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

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

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

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

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

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

Pt \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}=0\) vô n_o

(vì \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}< \dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\forall x\ge2\Rightarrow VT< 0\))

=> x - 2 = 0

<=> x = 2 (nhận)

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

\(\Leftrightarrow\dfrac{\left(4x+1\right)-\left(3x-2\right)}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{x+3}{5}=0\)

\(\Leftrightarrow\dfrac{x+3}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{x+3}{5}=0\)

\(\Leftrightarrow\left(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}\right)\left(x+3\right)=0\)

TH1:

x + 3 = 0

<=> x = - 3 (loại)

TH2:

\(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}=0\)

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

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

\(\Leftrightarrow\dfrac{4x+1-9}{\sqrt{4x+1}+3}+\dfrac{3x-2-4}{\sqrt{3x-2}+2}=0\)

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

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

Pt \(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}>0\forall x\ge\dfrac{2}{3}\) => vô n_o

=> x - 2 = 0

<=> x = 2 (nhận)

~ ~ ~

Vậy x = 2

\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)

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

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

\(\Leftrightarrow\sqrt{2\left(x+3\right)\left(x+1\right)}-\dfrac{\left(x+1\right)\left(3x+5\right)}{\sqrt{\left(x-1\right)\left(x+1\right)}+2\left(x+1\right)}=0\)

\(\Leftrightarrow\sqrt{x+1}\left[2\sqrt{x+3}-\dfrac{\sqrt{x+1}\left(3x+5\right)}{\sqrt{x+1}\left(\sqrt{x-1}+2\sqrt{x+1}\right)}\right]=0\)

\(\Leftrightarrow\sqrt{x+1}\left[2\sqrt{x+3}-\dfrac{3x+5}{\sqrt{x-1}+2\sqrt{x+1}}\right]=0\)

TH1

x + 1 = 0

<=> x = - 1 (loại)

TH2

\(2\sqrt{x+3}-\dfrac{3x+5}{\sqrt{x-1}+2\sqrt{x+1}}=0\)

mà \(2\sqrt{x+3}=\dfrac{4x+12}{2\sqrt{x+3}}>\dfrac{3x+5}{\sqrt{x-1}+2\sqrt{x+1}}\forall x\ge1\)

=> VT > 0

=> vô n_o

~ ~ ~

Vậy pt vô n_o

.

a)Dat \(x^2-4x+3=a;x^2-7x+6=b \Rightarrow a+b=2x^2-11x+9\)

....

a: Ta có: \(6-4x=5(x+3)+3\)

\(\Leftrightarrow6-4x-5x-12-3=0\)

\(\Leftrightarrow-9x=9\)

hay x=-1

b: Ta có: \(\dfrac{x+3}{2}-1=\dfrac{x-1}{3}+\dfrac{x+5}{6}\)

\(\Leftrightarrow15x+45-30=10x-30+5x+25\)

\(\Leftrightarrow15=-5\left(loại\right)\)

c: Ta có: \(\left(x-2\right)\left(2x+1\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow2\left(x-2\right)\left(x-1\right)=0\)

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

d: Ta có: \(\dfrac{2}{x^2-2x}+\dfrac{1}{x}=\dfrac{x+2}{x-2}\)

\(\Leftrightarrow2+x-2=x^2+2x\)

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

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

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

để mk làm cho ; bài này dùng liên hợp

pt<=> \(x+1-\sqrt{x^2-2x+5}+2x+4-2\sqrt{4x+5}+x^3-2x^2+2x-1=0\) ( ĐKXĐ: \(x\ge-\frac{5}{4}\))

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

<=>: \(\frac{x^2+2x+1-x^2+2x-5}{x+1+\sqrt{x^2-2x+5}}+\frac{4x^2+16x+16-16x-20}{2x+4+2\sqrt{4x+5}}+\left(x-1\right)\left(x^2-x+1\right)=0\)

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

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

<=> x=1 ( vì \(x\ge-\frac{5}{4}\)nên cái trong ngoặc thứ 2 khác 0)

vậy x=1 

Bài 2:
\(199-2x-x^2=200-(x^2+2x+1)=200-(x+1)^2\leq 200, \forall x\in\mathbb{Z}\)

\(\Rightarrow 4y^2=2+\sqrt{199-2x-x^2}\leq 2+\sqrt{200}\)

\(\Leftrightarrow y^2\leq \frac{2+\sqrt{200}}{4}< 9\)

\(\Rightarrow -3< y< 3\). Mà $y$ nguyên nên $y\in\left\{-2;-1;0;1;2\right\}$

Thay từng giá trị của $y$ vào PT ban đầu ta tìm được các cặp $(x,y)$ sau:

$(x,y)=(1,\pm 2); (-3,\pm 2); (13,\pm 1); (-15,\pm 1)$

Bài 1:

a) ĐKXĐ: \(x\geq \frac{-3}{2}\)

PT \(\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\)

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

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

Vì $(x+1)^2\geq 0; (\sqrt{2x+3}-1)^2\geq 0$ với mọi $x\geq \frac{-3}{2}$ nên để tổng của chúng bằng $0$ thì $(x+1)^2=(\sqrt{2x+3}-1)^2=0$

$\Leftrightarrow x=-1$

Vậy $x=-1$

b) ĐKXĐ: \(x^2-4x-8\geq 0\)

PT \(\Leftrightarrow 2(x^2-4x-8)-3\sqrt{x^2-4x-8}=2\)

Đặt \(\sqrt{x^2-4x-8}=a(a\geq 0)\) thì PT trở thành:

\(2a^2-3a=2\)

\(\Leftrightarrow 2a^2-3a-2=0\Leftrightarrow (a-2)(2a+1)=0\)

\(\Rightarrow a=2\) (do $a\geq 0$)

\(\Leftrightarrow x^2-4x-8=4\)

\(\Leftrightarrow x^2-4x-12=0\Leftrightarrow \left[\begin{matrix} x=6\\ x=-2\end{matrix}\right.\) (đều thỏa mãn)

Bài 2:
\(199-2x-x^2=200-(x^2+2x+1)=200-(x+1)^2\leq 200, \forall x\in\mathbb{Z}\)

\(\Rightarrow 4y^2=2+\sqrt{199-2x-x^2}\leq 2+\sqrt{200}\)

\(\Leftrightarrow y^2\leq \frac{2+\sqrt{200}}{4}< 9\)

\(\Rightarrow -3< y< 3\). Mà $y$ nguyên nên $y\in\left\{-2;-1;0;1;2\right\}$

Thay từng giá trị của $y$ vào PT ban đầu ta tìm được các cặp $(x,y)$ sau:

$(x,y)=(1,\pm 2); (-3,\pm 2); (13,\pm 1); (-15,\pm 1)$