Lời giải:

Nhớ không nhầm thì bạn đã đăng bài này rồi mà.

\(2\sqrt{2}m-\sqrt{2}-2m+1=3-m\)

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

\(\Leftrightarrow m(2\sqrt{2}-1)=2+\sqrt{2}\Rightarrow m=\frac{2+\sqrt{2}}{2\sqrt{2}-1}=\frac{6+5\sqrt{2}}{7}\)

Lời giải:
ĐKXĐ: $m\neq \frac{1}{2}$

Từ PT $\sqrt{2}-1=\frac{3-m}{2m-1}\Rightarrow (\sqrt{2}-1)(2m-1)=3-m$

$\Leftrightarrow 2+\sqrt{2}=m(2\sqrt{2}-1)$

$\Leftrightarrow m=\frac{2+\sqrt{2}}{2\sqrt{2}-1}=\frac{6+5\sqrt{2}}{7}$ (thỏa mãn)

Vậy...

\(2m^2-1=0\)

\(\Leftrightarrow m^2=\frac{1}{2}\)

\(\Leftrightarrow m=+\sqrt{\frac{1}{2}}\)

\(2m^2-9=0\)

\(\Leftrightarrow m^2=\frac{9}{2}\)

\(m=\pm\sqrt{\frac{9}{2}}\)

Vậy ....

a) \(2m^2-1=0\)

\(\Leftrightarrow2m^2=1\)

\(\Leftrightarrow m^2=\frac{1}{2}\)

\(\Leftrightarrow m=\sqrt{\frac{1}{2}}\)

b) \(2m^2-9=0\)

\(\Leftrightarrow2m^2=9\)

\(\Leftrightarrow m^2=\frac{9}{2}\)

\(\Leftrightarrow m=\sqrt{\frac{9}{2}}\)

bổ sung nha

Do m là mũ 2

a) \(m=\pm\sqrt{\frac{1}{2}}\)

b) \(m=\pm\sqrt{\frac{9}{2}}\)

ĐKXĐ: ...

\(\Leftrightarrow\left(\frac{x^2}{x+1}\right)^2+\frac{x^2}{x+1}-12=0\)

Đặt \(\frac{x^2}{x+1}=t\)

\(\Rightarrow t^2+t-12=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{x^2}{x+1}=3\\\frac{x^2}{x+1}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-3x-3=0\\x^2+4x+4=0\end{matrix}\right.\) \(\Rightarrow\) bấm casio

a/ Gọi pt đường thẳng có dạng \(y=ax+b\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=1\\2a+b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-2\\b=3\end{matrix}\right.\) \(\Rightarrow y=-2x+3\)

b/ Do đường thẳng d đi qua C và song song AB nên:

\(\left\{{}\begin{matrix}\left(m^2-3m\right).0+m^2-2m+2=2\\m^2-3m=-2\\m^2-2m+2\ne3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m^2-2m=0\\m^2-3m+2=0\\m^2-2m-1\ne0\end{matrix}\right.\) \(\Rightarrow m=2\)

ĐKXĐ : x> -2

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

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

=> 2x+\(\sqrt{6x^2+1}\) = x²+2x+1

=> \(\sqrt{6x^2+1}\) = x²+1

=> 6x² +1 = (x²+1)(x²+1)

=> 6x² +1 = x⁴+2x²+1

=> -x⁴+4x² = 0

=> x²(4-x²) = 0

=>x²(2-x)(2+x) = 0

=> x² =0, 2-x=0 , 2+x =0

=> x=0(TMĐKXĐ)

x=2(TMĐKXĐ)

x= -2 (KTMĐKXĐ)

Vậy ........

\(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

\(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(ĐK:x\ge3\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(c\right)\\x=6\left(c\right)\end{matrix}\right.\)

Vậy nghiệm của phương trình là \(S=\left\{3;6\right\}\)