Toán Lớp 8: 1. Chứng minh : (a – b)^3 + 3ab(a – b) = a^3 + b^3
2. Rút gọn: (x – 3)^3 – (x + 3)^3.
3. Cho a – b = 1.Chứng minh : a^3 – b^3 = 1 + 3ab.
4. Tìm x,biết : x3 – 3×2 + 3x – 1 = 0.
5. Chứng minh rằng biểu thức sau không phụ thuộc vào x:
(4x-1)^3-(4x-3)(16x^2+3)
6. Rút gọn biểu thức : (x + 5)^3 – x^3 – 125.
7. Tìm x, biết : (x – 2)^3 + 6(x + 1)^2 – x^3 + 12 = 0
8. Chứng minh rằng biểu thức sau không phụ thuộc vào x:
(x-1)^3-x^3+3x^2-3x-1
9. Tìm x,biết : x^3 + 6x^2 + 12x +8 = 0
10. Cho a +b +c = 0.Chứng minh : a^3 + b^3 + c^3 = 3abc.
11. Chứng minh rằng: (a + 2)^3 – (a +6)(a^2 +12) + 64 = 0,với mọi a.
12. Rút gọn biểu thức :
A = (m – n)(m2 + mn + n2) – (m + n)(m2 – mn + n2)
13. Chứng minh: (a – 1)(a – 2)(1 + a + a^2)(4 + 2a + a^2) = a^6 – 9a^3 + 8
14. Tìm x, biết : (x +2 )(x2 – 2x + 4) – x(x -3)(x + 3) = 26.
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1,\\
{\left( {a – b} \right)^3} + 3ab\left( {a – b} \right) = {a^3} – {b^3}\\
2,\\
– 18{x^2} – 54\\
3,\\
{a^3} – {b^3} = 1 + 3ab\\
4,\\
x = 1\\
5,\\
8\\
6,\\
15{x^2} + 75x\\
7,\\
x = – \dfrac{5}{{12}}\\
8,\\
– 2\\
9,\\
x = – 2\\
10,\\
{a^3} + {b^3} + {c^3} = 3abc\,\,\,\,\,\,\,\,\,\,\left( {a + b + c = 0} \right)\\
11,\\
{\left( {a + 2} \right)^3} – \left( {a + 6} \right)\left( {{a^2} + 12} \right) + 64 = 0\\
12,\\
A = – 2{n^3}\\
13,\\
\left( {a – 1} \right)\left( {a – 2} \right)\left( {1 + a + {a^2}} \right)\left( {4 + 2a + {a^2}} \right) = {a^6} – 9{a^3} + 8\\
14,\\
x = 2
\end{array}\)
1,\\
{\left( {a – b} \right)^3} + 3ab\left( {a – b} \right)\\
= \left( {{a^3} – 3{a^2}b + 3a{b^2} – {b^3}} \right) + \left( {3{a^2}b – 3a{b^2}} \right)\\
= {a^3} + \left( { – 3{a^2}b + 3a{b^2}} \right) + \left( {3a{b^2} – 3a{b^2}} \right) – {b^3}\\
= {a^3} + 0 + 0 – {b^3}\\
= {a^3} – {b^3}\\
2,\\
{\left( {x – 3} \right)^3} – {\left( {x + 3} \right)^3}\\
= \left( {{x^3} – 3.{x^2}.3 + 3.x{{.3}^2} – {3^3}} \right) – \left( {{x^3} + 3.{x^2}.3 + 3.x{{.3}^2} + {3^3}} \right)\\
= \left( {{x^3} – 9{x^2} + 27x – 27} \right) – \left( {{x^3} + 9{x^2} + 27x + 27} \right)\\
= {x^3} – 9{x^2} + 27x – 27 – {x^3} – 9{x^2} – 27x – 27\\
= – 18{x^2} – 54\\
3,\\
{a^3} – {b^3} = \left( {a – b} \right)\left( {{a^2} + ab + {b^2}} \right)\\
= 1.\left( {{a^2} + ab + {b^2}} \right)\\
= {a^2} + ab + {b^2}\\
= \left( {{a^2} – 2ab + {b^2}} \right) + 3ab\\
= {\left( {a – b} \right)^2} + 3ab\\
= {1^2} + 3ab\\
= 1 + 3ab\\
4,\\
{x^3} – 3{x^2} + 3x – 1 = 0\\
\Leftrightarrow {x^3} – 3.{x^2}.1 + 3.x{.1^2} – {1^3} = 0\\
\Leftrightarrow {\left( {x – 1} \right)^3} = 0\\
\Leftrightarrow x – 1 = 0\\
\Leftrightarrow x = 1\\
5,\\
{\left( {4x – 1} \right)^3} – \left( {4x – 3} \right)\left( {16{x^2} + 3} \right)\\
= \left[ {{{\left( {4x} \right)}^3} – 3.{{\left( {4x} \right)}^2}.1 + 3.4x{{.1}^2} – {1^3}} \right] – \left( {64{x^3} + 12x – 48{x^2} – 9} \right)\\
= \left( {64{x^3} – 48{x^2} + 12x – 1} \right) – \left( {64{x^3} + 12x – 48{x^2} – 9} \right)\\
= 64{x^3} – 48{x^2} + 12x – 1 – 64{x^3} – 12x + 48{x^2} + 9\\
= 8\\
6,\\
{\left( {x + 5} \right)^3} – {x^3} – 125\\
= \left( {{x^3} + 3.{x^2}.5 + 3.x{{.5}^2} + {5^3}} \right) – {x^3} – 125\\
= \left( {{x^3} + 15{x^2} + 75x + 125} \right) – {x^3} – 125\\
= 15{x^2} + 75x\\
7,\\
{\left( {x – 2} \right)^3} + 6.{\left( {x + 1} \right)^2} – {x^3} + 12 = 0\\
\Leftrightarrow \left( {{x^3} – 3.{x^2}.2 + 3.x{{.2}^2} – {2^3}} \right) + 6.\left( {{x^2} + 2.x.1 + {1^2}} \right) – {x^3} + 12 = 0\\
\Leftrightarrow \left( {{x^3} – 6{x^2} + 12x – 8} \right) + 6.\left( {{x^2} + 2x + 1} \right) – {x^3} + 12 = 0\\
\Leftrightarrow {x^3} – 6{x^2} + 12x – 8 + 6{x^2} + 12x + 6 – {x^3} + 12 = 0\\
\Leftrightarrow 24x + 10 = 0\\
\Leftrightarrow 24x = – 10\\
\Leftrightarrow x = – \dfrac{5}{{12}}\\
8,\\
{\left( {x – 1} \right)^3} – {x^3} + 3{x^2} – 3x – 1\\
= \left( {{x^3} – 3.{x^2}.1 + 3.x{{.1}^2} – {1^3}} \right) – {x^3} + 3{x^2} – 3x – 1\\
= \left( {{x^3} – 3{x^2} + 3x – 1} \right) – {x^3} + 3{x^2} – 3x – 1\\
= – 2\\
9,\\
{x^3} + 6{x^2} + 12x + 8 = 0\\
\Leftrightarrow {x^3} + 3.{x^2}.2 + 3.x{.2^2} + {2^3} = 0\\
\Leftrightarrow {\left( {x + 2} \right)^3} = 0\\
\Leftrightarrow x + 2 = 0\\
\Leftrightarrow x = – 2\\
10,\\
{a^3} + {b^3} + {c^3}\\
= \left( {{a^3} + 3{a^2}b + 3a{b^2} + {b^3}} \right) – 3{a^2}b – 3a{b^2} + {c^3}\\
= {\left( {a + b} \right)^3} – \left( {3{a^2}b + 3a{b^2}} \right) + {c^3}\\
= \left[ {{{\left( {a + b} \right)}^3} + {c^3}} \right] – \left( {3{a^2}b + 3a{b^2}} \right)\\
= \left[ {\left( {a + b} \right) + c} \right].\left[ {{{\left( {a + b} \right)}^2} – \left( {a + b} \right).c + {c^2}} \right] – 3ab\left( {a + b} \right)\\
= \left( {a + b + c} \right).\left[ {{{\left( {a + b} \right)}^2} – \left( {a + b} \right).c + {c^2}} \right] – 3ab\left( {a + b} \right)\\
= 0.\left[ {{{\left( {a + b} \right)}^2} – \left( {a + b} \right).c + {c^2}} \right] – 3ab\left( {a + b} \right)\\
= – 3ab\left( {a + b} \right)\\
a + b + c = 0 \Rightarrow a + b = – c\\
\Rightarrow – 3ab\left( {a + b} \right) = – 3ab.\left( { – c} \right) = 3abc\\
\Rightarrow {a^3} + {b^3} + {c^3} = 3abc\\
11,\\
{\left( {a + 2} \right)^3} – \left( {a + 6} \right)\left( {{a^2} + 12} \right) + 64\\
= \left( {{a^3} + 3.{a^2}.2 + 3.a{{.2}^2} + {2^3}} \right) – \left( {{a^3} + 12a + 6{a^2} + 72} \right) + 64\\
= \left( {{a^3} + 6{a^2} + 12a + 8} \right) – \left( {{a^3} + 12a + 6{a^2} + 72} \right) + 64\\
= {a^3} + 6{a^2} + 12a + 8 – {a^3} – 12a – 6{a^2} – 72 + 64\\
= 0\\
12,\\
A = \left( {m – n} \right)\left( {{m^2} + mn + {n^2}} \right) – \left( {m + n} \right).\left( {{m^2} – mn + {n^2}} \right)\\
= \left( {{m^3} – {n^3}} \right) – \left( {{m^3} + {n^3}} \right)\\
= {m^3} – {n^3} – {m^3} – {n^3}\\
= – 2{n^3}\\
13,\\
\left( {a – 1} \right)\left( {a – 2} \right)\left( {1 + a + {a^2}} \right)\left( {4 + 2a + {a^2}} \right)\\
= \left[ {\left( {a – 1} \right)\left( {1 + a + {a^2}} \right)} \right].\left[ {\left( {a – 2} \right)\left( {4 + 2a + {a^2}} \right)} \right]\\
= \left[ {\left( {a – 1} \right).\left( {{a^2} + a.1 + {1^2}} \right)} \right].\left[ {\left( {a – 2} \right).\left( {{a^2} + a.2 + {2^2}} \right)} \right]\\
= \left( {{a^3} – {1^3}} \right).\left( {{a^3} – {2^3}} \right)\\
= \left( {{a^3} – 1} \right)\left( {{a^3} – 8} \right)\\
= {a^6} – 8{a^3} – {a^3} + 8\\
= {a^6} – 9{a^3} + 8\\
14,\\
\left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right) – x\left( {x – 3} \right)\left( {x + 3} \right) = 26\\
\Leftrightarrow \left( {x + 2} \right).\left( {{x^2} – x.2 + {2^2}} \right) – x.\left( {{x^2} – {3^2}} \right) = 26\\
\Leftrightarrow \left( {{x^3} + {2^3}} \right) – \left( {{x^3} – 9x} \right) = 26\\
\Leftrightarrow {x^3} + 8 – {x^3} + 9x = 26\\
\Leftrightarrow 9x + 8 = 26\\
\Leftrightarrow 9x = 18\\
\Leftrightarrow x = 2
\end{array}\)