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

ĐKXĐ: x ≠ 2

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

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

<=> 2 + 3x - 6 = 3 - x

<=> 2 + 3x - 6 - 3 + x = 0

<=> 4x - 7 = 0

\(\Leftrightarrow x=\dfrac{7}{4}\)

Vậy:...

\(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\) (ĐKXĐ \(x\ne2\))

\(\Leftrightarrow\dfrac{1}{x-2}+\dfrac{3\left(x-2\right)}{x-2}=\dfrac{3-x}{x-2}\)

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

\(\Rightarrow3x-5=3-x\)

\(\Leftrightarrow3x+x=3+5\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\)

Mà \(x\ne2\) nên phương trình đề bài cho vô nghiệm

ĐKXĐ: x\(\ne\)2

Ta có: \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)

\(\Leftrightarrow\dfrac{1}{x-2}+\dfrac{3\left(x-2\right)}{x-2}=\dfrac{3-x}{x-2}\)

Suy ra: \(1+3x-6=3-x\)

\(\Leftrightarrow3x-5-3+x=0\)

\(\Leftrightarrow4x-8=0\)

\(\Leftrightarrow4x=8\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

3x + 2/2x - 3x +1/x = 1 

2/x=1

x = 1/2

\(\dfrac{2x}{x-3}+\dfrac{x}{x+3}=\dfrac{2x^2}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)

\(\Leftrightarrow\dfrac{2x\left(x+3\right)+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x^2}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow2x\left(x+3\right)+x\left(x-3\right)=2x^2\)

\(\Leftrightarrow2x^2+6x+x^2-3x-2x^2=0\)

\(\Leftrightarrow x^2+3x=0\)

\(\Leftrightarrow x\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)

Vậy ............................

ĐKXĐ: x khác 3 và x khác -3

\(\dfrac{2x}{x-3}+\dfrac{x}{x+3}=\dfrac{2x^2}{x^2-9}\)

\(\Leftrightarrow\dfrac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x^2}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow2x^2+6x+x^2-3x=2x^2\)

\(\Leftrightarrow x^2+3x=0\)

\(\Leftrightarrow x\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy......

Biến đổi A ta được :

\(A=x\left(x+11\right)\left(x+3\right)\left(x+8\right)+144\)

\(=\left(x^2+11x\right)\left(x^2+11x+24\right)+144\)

\(=\left(x^2+11x\right)^2+24\left(x^2+11x\right)+144\)

\(=\left(x^2+11x\right)^2+2.12.\left(x^2+11x\right)+12^2\)

\(=\left(x^2+11x+12\right)^2\) là một số chính phương \(\forall x\in Z\)

Vậy A là một số chính phương (đpcm)

Xin cảm ơn ạ.

\(A=x\left(x+3\right)\left(x+8\right)\left(x+11\right)+144\)

\(=x\left(x+11\right)\left(x+3\right)\left(x+8\right)+144\)

\(=\left(x^2+11\right)\left(x^2+11+24\right)+144\)

Đặt \(x^2+11=y\Rightarrow x^2+11+24=y+24\)

\(A=y\left(y+24\right)+144\)

\(=y^2+24y+144\)

\(=y^2+2.12y+144\)

=\(\left(y+12\right)^2\)

Có \(A=\left(y+12\right)^2\) là bình phương của 1 số => A là số chính phương

a ) \(\frac{3x+1}{5y+2}=\frac{6x+3}{10y+6}\)

\(\Leftrightarrow\left(3x+1\right).\left(10y+6\right)=\left(5y+2\right).\left(6x+3\right)\)

\(\Leftrightarrow30xy+18x+10y+6=30xy+15y+12x+6\)

\(\Leftrightarrow6x-5y=0\)

kHÔNG CÓ X,Y THÕA MÃN

cÂU B TƯƠNG TỰ

Giải :

3x + 1 = 2x + 10

=> 3x - 2x = 10 - 1

=>     x     = 9

Bài này ta áp dụng quy tắc chuyển vế đổi dấu ở lớp 6 nha bạn !!

3x + 1 = 2x + 10 <=> 2x +x + 1 = 2x + 10

                               <=>       x + 1    =  10

                               <=>         x          =  10 - 1 = 9

Vậy x thỏa mãn là : x = 9

1, <=> 13x = 19 <=x = 19/13 

2, <=> 14x = - 15 <=> x = -15/14 

3, <=> 8x = 11 <=> x = 11/8 

4, <=> 9 - 7x = 4x + 3 <=> 11x = 6 <=> x = 6/11

5, <=> 11-11x = 21 - 5x <=> 6x = - 10 <=> x = -5/3 

6, <=> -12 + 6x = 3 - x <=> 7x = 15 <=> x = 15/7 

7, <=> 40 + 15x + 6x - 16 = 0 <=> 21x = - 24 <=> x = -8/7 

8, <=> 6x - 3 - 3x + 1 = 0 <=> 3x - 2 = 0 <=> x = 2/3 

9, <=> -4x + 12 = 7x - 3 <=> 11x = 15 <=> x = 15/11 

10, <=> -5 - x - 3 = 2 - 5x <=> -8 - x = 2 - 5x <=> 4x = 10 <=> x = 5/2 

\(1,\Leftrightarrow5x+8x=16+3\)

\(\Leftrightarrow13x=19\)

\(\Leftrightarrow x=\dfrac{19}{13}\)

Vậy \(S=\left\{\dfrac{19}{13}\right\}\)

\(b,\Leftrightarrow-5x-9x=8+7\)

\(\Leftrightarrow-14x=15\)

\(\Leftrightarrow x=-\dfrac{15}{14}\)

Vậy \(S=\left\{-\dfrac{15}{14}\right\}\)

\(c,-5x-3x=7-18\)
\(\Leftrightarrow-8x=-11\)

\(\Leftrightarrow x=\dfrac{11}{8}\)

\(d\Leftrightarrow,7x-4x=3-9\)

\(\Leftrightarrow3x=-6\)

\(\Leftrightarrow x=-2\)

Vậy \(S=\left\{-2\right\}\)

\(5,\Leftrightarrow-11x+5x=21-11\)

\(\Leftrightarrow-6x=10\)

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

Vậy \(S=\left\{-\dfrac{5}{3}\right\}\)

\(6,\Leftrightarrow-14+6x=5-x-2\)

\(\Leftrightarrow6x+x=5+14-2\)

\(\Leftrightarrow7x=17\)

\(\Leftrightarrow x=\dfrac{17}{7}\)

Vậy \(S=\left\{\dfrac{17}{7}\right\}\)

\(7,40+15x+6x-16=0\)

\(\Leftrightarrow15x+6x=16-40\)

\(\Leftrightarrow21x=-24\)

\(\Leftrightarrow x=-\dfrac{24}{21}\)

Vậy \(S=\left\{-\dfrac{24}{21}\right\}\)

\(8,6x-3-3x+1=0\)

\(\Leftrightarrow6x-3x=3-1\)

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

Vậy \(S=\left\{\dfrac{2}{3}\right\}\)

Câu (9) và (10) bạn áp dụng như các câu trên, nhân các ngoặc và đổi dấu sau khi bỏ ngoặc hoặc chuyển vế.

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

\(\Rightarrow\left(5x^2+3x-2\right)^2-\left(4x^2-3x-2\right)^2=0\)

\(\Rightarrow\left[\left(5x^2+3x-2\right)-\left(4x^2-3x-2\right)\right]\left[\left(5x^2+3x-2\right)+\left(4x^2-3x-2\right)\right]=0\)

\(\Rightarrow\left(5x^2+3x-2-4x^2+3x+2\right)\left(5x^2+3x-2+4x^2-3x-2\right)=0\)

\(\Rightarrow\left(x^2+6x\right)\left(9x^2-4\right)=0\)

\(\Rightarrow x\left(x+6\right)\left[\left(3x\right)^2-2^2\right]=0\)

\(\Rightarrow x\left(x+6\right)\left(3x-2\right)\left(3x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+6=0\\3x-2=0\\3x+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\\3x=2\\3x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\\x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)