Bài 5: Phương trình chứa ẩn ở mẫu

a) x³ - 3x² + 3x - 1 = 0

<=> (x - 1)³ = 0

<=> x - 1 = 0

<=> x = 1

Vậy ....................

b) (2x - 5)² - (x + 2)² = 0

<=> (2x - 5 - x - 2)(2x - 5 + x + 2) = 0

<=> (x - 7)(3x - 3) = 0

<=> 3(x - 7)(x - 1) = 0

\(\Leftrightarrow \begin{bmatrix} x - 7 = 0 & & \\ x - 1 = 0 & & \end{bmatrix}\) pn bỏ dấu ngoặc bên phải nha

\(\Leftrightarrow \begin{bmatrix} x = 7 & & \\ x = 1 & & \end{bmatrix}\)

Vậy .............

a) x³ - 3x² + 3x - 1 = 0

⇔ (x - 1)³ = 0

⇔ x - 1 = 0

⇔ x = 1

Vậy tập nghiệm phương trình là S = \(\left\{1\right\}\)

b) (2x - 5)² - (x + 2)² = 0

⇔ (2x - 5 - x - 2)(2x - 5 + x + 2) = 0

⇔ (x - 7)(3x - 3) = 0

⇔ \(\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=7\\3x=3\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Vậy tập nghiệm phương trình là S = \(\left\{1;7\right\}\)

\(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}=5\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+.....+\dfrac{1}{x+99}-\dfrac{1}{x+100}=5\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+100}=5\)

\(\Leftrightarrow\dfrac{x+100-x-2}{\left(x+2\right)\left(x+100\right)}=5\)

\(\Leftrightarrow98=5\left(x+2\right)\left(x+100\right)\)

\(\Leftrightarrow98=5\left(x^2+102x+200\right)\)

\(\Leftrightarrow5x^2+510x+1000-98=0\)

\(\Leftrightarrow5x^2+510x+902=0\)

\(\Leftrightarrow5(x^2+102x+\dfrac{902}{5})=0\)

\(\Leftrightarrow(x^2+2.x.51+51^2-2601+\dfrac{902}{5})=0\)

\(\Leftrightarrow\left[\left(x+51\right)^2-\dfrac{12103}{5}\right]=0\)

\(\Leftrightarrow\left(x+51\right)^2=\dfrac{12103}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{12103}{5}-51\\x=-\dfrac{12103}{5}-51\end{matrix}\right.\)

Để mình giúp nha

\(x^2+\dfrac{1}{x^2}-\dfrac{9}{2}\left(x+\dfrac{1}{x}\right)+7=0\)

ĐKXD: x\(\ne0\)

\(\Leftrightarrow x^2+2+\dfrac{1}{x^2}-\dfrac{9}{2}\left(x+\dfrac{1}{x}\right)+5=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-\dfrac{9}{2}\left(x+\dfrac{1}{x}\right)+5=0\)

Đặt \(a=x+\dfrac{1}{x}\) khi đó phương trình trở thành

\(a^2-\dfrac{9}{2}a+5=0\)

\(\Leftrightarrow\left(a\right)^2-2.a.\dfrac{9}{4}+\left(\dfrac{9}{4}\right)^2-\dfrac{81}{16}+5=0\)

\(\Leftrightarrow\left(a+\dfrac{9}{4}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}a-\dfrac{9}{4}=\dfrac{1}{4}\\a-\dfrac{9}{4}=-\dfrac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{5}{2}\\a=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{5}{2}\\x+\dfrac{1}{x}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x^2+1}{x}=\dfrac{5}{2}\\\dfrac{x^2+1}{x}=2\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x\right)^2-2.x.\dfrac{5}{4}+\left(\dfrac{5}{4}\right)^2-\dfrac{25}{16}+1=0\\\left(x-1\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{5}{4}\right)^2-\dfrac{9}{16}=0\\x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=\dfrac{1}{2}\left(n\right)\\x=1\left(n\right)\end{matrix}\right.\)

Vậy S=\(\left\{1;2;\dfrac{1}{2}\right\}\)

\(mx+2=x-m\)

<=> \(mx-x=-2-m\)

<=> \(x\left(m-1\right)=-2-m\)

* Với : m # 1 , ta có :

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

Vậy , với m # 1 thì phương trình có nghiệm duy nhất : \(x=-\dfrac{m+2}{m-1}\)

* Với : m = 1 , ta có :

x( 1 - 1) = - 2 - 1

<=> 0x = -3 ( Vô lý )

Vậy , với m = 1 thì phương trình vô nghiệm

24: 

\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)

\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)

=>(x+10)(x-2)=0

=>x=-10 hoặc x=2

25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)

\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)

\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)

\(\Leftrightarrow x+5=0\)

hay x=-5