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

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

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

\(\Leftrightarrow-x=1\)

\(\Leftrightarrow x=-1\)

a/

⇔ (1−\(\sqrt{2}\)) (x² - 2x +3) = 0 (Đặt nhân tử chung)

⇔ 1- \(\sqrt{2}\) = 0 và x² -2x +3 = 0

b) nhân 6 với \(\sqrt{2}\)+1 là ra phương trình bậc 2

1. ĐKXĐ: $x\leq \frac{1}{2}$
PT \(\Leftrightarrow [(x^2-2)-(x-\sqrt{2})]\sqrt{1-2x}=0\)

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

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

Kết hợp đkxđ suy ra \(\left[\begin{matrix} x=1-\sqrt{2}\\ x=\frac{1}{2}\end{matrix}\right.\)

2. ĐKXĐ: $-1\leq x\leq 1$

Đặt $\sqrt{1+x}=a; \sqrt{1-x}=b(a,b\geq 0)$. Khi đó ta có:

$4a-\frac{a^2+b^2}{2}=\frac{3(a^2-b^2)}{2}+2b+ab=0$

$\Leftrightarrow 2a^2-b^2+ab-4a+2b=0$

$\Leftrightarrow (a+b-2)(2a-b)=0$

Xét 2 TH:

TH1: $a+b-2=0$

$\Leftrightarrow \sqrt{1-x}+\sqrt{1+x}=2$

$\Leftrightarrow 2+2\sqrt{1-x^2}=4$
$\Leftrightarrow \sqrt{1-x^2}=1$

$\Leftrightarrow x=0$ (tm)

TH2: $2a-b=0$

$\Leftrightarrow 2\sqrt{1+x}=\sqrt{1-x}$

$\Leftrightarrow 4(x+1)=1-x$

$\Leftrightarrow x=\frac{-3}{5}$ (tm)

Vậy.........

T sợ chỉ dám liên hợp thôi, nhường cách bình phương cho 1 ng` chăm chỉ :(

\(pt\Leftrightarrow6x+3x\sqrt{9x^2+3}+4x+2+\left(4x+2\right)\sqrt{x^2+x+1}=0\)

\(\Leftrightarrow2\left(5x+1\right)+\left(3x\sqrt{9x^2+3}+\dfrac{6\sqrt{21}}{25}\right)+\left(\left(4x+2\right)\sqrt{x^2+x+1}-\dfrac{6\sqrt{21}}{25}\right)=0\)

\(\Leftrightarrow2\left(5x+1\right)+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(5x+1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+1\right)\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}=0\)

\(\Leftrightarrow\left(5x+1\right)\left(2+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}\right)=0\)

\(\Rightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)

a/ ĐKXĐ: \(x\ge\frac{3}{4}\)

\(\Leftrightarrow6x+1+2\sqrt{5x^2+5x}=6x+1+2\sqrt{8x^2+10x-12}\)

\(\Leftrightarrow\sqrt{5x^2+5x}=\sqrt{8x^2+10x-12}\)

\(\Leftrightarrow5x^2+5x=8x^2+10x-12\)

\(\Leftrightarrow3x^2+5x-12=0\Rightarrow\left[{}\begin{matrix}x=-3< \frac{3}{4}\left(l\right)\\x=\frac{4}{3}\end{matrix}\right.\)

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

Đặt \(\sqrt{x^2+x+1}=t>0\)

\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)

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

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