VÌ \(\sqrt{x^2+48}-\sqrt{x^2+35}>0\)

=> \(x>\frac{3}{4}\)

Phương trình tương đương

\((x+6-\sqrt{x^2+48})+3\left(x-1\right)+\left(\sqrt{x^2+35}-6\right)=0\)

=> \(\frac{12\left(x-1\right)}{x+6+\sqrt{x^2+48}}+3\left(x-1\right)+\frac{x^2-1}{\sqrt{x^2+35}+6}=0\)

\(\hept{\begin{cases}x=1\\\frac{12}{x+6+\sqrt{x^2+48}}+3+\frac{x+1}{\sqrt{x^2+35}+6}=0\left(2\right)\end{cases}}\)

Phương trình (2) vô nghiệm do x>3/4=> VT>0

\(\sqrt{x^2+48}=4x-3+\sqrt{x^2+35}\Leftrightarrow\sqrt{x^2+48}-7=4x-4+\sqrt{x^2+35}-6\)

\(\Leftrightarrow\frac{x^2+48-49}{\sqrt{x^2+48}+7}=4x-4+\frac{x^2+35-36}{\sqrt{x^2+35}+6}\Leftrightarrow\frac{x^2-1}{\sqrt{x^2+48}+7}=4\left(x-1\right)+\frac{x^2-1}{\sqrt{x^2+35}+6}\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{x+1}{\sqrt{x^2+48}+7}-4-\frac{x+1}{\sqrt{x^2+35}+6}\right)=0\)\(\Leftrightarrow x-1=0\Leftrightarrow x=1\).

a)\(\sqrt{x^2+48}=4x-3+\sqrt{x^2+35}\)

\(\Leftrightarrow\sqrt{x^2+48}-7=4x-4+\sqrt{x^2+35}-6\)

\(\Leftrightarrow\dfrac{x^2+48-49}{\sqrt{x^2+48}+7}=4\left(x-1\right)+\dfrac{x^2+35-36}{\sqrt{x^2+35}+6}\)

\(\Leftrightarrow\dfrac{x^2-1}{\sqrt{x^2+48}+7}-4\left(x-1\right)-\dfrac{x^2-1}{\sqrt{x^2+35}+6}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x+1}{\sqrt{x^2+48}+7}-4-\dfrac{x+1}{\sqrt{x^2+35}+6}\right)=0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

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

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

\(\Leftrightarrow\left(\sqrt{x-1}+1-1\right)\left(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+1\right)+2\sqrt{x-1}-\left(1-x\right)=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+1\right)+2\sqrt{x-1}+x-1=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+3+\sqrt{x-1}\right)=0\)

Dễ thấy: \(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+3+\sqrt{x-1}>0\)

\(\Rightarrow\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)

a) Đk: \(\hept{\begin{cases}x^2-4x+1\ge0\\x+1\ge0\end{cases}}\)

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

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

\(\Leftrightarrow x^2-4x-x=0\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)thỏa mãn điều kiện

Vậy x=0 hoặc x=5

2)\(\sqrt{\left(x-1\right)\left(x-3\right)}+\sqrt{x-1}=0\)(1)

Đk: x>=3 hoặc x=1

pt  (1)<=> \(\sqrt{x-1}\left(\sqrt{x-3}+1\right)=0\)

<=> \(\sqrt{x-1}=0\)(vì\(\sqrt{x-3}+1>0\)mọi x )

<=> x-1=0

<=> x=1 ( thỏa mãn điều kiện)

Bài 1:

ĐK:...........

PT\((1)\Rightarrow x+y+2\sqrt{(x+y)(x-y)}+x-y=16\) (bình phương 2 vế)

\(\Leftrightarrow x+\sqrt{x^2-y^2}=8\)

\(\Leftrightarrow \sqrt{x^2-y^2}=8-x\Rightarrow \left\{\begin{matrix} 8-x\geq 0\\ x^2-y^2=(8-x)^2=x^2-16x+64\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\leq 8\\ y^2=16x-64\end{matrix}\right.\)

Thay vào PT(2) ta có:

\(x^2+16x-64=128\)

\(\Leftrightarrow x^2+16x-192=0\Rightarrow \left[\begin{matrix} x=8\\ x=-24\end{matrix}\right.\)

Nếu \(x=8\Rightarrow y^2=16x-64=64\Rightarrow y=\pm 8\) (thỏa mãn)

Nếu $x=-24\Rightarrow y^2=16x-64< 0$ (vô lý-loại)

Vậy $(x,y)=(8,\pm 8)$

Bài 2:

Ta thấy:

\(x^2-4x+11=(x^2-4x+4)+7=(x-2)^2+7\geq 0, \forall x\)

\(x^4-8x^2+21=(x^4-8x^2+16)+5=(x^2-4)^2+5\geq 5, \forall x\)

Do đó:

\((x^2-4x+11)(x^4-8x^2+21)\geq 7.5=35\)

Dấu "=" xảy ra khi \((x-2)^2=(x^2-4)^2=0\Leftrightarrow x=2\)

Vậy.......