\(a,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\\ \Leftrightarrow-2\sqrt{x-1}=-2\Leftrightarrow\sqrt{x-1}=1\\ \Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{1}{3}\sqrt{2x}-2\sqrt{2x}+3\sqrt{2x}=12\\ \Leftrightarrow\dfrac{4}{3}\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=9\\ \Leftrightarrow2x=81\Leftrightarrow x=\dfrac{81}{2}\left(tm\right)\)

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

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

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

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

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

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)

TH1 : \(x\le-3\) ( LĐ )

TH2 : \(x\ge0\)

BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)

\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)

\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge0\)

Vậy \(S=R/\left(-3;0\right)\)

Bài 4:

Để hai đường song song thì 2m-1=5

=>2m=6

=>m=3

Bài 3:

a: 4x-3y=2 và 4x+3y=-18

=>8x=-16 và 4x-3y=2

=>x=-2 và 3y=4x-2=4*(-2)-2=-10

=>x=-2; y=-10/3

b:\(A=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\left(x+16\right)^2}{\left(x-16\right)\left(\sqrt{x}+2\right)}\)

a. ĐKXĐ: \(x\ge-\frac{10}{3}\) 

Điều kiện có nghiệm : \(x^2+9x+20\ge0\Leftrightarrow\orbr{\begin{cases}x\ge-4\\x\le-5\end{cases}}\)

Kết hợp ta có điều kiện \(x\ge-\frac{10}{3}.\)

Từ phương trình ta có: \(x^2+9x+18=2\left(\sqrt{3x+10}-1\right)\)

\(\Leftrightarrow\left(x+3\right)\left(x+6\right)=2.\frac{3x+9}{\sqrt{3x+10}+1}\)

\(\Leftrightarrow\left(x+3\right)\left(x+6\right)=\frac{6\left(x+3\right)}{\sqrt{3x+10}+1}\)

\(\Leftrightarrow\left(x+3\right)\left(x+6-\frac{6}{\sqrt{3x+10}+1}\right)=0\)

TH1: x = - 3 (tm)

Th2: \(x+6-\frac{6}{\sqrt{3x+10}+1}=0\)

\(\Leftrightarrow\left(x+6\right)\sqrt{3x+10}+x+6-6=0\)

\(\Leftrightarrow\left(x+6\right)\sqrt{3x+10}+x=0\)

Đặt \(\sqrt{3x+10}=t\Rightarrow x=\frac{t^2-10}{3}\)

Vậy thì \(\left(\frac{t^2-10}{3}+6\right)t+\frac{t^2-10}{3}=0\)

\(\Leftrightarrow\frac{t^3+8t}{3}+\frac{t^2-10}{3}=0\Leftrightarrow t^3+t^2+8t-10=0\Leftrightarrow t=1\Leftrightarrow x=-3\left(tm\right).\)

Vậy pt có 1 nghiệm duy nhất x = - 3.

b. Nhân 2 vào hai vế của phương trình thứ nhất rồi trừ từng vế cho phương trình thứ hai, ta được:

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

\(\Leftrightarrow2x^2y^2-2x^2-y^3+2y^2-3=0\)

\(\Leftrightarrow2x^2\left(y^2-1\right)-\left(y+1\right)\left(y^2-3y+3\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(2x^2y-2x^2-y^2+3y-3\right)=0\)

Với y = - 1 ta có \(x^2-2x+1=0\Leftrightarrow x=1.\)

Với \(\left(2x^2+3\right)y-\left(2x^2+3\right)-y^2=0\Leftrightarrow\left(2x^2+3\right)\left(y-1\right)=y^2\)

\(\Rightarrow\frac{y^2}{y-1}-4x=-y^3\Rightarrow x=\frac{y^4-y^3+y^2}{4\left(y-1\right)}\)

Thế vào pt (1) : Vô nghiệm.

Vậy (x; y) = (1; -1)

Thank you bạn nha