1) ĐKXĐ: \(x\ge-5\)

\(pt\Leftrightarrow x+5=9\Leftrightarrow x=9-5=4\left(tm\right)\)

2) ĐKXĐ: \(x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\sqrt{x-3}=6\)

\(\Leftrightarrow2\sqrt{x-3}=6\Leftrightarrow\sqrt{x-3}=3\)

\(\Leftrightarrow x-3=9\Leftrightarrow x=12\left(tm\right)\)

3) ĐKXĐ: \(x\ge-1\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+1=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\)

tui uk.......u...a

Lmj thấy phương trình nào??

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)

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

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

\(\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\)

Cho mình sửa lại nhé:

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)

\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)

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

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

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

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

đây bạn nếu bạn ko hiểu thì lên mạng gõ cách lm bất phương trình mũ 2

nhows

1c

2c

3a

\(a.Hg\left(NO_3\right)_2-^{t^o}\rightarrow Hg+2NO_2+O_2\\ b.NaNO_3-^{t^o}\rightarrow NaNO_2+\dfrac{1}{2}O_2\\ c.2Zn\left(NO_3\right)_2-^{t^o}\rightarrow2ZnO+O_2+4NO_2\)

\(3x^4+4x^3-3x^2-2x+1=0\)

\(\Leftrightarrow3x^4+x^3-x^2+3x^3+x^2-x-3x^2-x+1=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2+x-1=0\left(1\right)\\3x^2+x-1=0\left(2\right)\end{cases}}\)

\(\Leftrightarrow x_{1,2}=\frac{-1\pm\sqrt{5}}{2}\left(tm\right)\)

\(x_{1,2}=\frac{-1\pm\sqrt{13}}{6}\left(tm\right)\)

\(\sqrt{x-9-6\sqrt{x-9}+9}=2\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-9}-3\right)^2}=2\)

\(\sqrt{x-9}=5\Rightarrow x-9=25\Rightarrow x=34\)

Điều kiện :x>9 phương trình <=> \(x-6\sqrt{x-9}=4=>x-6\sqrt{x-9}=4=>\left(x-9\right)-6\sqrt{x-9}+9=4=>\left(\sqrt{x-9}-3\right)^2=4\)

=>\(\orbr{\begin{cases}\sqrt{x-9}-3=-2\\\sqrt{x-9}-3=2\end{cases}=>\orbr{\begin{cases}\sqrt{x-9}=1\\\sqrt{x-9=5}\end{cases}=>\orbr{\begin{cases}x=10\\x=34\end{cases}}}}\)

câu trả lời của Nguyễn Trọng Kiên hình như sai rùi . Nó ra có 1 trường hợp là:

=>\(\hept{\begin{cases}5>0\\x-9=25\end{cases}}\)

ra vậy mới đúng chứ

ĐKXĐ:  \(x\ge1\)

\(\Rightarrow\left(\sqrt{x-1}+\sqrt{2x+1}\right)^2=1\Leftrightarrow x-1+2x+1+2\sqrt{\left(x-1\right)\left(2x+1\right)}=1\Leftrightarrow3x+2\sqrt{2x^2-x-1}=1\) \(\Leftrightarrow2\sqrt{2x^2-x-1}=1-3x\Rightarrow\left(2\sqrt{2x^2-x-1}\right)^2=\left(1-3x\right)^2\Leftrightarrow8x^2-4x-4=9x^2-6x+1\) \(\Leftrightarrow x^2-2x+5=0\Leftrightarrow\left(x-1\right)^2+4=0\Leftrightarrow\left(x-1\right)^2=-4\) vô lí vì VT\(\ge0\) mà VP<0 \(\Rightarrow\) ko có x Vậy...