a, \(\Leftrightarrow2x^2=72\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x=\pm6\)

Vậy ...

\(b,\Leftrightarrow\dfrac{3}{5}x-0,75=2\dfrac{4}{5}.\dfrac{3}{7}=\dfrac{6}{5}\)

\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{6}{5}+0,75=\dfrac{39}{20}\)

\(\Leftrightarrow x=\dfrac{39}{20}:\dfrac{3}{5}=\dfrac{13}{4}\)

Vậy ...

\(c,\Leftrightarrow2x=1\dfrac{5}{6}.\dfrac{6}{11}-\dfrac{3}{10}=\dfrac{7}{10}\)

\(\Leftrightarrow x=\dfrac{7}{10}:2=\dfrac{7}{20}\)

Vậy ...

\(d,\Leftrightarrow\dfrac{1}{x-7\dfrac{1}{3}}=1.5:2\dfrac{1}{4}=\dfrac{2}{3}\)

\(\Leftrightarrow x-7\dfrac{1}{3}=\dfrac{3}{2}\)

\(\Leftrightarrow x=\dfrac{3}{2}+7\dfrac{1}{3}=\dfrac{53}{6}\)

Vậy ...

a) 2x² - 72 = 0

\(\Rightarrow\) 2x² = 72

\(\Rightarrow\) x² = 36 = 6² = (- 6)²

\(\Rightarrow\) x = 6 hoặc x = - 6

Vậy x = 6 hoặc x = - 6

b) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(2\dfrac{4}{5}\)

\(\Rightarrow\)  (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(\dfrac{14}{5}\)

\(\Rightarrow\)  \(\dfrac{3}{5}\)x - \(\dfrac{3}{4}\) = \(\dfrac{6}{5}\)

\(\Rightarrow\) \(\dfrac{3}{5}\)x = \(\dfrac{39}{20}\)

\(\Rightarrow\) x = \(\dfrac{13}{4}\)

Vậy x = \(\dfrac{13}{4}\)

c) \(2x+\dfrac{3}{10}=1\dfrac{5}{6}.\dfrac{6}{11}\)

\(\Rightarrow\) \(2x+\dfrac{3}{10}=\dfrac{11}{6}.\dfrac{6}{11}=1\)

\(\Rightarrow\) \(2x=\dfrac{7}{10}\)

\(\Rightarrow\) \(x=\dfrac{7}{20}\)

Vậy \(x=\dfrac{7}{20}\)

d) \(2\dfrac{1}{4}:\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\Rightarrow\) \(\dfrac{9}{4}:\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}\)

\(\Rightarrow\) \(x-\dfrac{22}{3}=\dfrac{3}{2}\)

\(\Rightarrow\) \(x=\dfrac{53}{6}\)

Vậy \(x=\dfrac{53}{6}\)

1) (x+6)(3x-1)+x+6=0

⇔(x+6)(3x-1)+(x+6)=0

⇔(x+6)(3x-1+1)=0

⇔3x(x+6)=0

2) (x+4)(5x+9)-x-4=0

⇔(x+4)(5x+9)-(x+4)=0

⇔(x+4)(5x+9-1)=0

⇔(x+4)(5x+8)=0

3)(1-x)(5x+3)÷(3x-7)(x-1)

=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)

Nhiều vậy ai làm hết được :P

1)  \(\frac{3x-2}{3}-2=\frac{4x+1}{4}\)

\(\Leftrightarrow\frac{3x-8}{3}=\frac{4x-1}{4}\)

\(\Leftrightarrow4\left(3x-8\right)=3\left(4x-1\right)\)

\(\Leftrightarrow12x-32=12x-3\)(vô lí)

Vậy pt vô nghiệm

P/s: mấy câu sau tương tự thôi mà :)))

nhăm nhe 1 câu thôi 

\(10,\frac{3+5x}{5}-3=\frac{9x-3}{4}\)

\(\Leftrightarrow\frac{3+5x-15}{5}=\frac{9x-3}{4}\)

\(\Leftrightarrow\frac{-12+5x}{5}=\frac{9x-3}{4}\)

\(\Leftrightarrow\left(-12+5x\right)5=\left(9x-3\right)4\)

\(\Leftrightarrow-60+25x=36x-12\)

\(\Leftrightarrow26x-36x=-12+60\)

\(\Leftrightarrow-10x=48\)

\(\Leftrightarrow x=-4,8\)

mọi người giải giúp mình với ạ

1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}.3\)

\(\left|4-2x\right|=1\)

=>\(4-2x=\pm1\)

+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)

\(2x=4-1\) \(2x=4-\left(-1\right)\)

\(2x=3\) \(2x=4+1\)

\(x=3:2\) \(2x=5\)

\(x=1,5\) \(x=5:2\)

Vậy x=1,5 \(x=2,5\)

Vậy x=2,5

2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)

\(9:\left|x+\left(-1\right)\right|=-3\)

\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)

\(\left|x+\left(-1\right)\right|=-3\)

=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )

Vậy x = \(\varnothing\)

3) \(\left|\dfrac{3}{2}-x\right|+\left(-\dfrac{4}{5}\right)=-\dfrac{14}{5}\)

\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)-\left(-\dfrac{4}{5}\right)\)

\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)+\dfrac{4}{5}\)

\(\left|\dfrac{3}{2}-x\right|=-\dfrac{10}{5}\)

\(\left|\dfrac{3}{2}-x\right|=-2\)

=> \(\dfrac{3}{2}-x\) không có giá trị nào thỏa mãn ( Vì giá trị tuyệt đối không thể là số nguyên âm )

Vậy x=\(\varnothing\)