\right)-g\left(x\right)\) là hàm số bậc nhất nghịch biến

Sai thông cảm :>

\(b,x^2+2x-\sqrt{x^2+2x+1}-5=0\)

\(\Leftrightarrow x^2+2x+1-\sqrt{x^2+2x+1}-6=0\)

đặt \(\sqrt{x^2+2x+1}=a\left(a\ge0\right)\) pt trở thành :

\(a^2-a-6=0\Leftrightarrow\left(a-3\right)\left(a+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-2\left(loai\right)\end{cases}}\)

với a = 3 ta có \(\sqrt{x^2+2x+1}=3\Leftrightarrow x^2+2x+1=9\)

\(\Leftrightarrow x^2+2x-8=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=-4\\x=2\end{cases}}\)

a. \(\sqrt{\frac{1}{4}x^2+\frac{1}{4}x+1}-\sqrt{6-2\sqrt{5}}=0\)

\(\Leftrightarrow\sqrt{\frac{1}{4}x^2+\frac{1}{4}x+1}=\sqrt{6-2\sqrt{5}}\)

\(\Leftrightarrow\frac{1}{4}x^2+\frac{1}{4}x+1=6-2\sqrt{5}\)

\(\Leftrightarrow x^2+x+4=24-8\sqrt{5}\)

\(\Leftrightarrow x^2+x-20-8\sqrt{5}=0\)

\(\Delta=b^2-4ac=1^2-4\cdot\left(-20-8\sqrt{5}\right)=1+80+32\sqrt{5}=81+32\sqrt{5}>0\)

\(\Rightarrow x=\frac{-1\pm\sqrt{81+32\sqrt{5}}}{2}\)

đặt \(\hept{\begin{cases}\sqrt{4x^2+2x+3}=a\\\sqrt{x^2+1}=b\end{cases}\left(a;b\ge0\right)}\)

ta có : \(a^2-4b^2=4x^2+2x+3-4x^2-4=2x-1\)

pt trở thành \(a-2b=2\left(a^2-4b^2\right)\)

\(\Leftrightarrow2\left(a-2b\right)\left(a+2b\right)-\left(a-2b\right)=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a+4b-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=2b\\2a+4b=1\end{cases}}\)

với a = 2b ta có : \(\sqrt{4x^2+2x+3}=2\sqrt{x^2+1}\)

\(\Leftrightarrow4x^2+2x+3=4x^2+4\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

với 2a + 4b = 1 ta có \(2\sqrt{4x^2+2x+3}+4\sqrt{x^2+1}=1\)

có \(x^2+1\ge1\Rightarrow4\sqrt{x^2+1}\ge4>1\) và \(2\sqrt{4x^2+2x+3}\ge0\)

\(\Rightarrow VT>1\)

\(\Rightarrow pt.vo.nghiem\)

\(\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}+\frac{2+6\sqrt{3}}{\sqrt{3}}-\frac{13}{\sqrt{3}+4}\)

\(=\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{3}\left(\sqrt{3}+6\right)}{\sqrt{3}}-\frac{13\left(4-\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}\)

\(=-\sqrt{3}+\sqrt{3}+6-\frac{52-13\sqrt{3}}{16-3}\)

\(=6-\frac{52-13\sqrt{3}}{13}=\frac{78-52-13\sqrt{3}}{13}=\frac{26-13\sqrt{3}}{13}\)

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

ĐKXĐ: \(x\ge0;x\ne1;x\ne4\)

\(A=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{8\sqrt{x}}{x-1}\right):\frac{4\sqrt{x}-8}{1-x}\)

\(A=\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\frac{1-x}{4\sqrt{x}-8}\)

\(A=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{4\sqrt{x}-8}\)

\(A=\frac{-4\sqrt{x}}{4\sqrt{x}-8}=\frac{\sqrt{x}}{2-\sqrt{x}}\)