a. Nếu \(x\ge1\)thì: \(\hept{\begin{cases}x+3>0\\x-1\ge0\end{cases}}\)\(\Rightarrow x+3+x-1< 6\Leftrightarrow2x< 4\Leftrightarrow x< 2\)(Loại)

  nếu \(x\le-3\)thì \(\hept{\begin{cases}x+3\le0\\x-1< 0\end{cases}}\)\(\Rightarrow-x-3+1-x< 6\Leftrightarrow-2x< 8\Leftrightarrow x>-4\)\(\Rightarrow-4< x\le-3\)

Nếu \(-3< x< 1\)thì: \(\hept{\begin{cases}x+3>0\\x-1< 0\end{cases}}\)\(\Rightarrow x+3+1-x< 6\Leftrightarrow4< 6\)(luôn đúng)

có ai không giúp dùm với (hu hu hu)

k va kb nha

ĐK: \(x\ge0\)\(4\sqrt{x}-2\sqrt{9x}+16\sqrt{x}=5\)  5  (=) \(\sqrt{x}\left(4-2\sqrt{9}+16\right)=5\) (=) \(\sqrt{x}.14=5\)(=) x=\(\frac{25}{196}\)

ĐK: \(x\ge-5\)PT(=) \(\sqrt{5+x}\left(\sqrt{4}-3+\frac{4}{3}.3\right)=6\) (=) \(\sqrt{5+x}.3=6\) (=)\(\sqrt{5+x}=2\)(=) X = -1 (nhận)

\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)

\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)

\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)

\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)

\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)

Tick nha

2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)

\(\Leftrightarrow6x-18=15-5x\)

\(\Leftrightarrow11x=33\)

hay x=3

ĐK : \(x\ge-\frac{3}{2}\)

\(PT\Leftrightarrow x^2-4x+21-6\sqrt{2x+3}=0\)

\(\Leftrightarrow\left(2x+3-6\sqrt{2x+3}+9\right)+\left(x^2-6x+9\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x+3}-3=0\\x-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x+3=9\\x=3\end{cases}\Rightarrow}x=3\left(TM\right)}\)

Vậy nghiệm của PT là \(x=3\)

\(\Leftrightarrow x+66=0\)

hay x=-66

\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\\ \Rightarrow\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)=\left(\dfrac{x+5}{61}+1\right)+\left(\dfrac{x+7}{59}+1\right)\\ \Rightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\\ \Rightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\\ \Rightarrow x+66=0\\ \Rightarrow x=-66\)

\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\\ \dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\\ \left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\\ x+66=0\\ x=-66\)