\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)

\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

a: \(\left(a+1\right)^2-\left(a-1\right)^2\)

\(=\left(a+1-a+1\right)\left(a+1+a-1\right)\)

\(=4a\)

b: \(\left(x+5\right)^2-x^2\)

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

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

1) \(\left(a+1\right)^2-\left(a-1\right)^2\)

= \(\left(a+1-a+1\right)\left(a+1+a-1\right)\)

= \(2.2a\)

=\(4a\)

2)\(\left(x+5\right)^2-x^2\)

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

=5.\(\left(2x+5\right)\)

3)\(\left(x+2\right)^2-9x^2\)

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

=\(\left(x+2-3x\right)\left(x+2+3x\right)\)

4)\(\left(3x+1\right)^2-\left(5x-1\right)^2\)

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

=\(\left(-2x+2\right).8x\)

a, Thay x = 2 ta được 6 - 5 = 3 - 2 (luondung) 

Vậy x = 2 là nghiệm pt trên 

Thay x = 1 ta được 3 - 5 = 3 - 1 (voli) 

Vậy x = 1 ko phải là nghiệm pt trên 

b, Thay x = 2 ta được \(2m=m+6\Leftrightarrow m=6\)

a)thay x=2 ta có: 3.2 - 5 = 3 -2 

=>1=1(hợp lí)

vậy x =2 là 1 nghiệm của PT

thay x=1 ta có: 3.1 - 5 = 3 - 1 

=>-2=2(vô lí) vậy x = 1 không phải nghiệm của PT

b)thay x = 2, ta có:

2m=m+6

<=>m=6

vậy m = 6 khi x = 2

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

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

\(\Rightarrow x=\frac{5}{12}\)

b)\(x-\frac{2}{5}=\frac{5}{7}\)

\(\Rightarrow x=\frac{5}{7}+\frac{2}{5}\)

\(\Rightarrow x=1\frac{4}{35}\)

c)\(-x-\frac{2}{3}=-\frac{6}{7}\)

\(\Rightarrow-x=-\frac{6}{7}+\frac{2}{3}\)

\(\Rightarrow-x=-\frac{4}{21}\)

\(\Rightarrow x=\frac{4}{21}\)

d)\(\frac{4}{7}-x=\frac{1}{3}\)

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

\(\Rightarrow x=\frac{5}{21}\)

A) x+1/3=3/4

    x        =3/4-1/3

    x        =3/4+(-1/3)

    x        =5/12

B) x-2/5=5/7

    x       =5/7+2/5

    x       =39/35

C) -x-2/3=-6/7

     -x      =-6/7+2/3

     -x      =-4/21

     x       =4/21

D) 4/7-x=1/3

           x=4/7-1/3

           x=4/7+(-1/3)

           x=5/21

`@` `\text {Ans}`

`\downarorw`

`2^x \div 2 = 128?`

`=> 2^x = 128 * 2`

`=> 2^x = 256`

`=> 2^x = 2^8`

`=> x = 8`

Vậy, `x = 8`

______

`4^x \div 4^3 = 16`

`=> 4^x = 16 * 4^3`

`=> 4^x = 4^2 * 4^3`

`=> 4^x = 4^5`

`=> x = 5`

Vậy, ` x = 5`

________

`12^x \div 3 = 48 ?`

`=> 12^x = 48 * 3`

`=> 12^x = 144`

`=> 12^x = 12^2`

`=> x = 2`

Vậy, `x = 2`

________

`3^x - 5 = 81`

`3^x = 81 + 5`

`3^x = 86`

Bạn xem lại đề

`@` `\text {Kaizuu lv uuu}`

Giải:

Ta có: \(3\left(x-1\right)=2\left(y-2\right)\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}\)

\(4\left(y-2\right)=3\left(z-3\right)\Rightarrow\frac{y-2}{3}=\frac{z-3}{4}\)

\(\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6+z-3}{4+9+4}=\frac{\left(2x+3y+z\right)-\left(2+6+3\right)}{17}\)

\(=\frac{50-11}{17}=\frac{39}{17}\)

+) \(\frac{x-1}{2}=\frac{39}{17}\Rightarrow x-1=\frac{78}{17}\Rightarrow x=\frac{95}{17}\)

+) \(\frac{y-2}{3}=\frac{39}{17}\Rightarrow y-2=\frac{117}{17}\Rightarrow y=\frac{151}{17}\)

+) \(\frac{z-3}{4}=\frac{39}{17}\Rightarrow z-3=\frac{156}{17}\Rightarrow z=\frac{207}{17}\)

Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\frac{95}{17};\frac{151}{17};\frac{207}{17}\right)\)