Rút gọn các biểu thức
a) 2x(2x-1)²-3x(x+3)(x-3)-4x(x+1)²
b) (a-b+c) ²-(b-c)²+2ab-2ac
c) (3x+1)²-2(3x+1)(3x+5)+(3x+5)²
d) (2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)
e) (3+1)(3²+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
f) 10(11+1)(11²+1)(11^4+1)(11^8+1)(11^16+1)(11^32+1)-11^64
g) (a+b-c)²+(a-b+c) ²-2(b-c)²
h) (x-2)(x²-2x+4)(x+2)(x²+2x+4)
Rút gọn các biểu thức a) 2x(2x-1)²-3x(x+3)(x-3)-4x(x+1)² b) (a-b+c) ²-(b-c)²+2ab-2ac c) (3x+1)²-2(3x+1)(3x+5)+(3x+5)² d) (2+1)(2²+1)(2^4+1)(2^8+1)(2
By Sadie
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
2x{\left( {2x – 1} \right)^2} – 3x\left( {x + 3} \right)\left( {x – 3} \right) – 4x{\left( {x + 1} \right)^2}\\
= 2x.\left( {4{x^2} – 4x + 1} \right) – 3x.\left( {{x^2} – 9} \right) – 4x.\left( {{x^2} + 2x + 1} \right)\\
= \left( {8{x^3} – 8{x^2} + 2x} \right) – \left( {3{x^3} – 27x} \right) – \left( {4{x^3} + 8{x^2} + 4x} \right)\\
= 8{x^3} – 8{x^2} + 2x – 3{x^3} + 27x – 4{x^3} – 8{x^2} – 4x\\
= {x^3} – 16{x^2} + 25x\\
b,\\
{\left( {a – b + c} \right)^2} – {\left( {b – c} \right)^2} + 2ab – 2ac\\
= \left[ {\left( {a – b + c} \right) + \left( {b – c} \right)} \right].\left[ {\left( {a – b + c} \right) – \left( {b – c} \right)} \right] + 2a.\left( {b – c} \right)\\
= a.\left( {a – 2b + 2c} \right) + 2a\left( {b – c} \right)\\
= a.\left[ {\left( {a – 2b + 2c} \right) + 2.\left( {b – c} \right)} \right]\\
= a.a = {a^2}\\
c,\\
{\left( {3x + 1} \right)^2} – 2.\left( {3x + 1} \right).\left( {3x + 5} \right) + {\left( {3x + 5} \right)^2}\\
= {\left[ {\left( {3x + 1} \right) – \left( {3x + 5} \right)} \right]^2}\\
= {\left( { – 4} \right)^2}\\
= 16\\
d,\\
\left( {2 + 1} \right).\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right).\left( {{2^{16}} + 1} \right)\\
= 1.\left( {2 + 1} \right).\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right).\left( {{2^{16}} + 1} \right)\\
= \left( {2 – 1} \right).\left( {2 + 1} \right).\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right).\left( {{2^{16}} + 1} \right)\\
= \left( {{2^2} – 1} \right).\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right).\left( {{2^{16}} + 1} \right)\\
= \left( {{2^4} – 1} \right)\left( {{2^4} + 1} \right).\left( {{2^8} + 1} \right).\left( {{2^{16}} + 1} \right)\\
= \left( {{2^8} – 1} \right)\left( {{2^8} + 1} \right).\left( {{2^{16}} + 1} \right)\\
= \left( {{2^{16}} – 1} \right).\left( {{2^{16}} + 1} \right)\\
= {2^{32}} – 1\\
e,\\
\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right).\left( {{3^{16}} + 1} \right).\left( {{3^{32}} + 1} \right)\\
= \frac{1}{2}.\left( {3 – 1} \right).\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right).\left( {{3^{16}} + 1} \right).\left( {{3^{32}} + 1} \right)\\
= \frac{1}{2}.\left( {{3^2} – 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right).\left( {{3^{16}} + 1} \right).\left( {{3^{32}} + 1} \right)\\
= \frac{1}{2}.\left( {{3^4} – 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right).\left( {{3^{16}} + 1} \right).\left( {{3^{32}} + 1} \right)\\
= \frac{1}{2}.\left( {{3^8} – 1} \right).\left( {{3^8} + 1} \right).\left( {{3^{16}} + 1} \right).\left( {{3^{32}} + 1} \right)\\
= \frac{1}{2}.\left( {{3^{16}} – 1} \right).\left( {{3^{16}} + 1} \right).\left( {{3^{32}} + 1} \right)\\
= \frac{1}{2}.\left( {{3^{32}} – 1} \right).\left( {{3^{32}} + 1} \right)\\
= \frac{1}{2}.\left( {{3^{64}} – 1} \right)\\
f,\\
10.\left( {11 + 1} \right).\left( {{{11}^2} + 1} \right).\left( {{{11}^4} + 1} \right).\left( {{{11}^8} + 1} \right).\left( {{{11}^{16}} + 1} \right).\left( {{{11}^{32}} + 1} \right) – {11^{64}}\\
= \left( {11 – 1} \right).\left( {11 + 1} \right).\left( {{{11}^2} + 1} \right).\left( {{{11}^4} + 1} \right).\left( {{{11}^8} + 1} \right).\left( {{{11}^{16}} + 1} \right).\left( {{{11}^{32}} + 1} \right) – {11^{64}}\\
= \left( {{{11}^2} – 1} \right).\left( {{{11}^2} + 1} \right).\left( {{{11}^4} + 1} \right).\left( {{{11}^8} + 1} \right).\left( {{{11}^{16}} + 1} \right).\left( {{{11}^{32}} + 1} \right) – {11^{64}}\\
= \left( {{{11}^4} – 1} \right).\left( {{{11}^4} + 1} \right).\left( {{{11}^8} + 1} \right).\left( {{{11}^{16}} + 1} \right)\left( {{{11}^{32}} + 1} \right) – {11^{64}}\\
= \left( {{{11}^8} – 1} \right).\left( {{{11}^8} + 1} \right).\left( {{{11}^{16}} + 1} \right)\left( {{{11}^{32}} + 1} \right) – {11^{64}}\\
= \left( {{{11}^{16}} – 1} \right)\left( {{{11}^6} + 1} \right).\left( {{{11}^{32}} + 1} \right) – {11^{64}}\\
= \left( {{{11}^{32}} – 1} \right).\left( {{{11}^{32}} + 1} \right) – {11^{64}}\\
= \left( {{{11}^{64}} – 1} \right) – {11^{64}}\\
= – 1\\
g,\\
{\left( {a + b – c} \right)^2} + {\left( {a – b + c} \right)^2} – 2.{\left( {b – c} \right)^2}\\
= \left[ {{{\left( {a + b – c} \right)}^2} – {{\left( {b – c} \right)}^2}} \right] + \left[ {{{\left( {a – b + c} \right)}^2} – {{\left( {b – c} \right)}^2}} \right]\\
= \left[ {\left( {a + b – c} \right) – \left( {b – c} \right)} \right].\left[ {\left( {a + b – c} \right) + \left( {b – c} \right)} \right] + \left[ {\left( {a – b + c} \right) – \left( {b – c} \right)} \right].\left[ {\left( {a – b + c} \right) + \left( {b – c} \right)} \right]\\
= a.\left( {a + 2b – 2c} \right) + \left( {a – 2b + 2c} \right).a\\
= a.\left[ {\left( {a + 2b – 2c} \right) + \left( {a – 2b + 2c} \right)} \right]\\
= a.2a\\
= 2{a^2}\\
h,\\
\left( {x – 2} \right)\left( {{x^2} – 2x + 4} \right).\left( {x + 2} \right).\left( {{x^2} + 2x + 4} \right)\\
= \left[ {\left( {x – 2} \right)\left( {{x^2} + 2x + 4} \right)} \right].\left[ {\left( {x + 2} \right).\left( {{x^2} – 2x + 4} \right)} \right]\\
= \left( {{x^3} – 8} \right).\left( {{x^3} + 8} \right) = {x^6} – 64
\end{array}\)
Đáp án:
a) 2x(2x-1)²-3x(x+3)(x-3)-4x(x+1)²
=2x(4x²-4x+1)-3x(x²-9)-4x(x²+2x+1)
=8x³-8x²+2x-3x³+27x-4x³-8x²-4x
=x³-16x²+25x
=x(x²-16x+25)
b) (a-b+c) ²-(b-c)²+2ab-2ac
=a²+b²+c²-2ab-2bc+2ac-b²+2bc-c²+2ab-2ac
=a²
c)(3x+1)²-2(3x+1)(3x+5)+(3x+5)²
=9x²+6x+1-2(3x+3-2)(3x+3+2)+9x²+30x+25
=18x²+36x+26-2((3x+3)²-4)
=18x²+36x+26-2(9x²+18x+9-4)
=18x²+36x+26-2(9x²+18x+5)
=18x²+36x+26-18x²-36x-10
=16
d) (2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)
=1(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)
=(2-1)(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)
=(2²-1)(2²+1)(2^4+1)(2^8+1)(2^16+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
=(2^16-1)(2^16+1)
=2^32-1
Giải thích các bước giải:
chúc bn hk tốt