1.

ĐKXĐ: \(x\ge\dfrac{3+\sqrt{41}}{4}\)

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

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x}=a>0\\\sqrt{x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow a^2-3b^2-2ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(a-3b\right)=0\)

\(\Leftrightarrow a=3b\)

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

\(\Leftrightarrow x^2-x=9\left(x+1\right)\)

\(\Leftrightarrow...\) (bạn tự hoàn thành nhé)

2.

ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{x+1}=a\ge0\) pt trở thành:

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=x\left(x\ge0\right)\\2\sqrt{x+1}=-x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=x+1\\x^2=4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2-4x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=2-2\sqrt{2}\end{matrix}\right.\)

ĐK: \(-1\le x\le4\)

\(\sqrt{x+1}+\sqrt{4-x}=t\left(\sqrt{5}\le t\le\sqrt{10}\right)\Rightarrow\sqrt{-x^2+3x+4}=\dfrac{t^2-5}{2}\)

\(pt\Leftrightarrow t+\dfrac{t^2-5}{2}=5\)

\(\Leftrightarrow t^2+2t-15=0\)

\(\Leftrightarrow\left(t-3\right)\left(t+5\right)=0\)

\(\Leftrightarrow t=3\left(\text{Vì }\sqrt{5}\le t\le\sqrt{10}\right)\)

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

\(\Leftrightarrow5+2\sqrt{-x^2+3x+4}=9\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2-x+1=x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\1-x=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

b.

ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-3\end{matrix}\right.\)

Do \(\left\{{}\begin{matrix}\sqrt{x^2-3x+2}\ge0\\\sqrt{x^2+x-6}\ge0\end{matrix}\right.\) với mọi x thuộc TXĐ

\(\Rightarrow\sqrt{x^2-3x+2}+\sqrt{x^2+x-6}\ge0\)

Đẳng thức xảy ra khi:

\(\left\{{}\begin{matrix}x^2-3x+2=0\\x^2+x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x=2\) (thỏa mãn ĐKXĐ)

Vậy pt có nghiệm duy nhất \(x=2\)

c.

Với \(x< 1\Rightarrow\left\{{}\begin{matrix}x-1< 0\\\sqrt{x^4-2x^2+1}\ge0\end{matrix}\right.\) phương trình vô nghiệm

Với \(x\ge1\) pt tương đương:

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

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

\(\Leftrightarrow x^2-1=x-1\) (do \(x\ge1\Rightarrow x^2-1\ge0\Rightarrow\left|x^2-1\right|=x-1\))

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0< 1\left(loại\right)\\x=1\end{matrix}\right.\)

\(a,PT\Leftrightarrow x^2-3x+2+x^2-x\sqrt{3x-2}=0\left(x\ge\dfrac{2}{3}\right)\\ \Leftrightarrow\left(x^2-3x+2\right)+\dfrac{x\left(x^2-3x+2\right)}{x+\sqrt{3x-2}}=0\\ \Leftrightarrow\left(x^2-3x+2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\)

Vì \(x\ge\dfrac{2}{3}>0\Leftrightarrow1+\dfrac{x}{x+\sqrt{3x-2}}>0\)

Do đó \(x\in\left\{1;2\right\}\)

\(b,ĐK:0\le x\le4\\ PT\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{4-x}\\ \Leftrightarrow x-4\sqrt{x}+4=-\sqrt{4-x}\\ \Leftrightarrow\left(\sqrt{x}-2\right)^2=-\sqrt{4-x}\)

Vì \(VT\ge0\ge VP\Leftrightarrow VT=VP=0\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{4-x}=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)

Vậy PT có nghiệm \(x=4\)

Câu 4:

Giả sử điều cần chứng minh là đúng

\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:

\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)

Vậy điều cần chứng minh là đúng

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

⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)

⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)

⇔ x = 5

Vậy S = {5}

Bài 1:

ĐKĐB suy ra $x(x+1)+y(y+1)=3x^2+xy-4x+2y+2$

$\Leftrightarrow 2x^2+x(y-5)+(y-y^2+2)=0$

Coi đây là PT bậc 2 ẩn $x$

$\Delta=(y-5)^2-4(y-y^2+2)=(3y-3)^2$Do đó:

$x=\frac{y+1}{2}$ hoặc $x=2-y$. Thay vào một trong 2 phương trình ban đầu ta thu được:

$(x,y)=(\frac{-4}{5}, \frac{-13}{5}); (1,1)$