a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)

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

\(x=\dfrac{1}{2}+\dfrac{1}{3}\)

\(x=\dfrac{1}{5}\)

a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)

<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)

<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)

<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)

<=>\(4x-17=0\)

<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)

b) =>\(\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\3-\dfrac{1}{2}x=0\\1+\dfrac{3}{4}:x=0\end{matrix}\right.=>\left\{{}\begin{matrix}3x=\dfrac{1}{2}\\\dfrac{1}{2}x=3\\\dfrac{3}{4x}=-1\end{matrix}\right.=>\left\{{}\begin{matrix}x=\dfrac{1}{6}\\x=6\\x=\dfrac{-3}{4}\end{matrix}\right.\)

a/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

a) (x-2)³+6(x+1)²-x³+12=0

⇒ x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

⇒ x³-6x²+12x-8+6x²+12x+6-x³+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

a.

PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)

b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.

c.

PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$

$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý) 

Do đó pt vô nghiệm.

d.

PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$

$\Leftrightarrow 11x+3=-54x-97$

$\Leftrightarrow 65x=-100$

$\Leftrightarrow x=\frac{-20}{13}$

a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

\(\Leftrightarrow\left(3x-1\right)^3=1\)

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)

a) (x + 1/2) . (2/3 − 2x) = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)

\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)

\(\Rightarrow x=-\frac{2}{11}\)

c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)

\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)

\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)

\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)

\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)

\(\Rightarrow x=1\)

d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)

\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)

\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)

\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)

\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)

\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)

\(\Rightarrow x\approx28,7\) (số hơi lẻ)

e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)

Phạm Tuấn Kiệt câu a sao nhìn không đc vậy ???

Bài 1:

b) \(\left(x+\frac{1}{2}\right).\left(x-\frac{3}{4}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0-\frac{1}{2}\\x=0+\frac{3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{4}\right\}.\)

c) \(\left(2x-5\right)^4=81\)

\(\Rightarrow2x-5=\pm3\)

\(\Rightarrow\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=3+5=8\\2x=\left(-3\right)+5=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8:2\\x=2:2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

Vậy \(x\in\left\{4;1\right\}.\)

d) \(3^{x+1}+3^{x+3}=810\)

\(\Rightarrow3^x.3^1+3^x.3^3=810\)

\(\Rightarrow3^x.\left(3^1+3^3\right)=810\)

\(\Rightarrow3^x.30=810\)

\(\Rightarrow3^x=810:30\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

Vậy \(x=3.\)

Chúc bạn học tốt!

Chia nhỏ ra

a: =>1/2x=7/2-2/3=21/6-4/6=17/6

=>x=17/3

b: =>2/3:x=-7-1/3=-22/3

=>x=2/3:(-22/3)=-1/11

c: =>1/3x+2/5x-2/5=0

=>11/15x=2/5

hay x=6/11

d: =>2x-3=0 hoặc 6-2x=0

=>x=3/2 hoặc x=3

Giải:

a) \(\left(3\dfrac{1}{2}+2x\right).3\dfrac{2}{3}=5\dfrac{1}{3}\) 

     \(\left(\dfrac{7}{2}+2x\right).\dfrac{11}{3}=\dfrac{16}{3}\) 

                 \(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{11}{3}\) 

                 \(\dfrac{7}{2}+2x=\dfrac{16}{11}\) 

                         \(2x=\dfrac{16}{11}-\dfrac{7}{2}\) 

                         \(2x=\dfrac{-45}{22}\) 

                           \(x=\dfrac{-45}{22}:2\) 

                           \(x=\dfrac{-45}{44}\) 

b) \(3-\left(17-x\right)=-12\) 

       \(3-17+x=-12\) 

                     \(x=-12-3+17\) 

                     \(x=2\) 

c) \(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\) 

           \(\dfrac{2}{3}x=\dfrac{1}{10}-\dfrac{1}{2}\) 

           \(\dfrac{2}{3}x=\dfrac{-2}{5}\) 

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

              \(x=\dfrac{-3}{5}\) 

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

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

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

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

                 \(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\) 

Vì giá trị tuyệt đối của 1 số nguyên ko bao giờ là số âm nên \(x\in\varnothing\) 

e) \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\) 

                \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{1}{3}+\left(-1\right)\) 

                \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{-2}{3}\) 

                           \(-0,6x-\dfrac{1}{2}=\dfrac{-2}{3}:\dfrac{3}{4}\) 

                           \(-0,6x-\dfrac{1}{2}=\dfrac{-8}{9}\) 

                                   \(-0,6x=\dfrac{-8}{9}+\dfrac{1}{2}\) 

                                   \(-0,6x=\dfrac{-7}{18}\) 

                                           \(x=\dfrac{-7}{18}:-0.6\) 

                                           \(x=\dfrac{35}{54}\) 

f) \(\left(3x-1\right).\left(\dfrac{-1}{2}x+5\right)=0\) 

\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\\dfrac{-1}{2}x+5=0\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\) 

g) \(60\%.x+\dfrac{2}{3}=\dfrac{1}{3}.6\dfrac{1}{3}\) 

        \(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{1}{3}.\dfrac{19}{3}\) 

        \(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{19}{9}\) 

               \(\dfrac{3}{5}.x=\dfrac{19}{9}-\dfrac{2}{3}\) 

               \(\dfrac{3}{5}.x=\dfrac{13}{9}\) 

                    \(x=\dfrac{13}{9}:\dfrac{3}{5}\) 

                   \(x=\dfrac{65}{27}\) 

Chúc bạn học tốt!

f)câu khó nhất

=>3x-1=0 và -1/2x+5=0

   =>x=1/3 và x=10

\(b.3-\left(17-x\right)=-12\\ \Leftrightarrow17-x=15\\ \Rightarrow x=2\)

\(c.\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{2}{3}x=-\dfrac{2}{5}\\ \Rightarrow x=-\dfrac{3}{5}\)