Toán Lớp 7: 1:
(2x-1)^3=8
(2x-3)^2=9
2:
x^2=x^5
(3y-1)^10=(3y-1)^20
(x-5)^2=(1-3x)^2
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Toán Lớp 7: 1:
(2x-1)^3=8
(2x-3)^2=9
2:
x^2=x^5
(3y-1)^10=(3y-1)^20
(x-5)^2=(1-3x)^2
Comments ( 2 )
1){\left( {2x – 1} \right)^3} = 8\\
\Leftrightarrow {\left( {2x – 1} \right)^3} = {2^3}\\
\Leftrightarrow 2x – 1 = 2\\
\Leftrightarrow 2x = 3\\
\Leftrightarrow x = \dfrac{3}{2}\\
Vậy\,x = \dfrac{3}{2}\\
{\left( {2x – 3} \right)^2} = 9\\
\Leftrightarrow \left[ \begin{array}{l}
2x – 3 = 3\\
2x – 3 = – 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 6\\
2x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 0
\end{array} \right.\\
Vậy\,x = 3;x = 0\\
2)\\
{x^2} = {x^5}\\
\Leftrightarrow {x^5} – {x^2} = 0\\
\Leftrightarrow {x^2}\left( {{x^3} – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} = 0\\
{x^3} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
Vậy\,x = 0;x = 1\\
{\left( {3y – 1} \right)^{10}} = {\left( {3y – 1} \right)^{20}}\\
\Leftrightarrow {\left( {3y – 1} \right)^{20}} – {\left( {3y – 1} \right)^{10}} = 0\\
\Leftrightarrow {\left( {3y – 1} \right)^{10}}\left( {{{\left( {3y – 1} \right)}^{10}} – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
3y – 1 = 0\\
{\left( {3y – 1} \right)^{10}} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
y = \dfrac{1}{3}\\
3y – 1 = 1\\
3y – 1 = – 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
y = \dfrac{1}{3}\\
y = \dfrac{2}{3}\\
y = 0
\end{array} \right.\\
Vậy\,y = 0;y = \dfrac{1}{3};y = \dfrac{2}{3}\\
{\left( {x – 5} \right)^2} = {\left( {1 – 3x} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
x – 5 = 1 – 3x\\
x – 5 = 3x – 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
4x = 6\\
2x = – 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = – 2
\end{array} \right.\\
Vậy\,x = – 2;x = \dfrac{3}{2}
\end{array}$