(- (x - 3))/2 - 2 = 5(x + 2)/4

=> \(\dfrac{-\left(x-3\right)-4}{2}=\dfrac{5\left(x+2\right)}{4}\)

=> \(\dfrac{-2\left(x-3\right)-8}{4}=\dfrac{5\left(x+2\right)}{4}\)

=. -2x + 6 - 8 = 5x + 10

=> 7x = -12

=> x = -12/7

Các câu còn lại có cách làm tương tự là tính lần lượt trong ngoặc trước, quy đồng về cùng mẫu số để triệt tiêu mẫu và xử lý phần tử số có x như câu đầu tiên em nhé!

Chúc em học vui vẻ nha!

2) Ta có: \(\dfrac{2\left(2x+1\right)}{5}-\dfrac{6+x}{3}=\dfrac{5-4x}{15}\)

\(\Leftrightarrow\dfrac{6\left(2x+1\right)}{15}-\dfrac{5\left(6+x\right)}{15}=\dfrac{5-4x}{15}\)

\(\Leftrightarrow12x+6-30-5x-5+4x=0\)

\(\Leftrightarrow11x-29=0\)

\(\Leftrightarrow x=\dfrac{29}{11}\)

Vậy: \(S=\left\{\dfrac{29}{11}\right\}\)

b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1

=>-3/4x+1/6x=2+5-4/5=24/5

=>x=-288/35

c: =>6x^2+3x-30x-15=6x^2+10x-21x-35

=>-27x-15=-11x-35

=>-16x=-20

=>x=5/4

a)\(9x^2-6x-3=0\)

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

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

\(\Leftrightarrow\)\((3x-1)(x-1)=0\)

\(\Leftrightarrow\)\(\left[\begin{array}{} x=1\\ x=-\dfrac{1}{3} \end{array} \right.\)

a) \(9x^2-6x-3=0\)

\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) \(x^3+9x^2+27x+19=0\)

\(\Leftrightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)

\(\Leftrightarrow x=-1\)( do \(x^2+8x+19=\left(x+4\right)^2+3>0\))

c) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Leftrightarrow x\left(x^2-25\right)-x^3-8=3\)

\(\Leftrightarrow x^3-25x-x^3=8\Leftrightarrow-25x=11\Leftrightarrow x=-\dfrac{11}{25}\)

a) \(9x^2-6x-3=0\\ \Rightarrow\left(9x^2-9x\right)+\left(3x-3\right)=0\\ \Rightarrow9x\left(x-1\right)+3\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(9x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\9x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) \(x^3+9x^2+27x+19=0\\ \Rightarrow\left(x^3+x^2\right)+\left(8x^2+8x\right)+\left(19x+19\right)=0\\ \Rightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x^2+8x+19=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\\left(x^2+8x+16\right)+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\\left(x+4\right)^2+3=0\left(vôlí\right)\end{matrix}\right.\)

a. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$

$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$

$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$

Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$

$\Leftrightarrow \sqrt{x}=\frac{12}{5}$

$\Leftrightarrow x=5,76$ (thỏa mãn)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 2\sqrt{x-2}=8$

$\Leftrightarrow \sqrt{x-2}=4$

$\Leftrightarrow x=4^2+2=18$ (tm)

b. ĐKXĐ: $x^2\geq 5$

PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$

$\Leftrightarrow \sqrt{x^2-5}=0$

$\Leftrightarrow x=\pm \sqrt{5}$

a \(\dfrac{2}{3}x+\dfrac{1}{3}=\dfrac{1}{5}\\ \dfrac{2}{3}x=\dfrac{1}{5}-\dfrac{1}{3}\\ \dfrac{2}{3}x=\dfrac{-2}{15}\\ x=-\dfrac{2}{15}:\dfrac{2}{3}\\ x=-\dfrac{1}{5}\)   b) \(\dfrac{4}{5}-\dfrac{5}{3}x=-2\\ \dfrac{5}{3}x=\dfrac{4}{5}+2\\ \dfrac{5}{3}x=\dfrac{14}{5}\\ x=\dfrac{14}{5}:\dfrac{5}{3}\\ x=\dfrac{42}{25}\)c) \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\\ \dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}\\ \dfrac{5}{3}:x=\dfrac{3}{10}\\ x=\dfrac{5}{3}:\dfrac{3}{10}\\ x=\dfrac{50}{9}\)d) \(\dfrac{5}{7}:x-3=-\dfrac{2}{7}\\ \dfrac{5}{7}:x=3-\dfrac{2}{7}\\ \dfrac{5}{7}:x=\dfrac{19}{7}\\ x=\dfrac{5}{7}:\dfrac{19}{7}\\ x=\dfrac{5}{19}\)

a.

ĐKXĐ: \(x\ge-\dfrac{5}{3}\)

\(9x^2-3x-\left(3x+5\right)-\sqrt{3x+5}=0\)

Đặt \(\sqrt{3x+5}=t\ge0\)

\(\Rightarrow9x^2-3x-t^2-t=0\)

\(\Delta=9+36\left(t^2+t\right)=\left(6t+3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+6t+3}{18}=\dfrac{t+1}{3}\\x=\dfrac{3-6t-3}{18}=-\dfrac{t}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3x-1\\t=-3x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+5}=3x-1\left(x\ge\dfrac{1}{3}\right)\\\sqrt{3x+5}=-3x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+5=9x^2-6x+1\left(x\ge\dfrac{1}{3}\right)\\3x+5=9x^2\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

c.

ĐKXĐ: \(x\ge-5\)

\(x^2-3x+2-x-5-\sqrt{x+5}=0\)

Đặt \(\sqrt{x+5}=t\ge0\)

\(\Rightarrow-t^2-t+x^2-3x+2=0\)

\(\Delta=1+4\left(x^2-3x+2\right)=\left(2x-3\right)^2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1-x\left(x\le1\right)\\\sqrt{x+5}=x-2\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2-2x+1\left(x\le1\right)\\x+5=x^2-4x+4\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-\dfrac{8}{3}\)

\(\left(3x+2\right)^2-6-\sqrt{3x+8}=0\)

Đặt \(\sqrt{3x+8}=t\ge0\Rightarrow3x+2=t^2-6\)

\(\left(t^2-6\right)^2-6-t=0\)

\(\Leftrightarrow t^4-12t^2-t+30=0\)

\(\Leftrightarrow\left(t^2+t-5\right)\left(t^2-t-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+8}=3\\\sqrt{3x+8}=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)