Toán Lớp 8: 1Rút gọn biểu thức a) (3x+1)^2+(3x-1)^2-2(3x+1)(3x-1)
b) 8(3^2+1)(3^4+1)…(2^16+1)
c ) (2^2+1)(2^4+1)…(2^32+1)
2 Tìm x biết
a) x(2x-1)-2x+1=0
b) 3x(x-1)=x-1
c) 3(x+2)-x^2-2x=0
d) x^3+x=0
3 Phân tích thành nhân tử
a) 4x^3-x
b) 6x^2-12xy+6y^2-24z^2
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Comments ( 1 )
1)a){\left( {3x + 1} \right)^2} + {\left( {3x – 1} \right)^2} – 2\left( {3x + 1} \right)\left( {3x – 1} \right)\\
= {\left( {3x + 1 – 3x + 1} \right)^2}\\
= {2^2}\\
= 4\\
b)\\
8\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)…\left( {{3^{16}} + 1} \right)\\
= \left( {{3^2} – 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)…\left( {{3^{16}} + 1} \right)\\
= \left( {{3^4} – 1} \right)\left( {{3^4} + 1} \right)…\left( {{3^{16}} + 1} \right)\\
= \left( {{3^8} – 1} \right).\left( {{3^8} + 1} \right).\left( {{3^{16}} + 1} \right)\\
= \left( {{3^{16}} – 1} \right).\left( {{3^{16}} + 1} \right)\\
= {3^{32}} – 1\\
c)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)…\left( {{2^{32}} + 1} \right)\\
= \dfrac{1}{3}.\left( {{2^2} – 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)…\left( {{2^{32}} + 1} \right)\\
= \dfrac{1}{3}.\left( {{2^4} – 1} \right)\left( {{2^4} + 1} \right)…\left( {{2^{32}} + 1} \right)\\
= \dfrac{1}{3}\left( {{2^{32}} – 1} \right)\left( {{2^{32}} + 1} \right)\\
= \dfrac{1}{3}.\left( {{2^{64}} – 1} \right)\\
= \dfrac{1}{3}{.2^{64}} – \dfrac{1}{3}\\
2)\\
a)x\left( {2x – 1} \right) – 2x + 1 = 0\\
\Leftrightarrow \left( {2x – 1} \right)\left( {x – 1} \right) = 0\\
\Leftrightarrow x = \dfrac{1}{2};x = 1\\
Vậy\,x = \dfrac{1}{2};x = 1\\
b)3x\left( {x – 1} \right) = x – 1\\
\Leftrightarrow \left( {x – 1} \right)\left( {3x – 1} \right) = 0\\
\Leftrightarrow x = 1;x = \dfrac{1}{3}\\
Vậy\,x = 1;x = \dfrac{1}{3}\\
c)3\left( {x + 2} \right) – {x^2} – 2x = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {3 – x} \right) = 0\\
\Leftrightarrow x = – 2;x = 3\\
Vậy\,x = – 2;x = 3\\
d){x^3} + x = 0\\
\Leftrightarrow x\left( {{x^2} + 1} \right) = 0\\
\Leftrightarrow x = 0\\
Vậy\,x = 0\\
3)a)4{x^3} – x = x\left( {4{x^2} – 1} \right)\\
= x\left( {2x – 1} \right)\left( {2x + 1} \right)\\
b)6{x^2} – 12xy + 6{y^2} – 24{z^2}\\
= 6\left( {{x^2} – 2xy + {y^2} – 4{z^2}} \right)\\
= 6.\left( {{{\left( {x – y} \right)}^2} – {{\left( {2z} \right)}^2}} \right)\\
= 6.\left( {x – y – 2z} \right)\left( {x – y + 2z} \right)
\end{array}$