a) \(\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{\frac{37}{42}}{\frac{-37}{28}}=\frac{37}{42}.\frac{28}{-37}=\frac{-2}{3}\)

b) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)

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

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

Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)

a,\(\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}{3.\left(-\frac{1}{3}-\frac{3}{7}+\frac{1}{28}\right)}=\frac{-2}{3}\)

cách này k cần dùng máy tính (hok tốt)

b,\(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{3}\end{cases}}}\)

Vậy....

a) 2x^2 + 3 = 2x(x + 4) - 7

<=> 2x^2 + 3 = 2x^2 + 8x - 7

<=> 2x^2 - 2x^2 - 8x = - 7 - 3

<=> -8x = -10

<=> x = -10/-8 = 5/4

b) 4x^2 - 12x + 5 = 0

<=> 4x^2 - 2x - 10x + 5 = 0

<=> 2x(2x - 1) - 5(2x - 1) = 0

<=> (2x - 5)(2x - 1) = 0

<=> 2x - 5 = 0 hoặc 2x - 1 = 0

<=> x = 5/2 hoặc x = 1/2

c) |5 - 2x| = 1 - x
<=> \(\hept{\begin{cases}5-2x\text{ nếu }5-2x\ge0\Leftrightarrow x\ge\frac{5}{2}\\-\left(5-2x\right)\text{ nếu }5-2x< 0\Leftrightarrow x< \frac{5}{2}\end{cases}}\)

+) nếu x >= 5/2, ta có:

5 - 2x = 1 - x

<=> -2x + 1 = 1 - 5

<=> -x = -4

<=> x = 4 (tm)

+) nếu x < 5/2, ta có:

-(5 - 2x) = 1 - x

<=> -5 + 2x = 1 - x

<=> 2x + 1 = 1 + 5

<=> 3x = 6

<=> x = 2 (ktm)

d) \(\frac{2}{x-1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}-\frac{2x+3}{x^2+x+1}\) ; ĐKXĐ: x # 1 

<=> \(\frac{2}{x-1}=\frac{\left(2x-1\right)\left(2x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x+3}{x^2+x+1}\)

<=> \(\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(2x-1\right)\left(2x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

<=> 2(x^2 + x + 1) = (2x - 1)(2x + 1) - (2x + 3)(x - 1)

<=> 2x^2 + 2x + 2 = 2x^2 - x + 2

<=> 2x^2 - 2x^2 + 2x - x = 2 - 2

<=> x = 0

mạn phép vô đây để kiếm câu trả lời 

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

\(< =>2x^2+3=2x.x+4.2x-7\)

\(< =>2x^2+3=2x^2+8x-7\)

\(< =>2x^2+3-2x^2=8x-7\)

\(< =>\left(2x^2-2x^2\right)-8x=-7-3\)

\(< =>-8x=-10< =>8x=10\)

\(< =>x=10:8=\frac{10}{8}=\frac{5}{4}\)

\(4x^2-12x+5=0\)

\(< =>4x^2-2x-10x+5=0\)

\(< =>2x\left(2x-1\right)-5\left(2x-1\right)=0\)

\(< =>\left(2x-1\right)\left(2x-5\right)=0\)

\(< =>\orbr{\begin{cases}2x-1=0\\2x-5=0\end{cases}}< =>\orbr{\begin{cases}2x=1\\2x=5\end{cases}}\)

\(< =>\orbr{\begin{cases}x=1:2=\frac{1}{2}\\x=5:2=\frac{5}{2}\end{cases}}\)

1) ĐK: \(x\ge0\)

PT \(\Leftrightarrow\frac{2}{3}\sqrt{12x}+\sqrt{12x}-\frac{1}{3}\sqrt{3x}=9\)

\(\Leftrightarrow\frac{5}{3}\sqrt{12x}-\frac{1}{3}\sqrt{3x}=9\)

\(\Leftrightarrow3\sqrt{3x}=9\) \(\Leftrightarrow x=3\left(TM\right)\)

Vậy \(x=3\)

2) ĐK: \(x\ge0\)

PT \(\Leftrightarrow7\sqrt{2x}=14\) \(\Leftrightarrow x=2\left(TM\right)\)

Vậy \(x=2\)