Toán Lớp 8: Bài 1:Dùng HĐT Khai Triển
b, (x-3y)(x+3y)
c, (5-x)²
d,(x- 1212)
e,(x+ 1212 )
g, (y-2)³
h, 8x³ + 1
f, 27x³-y³
Bài 2 : Viết các đa thức sau dưới dạng một tích
a, (x² – 6x +9)
b, 4x² + 12xy +9y²
d) 8x’ +12x²y +6xy² + y³
e)x¹-9x² +27x-27
c, 64x² – 25
Bài 3: Rút gọn biểu thức
a, (x + y)² + (x – y)²
b, 2(x-y)(x+y)+(x+y)² + (x-y)²
c, x(x+4)(x-4)-(x² + 1)(x²-1)
d, (a+b-c)-(a-c)² – 2ab + 2ab
e) (x +y)³ – (x-y)³-2y³ g) (a+b- c)² + (a-b+c)² -2 (b- c)²
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Comments ( 2 )
1)b)\left( {x – 3y} \right)\left( {x + 3y} \right)\\
= {x^2} – {\left( {3y} \right)^2}\\
= {x^2} – 9{y^2}\\
c){\left( {5 – x} \right)^2} = {x^2} – 10x + 25\\
d)\left( {x – 1212} \right).\left( {x + 1212} \right)\\
= {x^2} – {1212^2}\\
= {x^2} – 1468944\\
g){\left( {y – 2} \right)^3} = {y^3} – 6{y^2} + 12y – 8\\
h)8{x^3} + 1 = \left( {2x + 1} \right)\left( {4{x^2} – 2x + 1} \right)\\
f)27{x^3} – {y^3}\\
= \left( {3x – y} \right)\left( {9{x^2} + 3xy + y} \right)\\
B2)\\
a)\left( {{x^2} – 6x + 9} \right) = {\left( {x – 3} \right)^2}\\
b)4{x^2} + 12xy + 9{y^2} = {\left( {2x + 3y} \right)^2}\\
d)8{x^3} + 12{x^2}y + 6x{y^2} + {y^3} = {\left( {2x + y} \right)^3}\\
e){x^3} – 9{x^2} + 27x – 27 = {\left( {x – 3} \right)^3}\\
c)64{x^2} – 25 = \left( {8x – 5} \right)\left( {8x + 5} \right)\\
B3)\\
a){\left( {x + y} \right)^2} + {\left( {x – y} \right)^2}\\
= {x^2} + 2xy + {y^2} + {x^2} – 2xy + {y^2}\\
= 2{x^2} + 2{y^2}\\
b)2\left( {x – y} \right)\left( {x + y} \right) + {\left( {x + y} \right)^2} + {\left( {x – y} \right)^2}\\
= {\left( {x + y + x – y} \right)^2}\\
= {\left( {2x} \right)^2}\\
= 4{x^2}\\
c)x\left( {x + 4} \right)\left( {x – 4} \right) – \left( {{x^2} + 1} \right)\left( {{x^2} – 1} \right)\\
= x\left( {{x^2} – 16} \right) – \left( {{x^4} – 1} \right)\\
= {x^3} – 16x – {x^4} + 1\\
= – {x^4} + {x^3} – 16x + 1\\
d)\left( {a + b – c} \right) – {\left( {a – c} \right)^2} – 2ab + 2ab\\
= a + b – c – {a^2} + 2ac + {c^2}\\
e){\left( {x + y} \right)^3} – {\left( {x – y} \right)^3} – 2{y^3}\\
= {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\\
– {x^3} + 3{x^2}y – 3x{y^2} + {y^3} – 2{y^3}\\
= 6{x^2}y\\
g){\left( {a + b – c} \right)^2} + {\left( {a – b + c} \right)^2} – 2{\left( {b – c} \right)^2}\\
= {a^2} + 2a\left( {b – c} \right) + {\left( {b – c} \right)^2}\\
+ {a^2} – 2a\left( {b – c} \right) + {\left( {b – c} \right)^2}\\
– 2{\left( {b – c} \right)^2}\\
= 2{a^2}
\end{array}$