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\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1-6x+1\right)^2\)
\(=2^2=4\)
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)\left(6x+1\right)+\left(6x-1\right)\left(6x-1\right)-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)\left(6x+1\right)+\left(6x-1\right)\left[\left(6x-1\right)-2\left(1+6x\right)\right]\)
\(=\left(6x+1\right)\left(6x+1\right)+\left(6x-1\right)\left(6x-1-2-12x\right)\)
\(=36x^2+12x+1+\left(6x-1\right)\left(-6x-3\right)\)
\(=36x^2+12x+1+\left(-36x^2-12x+3\right)\)
\(=36x^2+12x+1-36x^2-12x+3\)
\(=4\)
$(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$
$=(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2$
$=2^2$
$=4$
b)$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3\cdot\dfrac{2^4-1}{2^2-1}\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$
$=3\cdot\dfrac{2^{32}-1}{2^2-1}$
$=3\cdot\dfrac{2^{32}-1}{3}$
$=2^{32}-1$
$x(2x^2-3)-x^2(5x+1)+x^2$
$=2x^3-3x-5x^3-x^2+x^2$
$=2x^3-5x^3-3x$
$=-3x^3-3x$
$=-3x(x^2+1)$
d)$3x(x-2)-5x(1-x)-8(x^2-3)$
$=3x^2-6x-5x+5x^2-8x^2+24$
$=3x^2+5x^2-8x^2-11x+24$
$=-11x+24$
$=24-11x$
Bài 2
a. (x-2y)2 =2x-4y
b. (2x^2 +3)2 =4x^2+6
c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)
d. (2x-1)3 = 6x-3
Xin lỗi mik chỉ lm ổn bài 2 thôi!
$(x-2y)^2$
$=x^2-2\cdot x\cdot 2y+(2y)^2$
$=x^2-4xy+4y^2$
b)$(2x^2+3)^2$
$=(2x^2)^2+2\cdot2x^2\cdot3+3^2$
$=4x^4+12x^2+9$
c)$(x-2)(x^2+2x+4)$
$=x(x^2+2x+4)-2(x^2+2x+4)$
$=x^3+2x^2+4x-2x^2-4x-8$
$=x^3-8$
d)$(2x-1)^3$
$=(2x)^3-3(2x)^2\cdot1+3\cdot2x\cdot1^2-1^3$
$=8x^3-12x^2+6x-1$
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=36x^2+12x+1+36x^2-12x+1-72x^2+2\)
\(=4\)
c) \(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=-3x^3-3x\)
d) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
$(6x+1)^2-(6x-1)-2(1+6x)(6x-1)$
$=36x^2+12x+1-6x+1-2(6x+1)(6x-1)$
$=36x^2+6x+2-2(36x^2-1)$
$=36x^2+6x+2-72x^2+2$
$=-36x^2+6x+4$
b)$x(2x^2-3)-x^2(5x+1)+x^2$
$=2x^3-3x-5x^3-x^2+x^2$
$=2x^3-5x^3-3x$
$=-3x^3-3x$
$=-3x(x^2+1)$
$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3\cdot\dfrac{2^4-1}{2^2-1}\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$
$=3\cdot\dfrac{2^{32}-1}{2^2-1}$
$=3\cdot\dfrac{2^{32}-1}{3}$
$=2^{32}-1$
d)$3x(x-2)-5x(1-x)-8(x^2-3)$
$=3x^2-6x-5x+5x^2-8x^2+24$
$=3x^2+5x^2-8x^2-11x+24$
$=-11x+24$
$=24-11x$
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
$x^4-12x^3+12x^2-12x+111$ tại $x=11$
$=11^4-12\cdot11^3+12\cdot11^2-12\cdot11+111$
$=14641-15972+1452-132+111$
$=100$
$(6x+1)^2+(6x-1)^2-2(1-6x)(6x-1)$
$=(6x+1)^2+(6x-1)^2+2(6x-1)^2$
$=(6x+1)^2+3(6x-1)^2$
$=36x^2+12x+1+3(36x^2-12x+1)$
$=36x^2+12x+1+108x^2-36x+3$
$=144x^2-24x+4$
$=4(36x^2-6x+1)$
b)$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3\cdot\dfrac{2^4-1}{2^2-1}\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$
$=3\cdot\dfrac{2^{32}-1}{2^2-1}$
$=3\cdot\dfrac{2^{32}-1}{3}$
$=2^{32}-1$
\(a,=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-2b^3=6a^2b\\ b,=\left(6x+1-6x+1\right)^2=2^2=4\)