\(a,3\sqrt{x}-7=0\left(dk:x\ge0\right)\\ \Leftrightarrow3\sqrt{x}=7\\ \Leftrightarrow\sqrt{x}=\dfrac{7}{3}\\ \Leftrightarrow x=\dfrac{49}{9}\left(tmdk\right)\)

Vậy \(S=\left\{\dfrac{49}{9}\right\}\)

\(b,\sqrt{x-2}+\sqrt{4x-8}=3\left(dk:x\ge2\right)\\ \Leftrightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}=3\\ \Leftrightarrow\sqrt{x-2}+2\sqrt{x-2}=3\\ \Leftrightarrow3\sqrt{x-2}=3\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\\ \Leftrightarrow x=3\left(tmdk\right)\)

Vậy \(S=\left\{3\right\}\)

a: =>3*căn x=7

=>căn x=7/3

=>x=49/9

b: =>3*căn x-2=3

=>căn x-2=1

=>x-2=1

=>x=3

a) x² - 4x = 0

<=> x(x - 4) = 0

<=> x = 0 hoặc x - 4 = 0

<=> x = 0 hoặc x = 4

b) 4x - (3 + 5x) = 14

<=> 4x - 3 - 5x = 14

<=> - x = 14 +3

<=> - x = 17

=> x = - 17

c)\(\frac{-7^5.3^3}{7^2.\left(-3\right)^3}=\frac{-7^5.3^3}{-7^2.3^3}=\frac{7^5}{7^2}=7^3=343\)

a: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-4x+4}=7\)

=>\(\sqrt{\left(x-2\right)^2}=7\)

=>|x-2|=7

=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)

b: ĐKXĐ: x>=-3

\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)

=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)

=>\(3\sqrt{x+3}=6\)

=>\(\sqrt{x+3}=2\)

=>x+3=4

=>x=1(nhận)

Nhanh giúp mk nha cảm ơn các bạn nhiều

Xét : \(x>1\) ta có :

\(\left(4x+3\right)-\left(x-1\right)=7\Leftrightarrow3x+4=7\Rightarrow x=1\) (loại)

Xét \(-\frac{3}{4}\le x\le1\) ta có :

\(\left(4x+3\right)-\left(1-x\right)=7\Leftrightarrow5x+2=7\Rightarrow x=1\) (TM)

Xét \(x< -\frac{3}{4}\) ta có :

\(\left(-4x-3\right)-\left(1-x\right)=7\Leftrightarrow-3x-4=7\Rightarrow x=-\frac{11}{3}\) (TM)

Vậy \(x=\left\{-\frac{11}{3};1\right\}\)

\(4x^2-9+\left(2x-3\right)\left(x+7\right)=0\)

\(\Rightarrow\left(4x^2-9\right)+\left(2x-3\right)\left(x+7\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(2x+3\right)+\left(2x-3\right)\left(x+7\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(2x+3+x+7\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(3x+10\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\3x+10=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

1) \(\Rightarrow x^2+4x+4-x^2+1=9\)

\(\Rightarrow4x=4\Rightarrow x=1\)

2) \(\Rightarrow x\left(2x+7\right)+2\left(2x+7\right)=0\)

\(\Rightarrow\left(2x+7\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-2\end{matrix}\right.\)

3) \(\Rightarrow x^3+3x^2+3x+1-x^3-3x^2=2\)

\(\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\)