Toán Lớp 8: 4/ Tìm x biết
a. x^2-1/4x=0
b. (2x-1)^2-49=0
c. 3x^3-12x^2+12x=0
d. (2x+1)^2-x(2x+1)=0
e. 2(x-11)^2-x+1=0
giúp mik với huhuhu
xin lỗi admin
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: 4/ Tìm x biết
a. x^2-1/4x=0
b. (2x-1)^2-49=0
c. 3x^3-12x^2+12x=0
d. (2x+1)^2-x(2x+1)=0
e. 2(x-11)^2-x+1=0
giúp mik với huhuhu
xin lỗi admin
Comments ( 1 )
a,\\
\left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{4}
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = 4\\
x = – 3
\end{array} \right.\\
c,\\
\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
d,\\
\left[ \begin{array}{l}
x = – \dfrac{1}{2}\\
x = – 1
\end{array} \right.\\
e,\\
\left[ \begin{array}{l}
x = 1\\
x = \dfrac{3}{2}
\end{array} \right.
\end{array}\)
a,\\
{x^2} – \dfrac{1}{4}x = 0\\
\Leftrightarrow x.\left( {x – \dfrac{1}{4}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x – \dfrac{1}{4} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{4}
\end{array} \right.\\
b,\\
{\left( {2x – 1} \right)^2} – 49 = 0\\
\Leftrightarrow {\left( {2x – 1} \right)^2} – {7^2} = 0\\
\Leftrightarrow \left[ {\left( {2x – 1} \right) – 7} \right].\left[ {\left( {2x – 1} \right) + 7} \right] = 0\\
\Leftrightarrow \left( {2x – 8} \right).\left( {2x + 6} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x – 8 = 0\\
2x + 6 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = – 3
\end{array} \right.\\
c,\\
3{x^3} – 12{x^2} + 12x = 0\\
\Leftrightarrow 3x.\left( {{x^2} – 4x + 4} \right) = 0\\
\Leftrightarrow 3x.\left( {{x^2} – 2.x.2 + {2^2}} \right) = 0\\
\Leftrightarrow 3x.{\left( {x – 2} \right)^2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{\left( {x – 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x – 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
d,\\
{\left( {2x + 1} \right)^2} – x.\left( {2x + 1} \right) = 0\\
\Leftrightarrow \left( {2x + 1} \right).\left[ {\left( {2x + 1} \right) – x} \right] = 0\\
\Leftrightarrow \left( {2x + 1} \right).\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 1 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{1}{2}\\
x = – 1
\end{array} \right.\\
e,\\
2.{\left( {x – 1} \right)^2} – x + 1 = 0\\
\Leftrightarrow 2.{\left( {x – 1} \right)^2} – \left( {x – 1} \right) = 0\\
\Leftrightarrow \left( {x – 1} \right).\left[ {2.\left( {x – 1} \right) – 1} \right] = 0\\
\Leftrightarrow \left( {x – 1} \right).\left( {2x – 2 – 1} \right) = 0\\
\Leftrightarrow \left( {x – 1} \right)\left( {2x – 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 1 = 0\\
2x – 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{3}{2}
\end{array} \right.
\end{array}\)