Toán Lớp 8: Bài 1: phân tích đa thức thành nhân tử
A= y^2+x^2-16-2xy
B= x^2-8-3 (2-x)
Bài 2: tìm x biết
a, x(5-6x)+(2x-1)(3x+4)=6
B, x^2 (x-2021)-x+2021=0
c, 2x^2-3x-5=0
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Toán Lớp 8: Bài 1: phân tích đa thức thành nhân tử
A= y^2+x^2-16-2xy
B= x^2-8-3 (2-x)
Bài 2: tìm x biết
a, x(5-6x)+(2x-1)(3x+4)=6
B, x^2 (x-2021)-x+2021=0
c, 2x^2-3x-5=0
Comments ( 1 )
1)A = {y^2} + {x^2} – 16 – 2xy\\
= {x^2} – 2xy + {y^2} – 16\\
= {\left( {x – y} \right)^2} – 16\\
= \left( {x – y – 4} \right)\left( {x – y + 4} \right)\\
B = {x^2} – 8 – 3\left( {2 – x} \right)\\
= \left( {x – 2} \right)\left( {{x^2} + 2x + 4} \right) + 3\left( {x – 2} \right)\\
= \left( {x – 2} \right)\left( {{x^2} + 2x + 4 + 3} \right)\\
= \left( {x – 2} \right)\left( {{x^2} + 2x + 7} \right)\\
B2)\\
a)x\left( {5 – 6x} \right) + \left( {2x – 1} \right)\left( {3x + 4} \right) = 6\\
\Leftrightarrow 5x – 6{x^2} + 6{x^2} + 8x – 3x – 4 = 6\\
\Leftrightarrow 10x = 10\\
\Leftrightarrow x = 1\\
Vậy\,x = 1\\
b){x^2}\left( {x – 2021} \right) – x + 2021 = 0\\
\Leftrightarrow \left( {x – 2021} \right)\left( {{x^2} – 1} \right) = 0\\
\Leftrightarrow \left( {x – 2021} \right)\left( {x – 1} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow x = 2021;x = 1;x = – 1\\
Vậy\,x = 2021;x = 1;x = – 1\\
c)2{x^2} – 3x – 5 = 0\\
\Leftrightarrow 2{x^2} – 5x + 2x – 5 = 0\\
\Leftrightarrow \left( {2x – 5} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow x = \dfrac{5}{2};x = – 1\\
Vậy\,x = \dfrac{5}{2};x = – 1
\end{array}$