bt em gửi cô Thương

1)\(ĐKXĐ\hept{\begin{cases}x\ne1\\x\ne3\end{cases}}\)

 \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)

\(\Leftrightarrow\frac{x+5}{x-1}-\frac{x+1}{x-3}+\frac{8}{x^2-4x+3}=0\)

\(\Leftrightarrow\frac{x+5}{x-1}-\frac{x+1}{x-3}+\frac{8}{x^2-x-3x+3}=0\)

\(\Leftrightarrow\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}+\frac{8}{x\left(x-1\right)-3\left(x-1\right)}=0\)

\(\Leftrightarrow\frac{x^2+2x-15}{\left(x-1\right)\left(x-3\right)}-\frac{x^2-1}{\left(x-3\right)\left(x-1\right)}+\frac{8}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\frac{2x-6}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow2x-6=0\)

\(\Leftrightarrow x=3\)( tm)

Vậy nghiemj của pt x=3

2)\(x^3-x^2-9x+9=0\)

\(\Leftrightarrow x^2\left(x-1\right)-9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)hoặc x+3=0

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

Vậy tập hợp nghiệm \(S=\left\{1;3;-3\right\}\)

Điều kiện: $ - \frac{1}{3} \le x \le 6$
Ta nhẩm thấy x = 5 là nghiệm của PT, thêm bớt và trục căn thức ta có:
Phương trình $ \Leftrightarrow \left( {\sqrt {3x + 1} - 4} \right) - \left( {\sqrt {6 - x} - 1} \right) + \left( {3{x^2} - 14x - 5} \right) = 0$

$ \Leftrightarrow \frac{{3\left( {x - 5} \right)}}{{\sqrt {3x + 1} + 4}} + \frac{{x - 5}}{{\sqrt {6 - x} + 1}} + \left( {3x + 1} \right)\left( {x - 5} \right) = 0$

$ \Leftrightarrow \left( {x - 5} \right)\left[ {\frac{3}{{\sqrt {3x + 1} + 4}} + \frac{1}{{\sqrt {6 - x} + 1}} + \left( {3x + 1} \right)} \right] = 0 \Leftrightarrow \left( {x - 5} \right)g\left( x \right) = 0$

Với điều kiện trên ta thấy g(x) > 0 vậy x = 5 là nghiệm của PT.

3) \(\frac{x-2}{x-5}-\frac{5}{x^2-5x}=\frac{1}{x}\)

\(\Leftrightarrow\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)

\(\Leftrightarrow\frac{x.\left(x-2\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x-5\right)}{x.\left(x-5\right)}\)

Mc: \(x.\left(x-5\right)\)

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

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

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

\(\Leftrightarrow\) \(x\) . (\(x\) - 3) = 0

\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) - 3 = 0

\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) = 3

Vậy \(x\) = 0 hoặc \(x\) = 3

\(x-5\ne0\Rightarrow x\ne5\)

\(x^2-5\ne0\Rightarrow x\ne5\) và \(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)

\(x\ne0\)

Vậy S = {3}

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

\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x.\left(x+7\right)}\)

\(\Leftrightarrow\frac{x.\left(x-4\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)

Mc: \(x.\left(x+7\right)\)

\(\Leftrightarrow x^2-4x-x-7=-7\)

\(\Leftrightarrow x^2-4x-x=-7+7\)

\(\Leftrightarrow\) \(x^2-5x=0\)

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

\(\Leftrightarrow x=0\) hoặc \(x-5=0\)

\(\Leftrightarrow x=0\) hoặc \(x=5\)

Vậy \(x=0\) hoặc \(x=5\)

\(x+7\ne0\Rightarrow x\ne-7\)

\(x^2+7\ne0\Rightarrow x\ne-7\) và \(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-7\end{matrix}\right.\)

\(x\ne0\)

Vậy S = {5}

5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)

\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x^2-4\ne0\end{matrix}\right.\Rightarrow TXĐ\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

Mc : \(\left(x-2\right).\left(x+2\right)\)

\(\Leftrightarrow\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)

\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)

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

\(\Leftrightarrow2x^2-8x+8=0\)

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

\(\Leftrightarrow\) \(2x.\left(x-2\right)-4.\left(x-2\right)=0\)

\(\Leftrightarrow\left(2x-4\right).\left(x-2\right)=0\)

\(\Leftrightarrow2x-4=0\) hoặc \(x-2=0\)

\(\Leftrightarrow x=2\) hoặc \(x=2\)

\(\Leftrightarrow x=2\) (Loại) hoặc x = 2 (Loại)

Vậy S = \(\left\{\varnothing\right\}\)

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

\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)

MC: \(\left(x-1\right).\left(x+1\right)\)

\(\Leftrightarrow x^2+x+x+1-x^2+x+x-1=4\)

\(\Leftrightarrow x^2-x^2+x+x+x+x+1-1-4=0\)

\(\Leftrightarrow4x-4=0\)

\(\Leftrightarrow4.\left(x-1\right)=0\)

\(\Leftrightarrow\) 4 = 0 hoặc \(x-1=0\)

\(\Leftrightarrow\) 4 = 0 hoặc \(x=1\)

\(\Leftrightarrow\) 4 = 0 (Loại) hoặc \(x=1\) (Loại)

Vậy S = \(\left\{\varnothing\right\}\)

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

\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\)

\(Mc:\left(x-1\right).\left(x+1\right)\)

\(\Leftrightarrow\) \(x^2+x+x+1-4x=x^2-x-x+1\)

\(\Leftrightarrow x^2-x^2+x+x-4x+x+x=-1+1\)

\(\Leftrightarrow0=0\) (Nhận)

Vậy S = \(\left\{x\in R;x\ne\pm1\right\}\)

1) \(\frac{3x-1}{4}+\frac{2x-3}{3}=\frac{x-1}{2}\) Mc : 12

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

\(\Leftrightarrow\) 9x - 3 + 8x - 12 = 6x - 6

\(\Leftrightarrow\) 9x + 8x - 6x = 3 + 12 - 6

\(\Leftrightarrow\) 11x = 9

\(\Leftrightarrow\) x = 0,8

Vậy S = {0,8}

2) \(\frac{x+1}{2}-\frac{x+3}{12}=3-\frac{5-3x}{3}\) Mc : 12

\(\Leftrightarrow\) \(\frac{6.\left(x+1\right)}{12}-\frac{x+3}{12}=\frac{12.3}{12}-\frac{4.\left(5-3x\right)}{12}\)

\(\Leftrightarrow\) 6x + 6 - x + 3 = 36 - 20 - 12x

\(\Leftrightarrow\) 6x - x + 12x = -6 - 3 + 36 - 20

\(\Leftrightarrow\) 17x = 7

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

Vậy S = {\(\frac{7}{17}\)}

3) x - \(\frac{x+1}{3}\) = \(\frac{2x-1}{5}\) Mc : 15

\(\Leftrightarrow\) \(\frac{15.x}{15}-\frac{5.\left(x+1\right)}{15}=\frac{3.\left(2x-1\right)}{15}\)

\(\Leftrightarrow\) 15x - 5x - 5 = 6x - 3

\(\Leftrightarrow\) 15x - 5x - 6x = 5 - 3

\(\Leftrightarrow\) 4x = 2

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

Vậy S = {\(\frac{1}{2}\)}

4) \(\frac{2x+7}{3}-\frac{x-2}{4}=-2\) Mc : 12

\(\Leftrightarrow\) \(\frac{4.\left(2x+7\right)}{12}-\frac{3.\left(x-2\right)}{12}=\frac{12.\left(-2\right)}{12}\)

\(\Leftrightarrow\) 8x + 28 -3x + 6 = -24

\(\Leftrightarrow\) 8x - 3x = -28 - 6 -24

\(\Leftrightarrow\) 5x = -58

\(\Leftrightarrow\) x = -11,6

Vậy S = {-11,6}

5) \(\frac{2x-3}{4}-\frac{4x-5}{3}=\frac{5-x}{6}\) Mc : 12

\(\Leftrightarrow\) \(\frac{3.\left(2x-3\right)}{12}-\frac{4.\left(4x-5\right)}{12}=\frac{2.\left(5-x\right)}{12}\)

\(\Leftrightarrow\) 6x - 9 - 16x + 20 = 10 - 2x

\(\Leftrightarrow\) 6x - 16x + 2x = 9 - 20 + 10

\(\Leftrightarrow\) -8x = -1

\(\Leftrightarrow\) x = \(\frac{1}{8}\)

Vậy S = {\(\frac{1}{8}\)}

6) \(\frac{12x+1}{4}=\frac{9x+1}{3}-\frac{3-5x}{12}\) Mc : 12

\(\Leftrightarrow\frac{3.\left(12x+1\right)}{12}=\frac{4.\left(9x+1\right)}{12}-\frac{3-5x}{12}\)

\(\Leftrightarrow\) 36x + 3 = 36x + 4 - 3 + 5x

\(\Leftrightarrow\) 36x - 36x - 5x = -3 + 4 - 3

\(\Leftrightarrow\) -5x = -2

\(\Leftrightarrow x=\frac{2}{5}\)

7) \(\frac{x+6}{4}\) - \(\frac{x-2}{6}-\frac{x+1}{3}=0\) Mc : 12

\(\Leftrightarrow\) \(\frac{3.\left(x+6\right)}{12}-\frac{2.\left(x-2\right)}{12}-\frac{4.\left(x+1\right)}{12}=0\)

\(\Leftrightarrow\) 3x + 18 - 2x + 4 - 4x - 4 = 0

\(\Leftrightarrow\) 3x - 2x - 4x = -18 - 4 + 4

\(\Leftrightarrow\) -3x = -18

\(\Leftrightarrow\) x = 6

Vậy S = {6}

8) x\(^2\) - x - 6 = 0

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

\(\Leftrightarrow\) x.(x + 2) - 3.(x + 2) = 0

\(\Leftrightarrow\) (x - 3).(x + 2) = 0

\(\Leftrightarrow\) x - 3 = 0 hoặc x + 2 = 0

\(\Leftrightarrow\) x = 3 hoặc x = -2

Vậy S = {3; -2}

\(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{3}{2}\Rightarrow2\left(x+y\right)=3xy\)

\(x^2+y^2=5\Leftrightarrow\left(x+y\right)^2-2xy=5\)

Đặt x+y=u; xy=v, ta có hệ

\(\int^{2\left(x+y\right)-3xy=0}_{\left(x+y\right)^2-2xy=5}\Leftrightarrow\int^{2u-3v=0}_{u^2-2v=5}\Leftrightarrow u=3;v=2\)hoặc \(u=-\frac{5}{3};v=-\frac{10}{9}\)

đến đây dùng viet, x và y là nghiệm của 2 phương trình \(X^2-3X+2=0\) hoặc \(X^2+\frac{5}{3}X-\frac{10}{9}=0\). Giải ra được nghiệm (x;y) là   \(\left(1;2\right),\left(2;1\right),\left(\frac{-5+\sqrt{65}}{6};\frac{-5-\sqrt{65}}{6}\right),\left(\frac{-5-\sqrt{65}}{6};\frac{-5+\sqrt{65}}{6}\right)\)

1.     \(\frac{x^3-10x^2+25x}{x^2-5x}\)\(=0\)  ( đkxđ: \(x\ne0;5\))

<=> \(\frac{x\left(x-5\right)^2}{x\left(x-5\right)}=0\)<=>   \(x-5=0\)<=> vô n_o

_{2.       \(A=\)\(\frac{2x^2-2}{x^3-x^2-4x+4}\)\(=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x+2\right)}\)}   ( a, đkxđ: \(x\ne1;\pm2\))

b, \(A=0\)<=> \(\frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}=0\)<=> \(x=-1\)( TM) . Vậy \(A=0\Leftrightarrow x=-1\)

3.    \(B=\frac{3x^2-12}{\left(x-3\right)\left(x^2+4x+4\right)}\)\(=\frac{3\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+2\right)^2}\) ( a,  đkxđ: \(x\ne3,-2\))

\(b,B=0\Leftrightarrow\frac{3\left(x-2\right)}{\left(x-3\right)\left(x+2\right)}=0\Leftrightarrow x=2\left(tm\right)\). Vậy \(B=0\Leftrightarrow x=2\)

\(\frac{1}{x}+\frac{1}{y}=\frac{1}{5}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{1}{5}\)

\(\Leftrightarrow5x+5y=xy\)

\(\Leftrightarrow5x+5y-xy=0\)

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

\(\Leftrightarrow x\left(5-y\right)+5y-25=-25\)

\(\Leftrightarrow x\left(5-y\right)-5\left(5-y\right)=-25\)

\(\Leftrightarrow\left(5-y\right)\left(x-5\right)=-25\)

nốt :V

Giải phương trình:

1.\(\frac{x-5}{x-5}+\frac{x-6}{x-5}+\frac{x-7}{x-5}+...+\frac{1}{x-5}=4\left(x\in N\right)\)

2.\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)

3.\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{x\left(x+2\right)}\right)=\frac{31}{16}\left(x\in N\right)\)

4.\(8\left(x^2+\frac{1}{x^2}\right)-34\left(x+\frac{1}{x}\right)+51=0\)

5.\(6x^4-5x^3-38x^2-5x+6=0\)