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)
Đáp án:
$a) x^3-16x^2+25x\\
b) a^2\\
c)16\\
d) 2^{32}-1\\
e)
\dfrac{1}{2}.(3^{64}-1)\\
f)
-1\\
g) 2a^2-c^2\\
h)
x^6-64$
Giải thích các bước giải:
$a) 2x(2x-1)^2-3x(x+3)(x-3)-4x(x+1)^2\\
=2x(4x^2-4x+1)-3x(x^2-9)-4x(x^2+2x+1)\\
=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\\
=(8x^3-3x^3-4x^3)+(-8x^2-8x^2)+(2x+27x-4x)\\
=x^3-16x^2+25x\\
b) (a-b+c)^2-(b-c)^2+2ab-2ac\\
=a^2-2a.(b-c)+(b-c)^2-(b-c)^2+2ab-2ac\\
=a^2-2ab+2ac+2ab-2ac\\
=a^2\\
c) (3x+1)^2-2(3x+1)(3x+5)+(3x+5)^2\\
=(3x+1-3x-5)^2\\
=(-4)^2=16\\
d) (2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)\\
=3.(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)\\
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)\\
=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)\\
=(2^8-1)(2^8+1)(2^{16}+1)\\
=(2^{16}-1)(2^{16}+1)\\
=2^{32}-1\\
e)
(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\\
=\dfrac{1}{2}.2.4.(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\\
=\dfrac{1}{2}.(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\\
=\dfrac{1}{2}.(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\\
=\dfrac{1}{2}.(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)\\
=\dfrac{1}{2}.(3^{16}-1)(3^{16}+1)(3^{32}+1)\\
=\dfrac{1}{2}.(3^{32}-1)(3^{32}+1)\\
=\dfrac{1}{2}.(3^{64}-1)\\
f)
10(11+1)(11^2+1)(11^4+1)(11^8+1)(11^{16}+1)(11^{32}+1)-11^{64}\\
=(11-1)(11+1)(11^2+1)(11^4+1)(11^8+1)(11^{16}+1)(11^{32}+1)-11^{64}\\
=(11^2-1)(11^2+1)(11^4+1)(11^8+1)(11^{16}+1)(11^{32}+1)-11^{64}\\
=(11^4-1)(11^4+1)(11^8+1)(11^{16}+1)(11^{32}+1)-11^{64}\\
=(11^8-1)(11^8+1)(11^{16}+1)(11^{32}+1)-11^{64}\\
=(11^{16}-1)(11^{16}+1)(11^{32}+1)-11^{64}\\
=(11^{32}-1)(11^{32}+1)-11^{64}\\
=11^{64}-1-11^{64}\\
=-1\\
g) (a+b-c)^2+(a-b+c)^2-2(b-c)^2\\
=(a+b)^2-2(a+b).c+c^2+(a-b)^2+2.(a-b).c+c^2-2(b^2-2bc+c^2)\\
=a^2+2ab+b^2-2ac-2bc+a^2-2ab+b^2+2ac-2bc+c^2-2b^2+4bc-2c^2\\
=(a^2+a^2)+(2ab-2ab)+(b^2+b^2-2b^2)+(-2ac+2ac)+(-2bc-2bc+4bc)+(c^2-2c^2)\\
=2a^2-c^2\\
h)
(x-2)(x^2-2x+4)(x+2)(x^2+2x+4)\\
=(x^3-2^3)(x^3+2^3)\\
=(x^3)^2-8^2\\
=x^6-64$
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
e) (3+1)(3²+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
=$\frac{1}{2}$(3-1)(3+1)(3²+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
=$\frac{1}{2}$(3²-1) (3²+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
=$\frac{1}{2}$(3^4-1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
=$\frac{1}{2}$(3^8-1)(3^8+1)(3^16+1)(3^32+1)
=$\frac{1}{2}$(3^16-1)(3^16+1)(3^32+1)
=$\frac{1}{2}$(3^32-1)(3^32+1)
=$\frac{1}{2}$(3^64-1)
h) (x-2)(x²-2x+4)(x+2)(x²+2x+4)
=((x-2)(x²+2x+4))((x+2)(x²-2x+4))
=(x³-8)(x³+8)
=x^6-64
Giải thích các bước giải:
chúc bn hk tốt