Toán Lớp 6: a, 27.(x-1)^4 – x – 1 = 0
b, (x-2)^2 – 4.(2-x)^4=0
c, 4.(x-3) – (3-x)^3=0
hứa vote 5 sao và ctlhn
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Toán Lớp 6: a, 27.(x-1)^4 – x – 1 = 0
b, (x-2)^2 – 4.(2-x)^4=0
c, 4.(x-3) – (3-x)^3=0
hứa vote 5 sao và ctlhn
HELLLLLLLLP
Comments ( 2 )
a)27{\left( {x – 1} \right)^4} – x + 1 = 0\\
\Leftrightarrow \left( {x – 1} \right)\left( {27{{\left( {x – 1} \right)}^3} – 1} \right) = 0\\
\Leftrightarrow \left( {x – 1} \right).\left[ {{{\left( {3x – 3} \right)}^3} – 1} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 1 = 0\\
3x – 3 = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{4}{3}
\end{array} \right.\\
Vậy\,x = 1;x = \dfrac{4}{3}\\
b){\left( {x – 2} \right)^2} – 4{\left( {2 – x} \right)^4} = 0\\
\Leftrightarrow {\left( {x – 2} \right)^2} – 4{\left( {x – 2} \right)^4} = 0\\
\Leftrightarrow {\left( {x – 2} \right)^2}\left( {1 – 4{{\left( {x – 2} \right)}^2}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 2 = 0\\
4{\left( {x – 2} \right)^2} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x – 2 = \dfrac{1}{2}\\
x – 2 = – \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \dfrac{5}{2}\\
x = \dfrac{3}{2}
\end{array} \right.\\
Vậy\,x = 2;x = \dfrac{5}{2};x = \dfrac{3}{2}\\
c)4\left( {x – 3} \right) – {\left( {3 – x} \right)^3} = 0\\
\Leftrightarrow 4\left( {x – 3} \right) + {\left( {x – 3} \right)^3} = 0\\
\Leftrightarrow \left( {x – 3} \right)\left( {4 + {{\left( {x – 3} \right)}^2}} \right) = 0\\
\Leftrightarrow x – 3 = 0\\
\Leftrightarrow x = 3\\
Vậy\,x = 3
\end{array}$