Toán Lớp 8: 1) Phân tích đa thức thành nhân tử.
a) 14x^2y – 21xy^2 + 28x^2y^2
b) 6x^2 – 9x^3
c) 2x(x + 1) + 2(x + 1)
d) 16a^2 – 9b^2
e) 9x^2 + 6x + 1
f) 9x^2-6xy+y^2
g) 7x^2 – 63y^2
h) 10x(x – y) – 8(y – x)
2) Tìm x
a) x^2 – 7x = 0
b) -3x^2 + 5x = 0
c) x(x + 4 ) – 5( x + 4) = 0
d) 3x(x – 7) – 2(x – 7) = 0
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1)a)\\
14{x^2}y – 21x{y^2} + 28{x^2}{y^2}\\
= 7xy\left( {2x – 3y + 4xy} \right)\\
b)6{x^2} – 9{x^3}\\
= 3{x^2}\left( {2 – 3x} \right)\\
c)2x\left( {x + 1} \right) + 2\left( {x + 1} \right)\\
= 2\left( {x + 1} \right)\left( {x + 1} \right)\\
= 2{\left( {x + 1} \right)^2}\\
d)16{a^2} – 9{b^2}\\
= \left( {4a – 3b} \right)\left( {4a + 3b} \right)\\
e)9{x^2} + 6x + 1\\
= {\left( {3x + 1} \right)^2}\\
f)9{x^2} – 6xy + {y^2}\\
= {\left( {3x – y} \right)^2}\\
g)7{x^2} – 63{y^2}\\
= 7\left( {{x^2} – 9{y^2}} \right)\\
= 7\left( {x – 3y} \right)\left( {x + 3y} \right)\\
h)10x\left( {x – y} \right) – 8\left( {y – x} \right)\\
= 2\left( {x – y} \right)\left( {5x + 4} \right)\\
B2)\\
a){x^2} – 7x = 0\\
\Leftrightarrow x\left( {x – 7} \right) = 0\\
\Leftrightarrow x = 0;x = 7\\
Vậy\,x = 0;x = 7\\
b)\\
– 3{x^2} + 5x = 0\\
\Leftrightarrow x\left( { – 3x + 5} \right) = 0\\
\Leftrightarrow x = 0;x = \dfrac{5}{3}\\
Vậy\,x = 0;x = \dfrac{5}{3}\\
c)x\left( {x + 4} \right) – 5\left( {x + 4} \right) = 0\\
\Leftrightarrow \left( {x + 4} \right)\left( {x – 5} \right) = 0\\
\Leftrightarrow x = – 4;x = 5\\
Vậy\,x = – 4;x = 5\\
d)3x\left( {x – 7} \right) – 2\left( {x – 7} \right) = 0\\
\Leftrightarrow \left( {x – 7} \right)\left( {3x – 2} \right) = 0\\
\Leftrightarrow x = 7;x = \dfrac{2}{3}\\
Vậy\,x = 7;x = \dfrac{2}{3}
\end{array}$