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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)$(2x-1)^3$
$=(2x)^3-3(2x)^2\cdot1+3\cdot2x\cdot1^2-1^3$
$=8x^3-12x^2+6x-1$
$\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}$
$=\dfrac{5(x+2)}{4(x-2)}\cdot\dfrac{-2(x-2)}{x+2}$
$=-\dfrac{10}{4}$
$=-\dfrac{5}{2}$
$\dfrac{6x^2y^3}{8x^3y^2}$
$=\dfrac{3y}{4x}$
b)$\dfrac{x^3-x}{3x+3}$
$=\dfrac{x(x^2-1)}{3(x+1)}$
$=\dfrac{x(x-1)(x+1)}{3(x+1)}$
$=\dfrac{x(x-1)}{3}$
c)$\dfrac{x^2+3xy}{x^2-9y^2}$
$=\dfrac{x(x+3y)}{(x-3y)(x+3y)}$
$=\dfrac{x}{x-3y}$
d)$\dfrac{x^2+4x+4}{3x+6}$
$=\dfrac{(x+2)^2}{3(x+2)}$
$=\dfrac{x+2}{3}$
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)$3x(x-2)-5x(1-x)-8(x^2-3)$
$=3x^2-6x-5x+5x^2-8x^2+24$
$=3x^2+5x^2-8x^2-11x+24$
$=-11x+24$
$=24-11x$
a) ( 6x + 1 )2 + ( 6x - 1 )2 - 2( 1 + 6x )( 6x - 1 )
= ( 6x + 1 )2 - 2( 1 + 6x )( 6x - 1 ) + ( 6x - 1 )2
= ( 6x + 1 - 6x + 1 )2 = 22 = 4
$(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$
$=(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2$
$=2^2$
$=4$
2.$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3\cdot\dfrac{2^4-1}{2^2-1}\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$
$=3\cdot\dfrac{2^{32}-1}{2^2-1}$
$=3\cdot\dfrac{2^{32}-1}{3}$
$=2^{32}-1$
Câu 1 :
\(\left(x-2\right)^2=x^2-4x+4\)
Câu 2:
\(2x^2\left(4x-5x^3\right)+10x^5-5x^3\)
\(=8x^3-10x^5+10x^5-5x^3\)
\(=3x^3\)
\(\left(x-2\right)\left(x^2-2x+4\right)+\left(x-4\right)\left(x-2\right)\)
\(=x^3-4x^2+8x-8+x^2-6x+8\)
\(=x^3-3x^2+2x\)
Còn lại tự làm nha dài lắm
a) (6x+1)2 + (6x-1)2 - 2(1+6x)(6x-1)
= (6x+1+6x-1)2
=144x2
b) x(2x2 -3) - x2(5x+1) +x2
=2x3 - 3x - 5x3 -x2+x2
=-3x3-3x
=-3x(x2+1)
c) 3(22+1)(24+1)(28+1)(216+1)
= (22-1)(22+1)(24+1)(28+1)(216+1)
= (24-1)(24+1)(28+1)(216+1)
= (28-1)(28+1)(216+1)
= (216-1)(216+1)
= 232 -1
d) 3x(x-2) - 5x(1-x) - 8(x2 -3)
= 3x2-6x - 5x + 5x2 - 8x2 +24
= -11x +24
$(x-2)(x^2+2x+4)$
$=x(x^2+2x+4)-2(x^2+2x+4)$
$=x^3+2x^2+4x-2x^2-4x-8$
$=x^3-8$
$(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$
$=(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2$
$=2^2$
$=4$
b)$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3\cdot\dfrac{2^4-1}{2^2-1}\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$
$=3\cdot\dfrac{2^{32}-1}{2^2-1}$
$=3\cdot\dfrac{2^{32}-1}{3}$
$=2^{32}-1$
$x(2x^2-3)-x^2(5x+1)+x^2$
$=2x^3-3x-5x^3-x^2+x^2$
$=2x^3-5x^3-3x$
$=-3x^3-3x$
$=-3x(x^2+1)$
d)$3x(x-2)-5x(1-x)-8(x^2-3)$
$=3x^2-6x-5x+5x^2-8x^2+24$
$=3x^2+5x^2-8x^2-11x+24$
$=-11x+24$
$=24-11x$
$101^2$
$=(100+1)^2$
$=100^2+2\cdot100\cdot1+1^2$
$=10000+200+1$
$=10201$
b)$97\cdot103$
$=(100-3)(100+3)$
$=100^2-3^2$
$=10000-9$
$=9991$
Bài 3.
c)$77^2+23^2+77\cdot46$
$=77^2+23^2+2\cdot77\cdot23$
$=(77+23)^2$
$=100^2$
$=10000$
d)$105^2-5^2$
$=(105-5)(105+5)$
$=100\cdot110$
$=11000$
e)$A=(x-y)(x^2+xy+y^2x)+2y^3$
Tại $x=\dfrac{2}{3}$ và $y=\dfrac{1}{3}$:
$A=\left(\dfrac{2}{3}-\dfrac{1}{3}\right)\left[\left(\dfrac{2}{3}\right)^2+\dfrac{2}{3}\cdot\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\cdot\dfrac{2}{3}\right]+2\left(\dfrac{1}{3}\right)^3$
$=\dfrac{1}{3}\left(\dfrac{4}{9}+\dfrac{2}{9}+\dfrac{2}{27}\right)+\dfrac{2}{27}$
$=\dfrac{1}{3}\cdot\dfrac{20}{27}+\dfrac{2}{27}$
$=\dfrac{20}{81}+\dfrac{6}{81}$
$=\dfrac{26}{81}$
Vậy $A=\dfrac{26}{81}$.