Toán Lớp 8: bài 1 :tìm x phân tích đa thức thành nhân tử bằng phương pháp nhóm hạng tử
a)(x-2)^2-(x-3)(x+3)=0
b)(3x-1)^2-16=0
c)5x^3-35x=0
d)(x-3)^2=4x^2-20x+25
e)x+x^2-x^3-x^4=0
f)(7x-5)^2-2(7x-5)(X+3)+(x+3)^2=0
hứa vote 5 sao hơi nhiều mng thông cảm ạ
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a){\left( {x – 2} \right)^2} – \left( {x – 3} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow {x^2} – 4x + 4 – \left( {{x^2} – 9} \right) = 0\\
\Leftrightarrow {x^2} – 4x + 4 – {x^2} + 9 = 0\\
\Leftrightarrow 4x = 13\\
\Leftrightarrow x = \dfrac{{13}}{4}\\
Vậy\,x = \dfrac{{13}}{4}\\
b){\left( {3x – 1} \right)^2} – 16 = 0\\
\Leftrightarrow \left( {3x – 1 – 4} \right)\left( {3x – 1 + 4} \right) = 0\\
\Leftrightarrow \left( {3x – 5} \right)\left( {3x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 5\\
3x = – 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{3}\\
x = – 1
\end{array} \right.\\
Vậy\,x = \dfrac{5}{3};x = – 1\\
c)5{x^3} – 35x = 0\\
\Leftrightarrow 5x\left( {{x^2} – 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 7
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = – \sqrt 7 \\
x = \sqrt 7
\end{array} \right.\\
Vậy\,x = 0;x = \sqrt 7 ;x = – \sqrt 7 \\
d){\left( {x – 3} \right)^2} = 4{x^2} – 20x + 25\\
\Leftrightarrow {\left( {x – 3} \right)^2} = {\left( {2x – 5} \right)^2}\\
\Leftrightarrow {\left( {x – 3} \right)^2} – {\left( {2x – 5} \right)^2} = 0\\
\Leftrightarrow \left( {x – 3 – 2x + 5} \right)\left( {x – 3 + 2x – 5} \right) = 0\\
\Leftrightarrow \left( {2 – x} \right)\left( {3x – 8} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \dfrac{8}{3}
\end{array} \right.\\
Vậy\,x = 2;x = \dfrac{8}{3}\\
e)x + {x^2} – {x^3} – {x^4} = 0\\
\Leftrightarrow x\left( {1 + x} \right) – {x^3}\left( {1 + x} \right) = 0\\
\Leftrightarrow \left( {1 + x} \right).x.\left( {1 – {x^2}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
1 + x = 0\\
x = 0\\
1 – {x^2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
x = 0\\
x = 1
\end{array} \right.\\
Vậy\,x = 0;x = 1;x = – 1\\
f){\left( {7x – 5} \right)^2} – 2\left( {7x – 5} \right)\left( {x + 3} \right) + {\left( {x + 3} \right)^2} = 0\\
\Leftrightarrow {\left( {7x – 5 – x – 3} \right)^2} = 0\\
\Leftrightarrow {\left( {6x – 8} \right)^2} = 0\\
\Leftrightarrow 6x – 8 = 0\\
\Leftrightarrow x = \dfrac{4}{3}\\
Vậy\,x = \dfrac{4}{3}
\end{array}$