\(a,3^x>\dfrac{1}{243}\\ \Leftrightarrow3^x>3^{-5}\\ \Leftrightarrow x>-5\\ b,\left(\dfrac{2}{3}\right)^{3x-7}\le\dfrac{3}{2}\\ \Leftrightarrow3x-7\le1\\ \Leftrightarrow3x\le8\\ \Leftrightarrow x\le\dfrac{8}{3}\\ c,4^{x+3}\ge32^x\\ \Leftrightarrow2^{2x+6}\ge2^{5x}\\ \Leftrightarrow2x+6\ge5x\\ \Leftrightarrow3x\le6\\ \Leftrightarrow x\le2\)

d, Điều kiện: x > 1

\(log\left(x-1\right)< 0\\ \Leftrightarrow x-1< 1\\ \Leftrightarrow1< x< 2\)

e, Điều kiện: \(x>\dfrac{1}{2}\)

\(log_{\dfrac{1}{5}}\left(2x-1\right)\ge log_{\dfrac{1}{5}}\left(x+3\right)\\ \Leftrightarrow2x-1\ge x+3\\ \Leftrightarrow x\ge4\)

f, Điều kiện: x > 4

\(ln\left(x+3\right)\ge ln\left(2x-8\right)\\ \Leftrightarrow x+3\ge2x-8\\\Leftrightarrow4< x\le11\)

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

\(x^2\left(x-3\right)=0\)

\(\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=3\left(TM\right)\end{cases}}}\)

\(2,3x^3-48x=0\)

\(3x\left(x^2-16\right)=0\)

\(\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x^2=16\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=\pm4\left(TM\right)\end{cases}}}}\)

\(3,5x\left(x-1\right)=x-1\)

\(5x^2-5x=x-1\)

\(5x^2-6x+1=0\)

\(5x^2-5x-x+1=0\)

\(5x\left(x-1\right)-\left(x-1\right)=0\)

\(\left(5x-1\right)\left(x-1\right)=0\)

\(\orbr{\begin{cases}5x-1=0\\x-1=0\end{cases}\orbr{\begin{cases}x=\frac{1}{5}\left(TM\right)\\x=1\left(TM\right)\end{cases}}}\)

\(4,2\left(x+5\right)-x^2-5x=0\)

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

\(-x^2-3x+10=0\)

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

\(-x\left(x+5\right)+2\left(x+5\right)=0\)

\(\left(x+5\right)\left(2-x\right)=0\)

\(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\orbr{\begin{cases}x=-5\left(TM\right)\\x=2\left(TM\right)\end{cases}}}\)

\(5,2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(2x^2-10x-3x-2x^2=26\)

\(-13x-26=0\)

\(-13\left(x+2\right)=0\)

\(x=-2\left(TM\right)\)

Trả lời:

1, \(x^3-3x^2=0\)

\(\Leftrightarrow x^2\left(x-3\right)=0\)

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

Vậy x = 0; x = 3 là nghiệm của pt.

2, \(3x^3-48x=0\)

\(\Leftrightarrow3x\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}}\)

Vậy x = 0; x = 4; x = - 4 là nghiệm của pt.

3, \(5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}}\)

Vậy x = 1; x = 1/5 là nghiệm của pt.

4, \(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}}\)

Vậy x = - 5; x = 2 là nghiệm của pt.

5, \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=-2\)

Vậy x = - 2 là nghiệm của pt.

1) \(\left(\frac{2x}{3}-3\right):\left(-10\right)=\frac{2}{5}\)

\(\Leftrightarrow-\frac{\frac{2x}{3}-3}{10}=\frac{2}{5}\)

\(\Leftrightarrow-\left(\frac{\frac{2x}{3}}{10}-\frac{3}{10}\right)=\frac{2}{5}\)

\(\Leftrightarrow-\left(\frac{2x}{3\times10}-\frac{3}{10}\right)=\frac{2}{5}\)

\(\Leftrightarrow-\left(\frac{2x}{30}-\frac{3}{10}\right)=\frac{2}{5}\)

\(\Leftrightarrow-\frac{x}{15}+\frac{3}{10}=\frac{2}{5}\)

\(\Leftrightarrow\frac{3}{10}-\frac{x}{15}=\frac{2}{5}\)

\(\Leftrightarrow-\frac{x}{15}=\frac{2}{5}-\frac{3}{10}\)

\(\Leftrightarrow-\frac{x}{15}=\frac{1}{10}\)

\(\Leftrightarrow-x=\frac{15}{10}\)

\(\Leftrightarrow-x=\frac{3}{2}\)

\(\Leftrightarrow x=-\frac{3}{2}\)

Vậy \(x=-\frac{3}{2}\)

2) \(\left|2x-1\right|+1=4\)

\(\Leftrightarrow\left|2x-1\right|=3\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

Vậy \(x\in\left\{2;-1\right\}\)

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

\(\Leftrightarrow\left(x+1\right).4=\left(x-2\right).3\)

\(\Leftrightarrow4x+4=3x-6\)

<=>4x-3x=-6-4

<=>x=-10

2.\(\frac{52}{2x-1}=\frac{13}{30}\)

<=>52.30=(2x-1).13

<=>1560=26x-13

<=>-26x=-13-1560

<=>-26x=-1573

<=>x=60,5

3.\(\frac{2x-3}{x+1}=\frac{4}{7}\)

<=>(2x-3).7=(x+1).4

<=>14x-21=4x+4

<=>14x-4x=4+21

<=>10x=25

<=>x=2,5

4.\(\frac{2x+3}{42}=\frac{3x-1}{32}\)

<=>(2x+3).32=42(3x-1)

<=>64x+96=126x-42

<=>64x-126x=-42-96

<=>-62x=-138

<=>x=69/31

cho mk hỏi 

10x=25 số 10 ở đâu vậy bn 

4. ( x - 250 ) : 6 = 64 - 12

( x- 250 ) : 6 = 52 

x - 250 = 312

x = 562 

5. 10x = 1030 

=> x = 103

6. 30x = 120 

x = 4

7. \(x=2023\)

\(8.165-\left(35:x+3\right).19=13\)

\(\left(35:x+3\right).19=152\)

\(35:x+3=8\)

\(35:x=5\)

\(x=7\)

4) \(\left(x-250\right)\div6=4^3-2^2\times3\) 

    \(\left(x-250\right)\div6=64-4\times3\) 

      \(\left(x-250\right)\div6=64-12=52\) 

                \(x-250=52\times6=312\) 

                         \(x=312+250\) 

                           \(x=562\) 

5) \(2x+3x+5x=1030\) 

      \(x\left(2+3+5\right)=1030\) 

                   \(10x=1030\) 

                        \(x=1030\div10\) 

                        \(x=103\) 

6) \(15x-35x+50x=120\) 

       \(x\left(15-35+50\right)=120\) 

                           \(30x=120\) 

                                \(x=120\div30\) 

                                \(x=4\) 

7) \(\dfrac{1}{2}x+\dfrac{1}{6}x+\dfrac{1}{3}x=2023\) 

     \(x\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)=2023\)  

                       \(x\times1=2023\) 

                               \(x=2023\) 

8) \(165-\left(35\div x+3\right)\times19=13\) 

              \(\left(35\div x+3\right)\times19=165-13\) 

                 \(\left(35\div x+3\right)\times19=152\) 

                          \(35\div x+3=152\div19=8\) 

                                   \(35\div x=8-3=5\) 

                                             \(x=35\div5\) 

                                             \(x=7\)

a,    x=10

b,   x= 2

A/\(\left(2,8x-32\right):\frac{2}{3}=-90\)

\(\left(\frac{28}{10}x-32\right)=\frac{-90}{1}.\frac{2}{3}\)

\(\left(\frac{14}{5}x-32\right)=\frac{-30}{1}.\frac{2}{1}\)

\(\left(\frac{14}{5}x-32\right)=-60\)

\(\frac{14}{5}x=-60+32\)

\(\frac{14}{5}x=-28\)

\(x=\frac{-28}{1}:\frac{14}{5}\)

\(x=\frac{-28}{1}.\frac{5}{14}\)

\(x=\frac{-2}{1}.\frac{5}{1}=-10\)

B/\(\left(4,5-2x\right).1\frac{4}{7}=\frac{11}{14}\)

\(\left(\frac{45}{10}-2x\right).\frac{11}{7}=\frac{11}{14}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}:\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}.\frac{7}{11}\)

\(\left(\frac{9}{2}-2x\right)=\frac{1}{2}.\frac{1}{1}=\frac{1}{2}\)

\(2x=\frac{9}{2}-\frac{1}{2}\)

\(2x=\frac{8}{2}\)

\(x=\frac{8}{2}:\frac{2}{1}=\frac{8}{2}.\frac{1}{2}\)

\(x=\frac{4}{2}.\frac{1}{1}=\frac{4}{2}=2\)