a.

ĐKXĐ: \(x\ge3\)

(Tốt nhất bạn kiểm tra lại đề cái căn đầu tiên của \(\sqrt{x-3}\) là căn bậc 2 hay căn bậc 3). Vì nhìn ĐKXĐ thì thấy căn bậc 2 là không hợp lý rồi đó

Pt tương đương:

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

Do \(x\ge3\Rightarrow x-2>0\Rightarrow\left(x+1\right)\left(x-2\right)>0\)

\(\Rightarrow\sqrt{x-3}+\sqrt[3]{x^2+1}+\left(x+1\right)\left(x-2\right)>0\)

Pt vô nghiệm

b.

ĐKXĐ: \(x\ge-\dfrac{3}{2}\)

Pt: \(2x+3-\sqrt{2x+3}-\left(4x^2-6x+2\right)=0\)

Đặt \(\sqrt{2x+3}=t\ge0\) ta được:

\(t^2-t-\left(4x^2-6x+2\right)=0\)

\(\Delta=1+4\left(4x^2-6x+2\right)=\left(4x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t_1=\dfrac{1+4x-3}{2}=2x-1\\t_2=\dfrac{1-4x+3}{2}=2-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+3}=2x-1\left(x\ge\dfrac{1}{2}\right)\\\sqrt{2x+3}=2-2x\left(x\le1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4x^2-4x+1\left(x\ge\dfrac{1}{2}\right)\\2x+3=4x^2-8x+4\left(x\le1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{17}}{4}\\x=\dfrac{5-\sqrt{21}}{4}\end{matrix}\right.\)

c.

ĐKXĐ: \(x\ge-1\)

\(\Leftrightarrow2\left(x^2-4x+4\right)+2\left(x+1\right)-5\left(x-2\right)\sqrt{x+1}=0\)

\(\Leftrightarrow2\left(x-2\right)^2-5\left(x-2\right)\sqrt{x+1}+2\left(x+1\right)=0\)

Đặt \(\left\{{}\begin{matrix}x-2=a\\\sqrt{x+1}=b\end{matrix}\right.\) ta được:

\(2a^2-5ab+2b^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=x-2\left(x\ge2\right)\\\sqrt{x+1}=2x-4\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-4x+4\\x+1=4x^2-16x+16\end{matrix}\right.\) (\(x\ge2\))

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\) (đã loại nghiệm)

a: ĐKXĐ: x>=-3/2

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

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

=>\(x^2-2x+1=0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1(nhận)

b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))

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

=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>x=4/3(nhận) hoặc x=-2(loại)

c:

Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)

ĐKXĐ: \(x>=-3\)

\(\sqrt{4x+12}=\sqrt{9x+27}-5\)

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

=>\(-\sqrt{x+3}=-5\)

=>x+3=25

=>x=22(nhận)

d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)

=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)

=>\(4x^2-6x+1=4x^2-20x+25\)

=>\(-6x+20x=25-1\)

=>\(14x=24\)

=>x=12/7(nhận)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

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:

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