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<=>\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
<=>\(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)
Vì \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)
=>x+100=0
<=>x=-100
k nha bạn
\(\Leftrightarrow\frac{37x+1648}{1026}=\frac{11x+556}{272}\Rightarrow\left(37x+1648\right)272=1026\left(11x+556\right)\)
<=>(37x+1648)272=272(37x+1648)
=>272(37x+1648)=1026(11x+556)
=>10064x+448256=11286x+570456
<=>-1222x=122200
=>x=122200:-1222
=>x=-100 ( dễ hiểu chưa hả )
\(8,=\left(2x-3\right)\left(2x+3\right)\\ 9,=\left(1-5a^2\right)\left(1+5a^2\right)\)
8) \(-9+4x^2=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)
9) \(1-25a^4=1-\left(5a^2\right)^2=\left(1-5a^2\right)\left(1+5a^2\right)\)
Câu 3:
a: \(25x-30x^2=5x\left(5-6x\right)\)
b: \(8x\left(x+4y\right)+16\left(x+4y\right)=8\left(x+4y\right)\left(x+2\right)\)
c: \(24x^8y^7-15x^6y^5+18x^6y^3\)
\(=3x^6y^3\left(8x^2y^4-5y^2+6\right)\)
câu 1:
a) 4x( 5x + 9 )
= 4x5x + 4x9
= 20x2+36x
b) (3x - 2) ( x+7 )
= 3xx + 3x7 - 2x - 2 . 7
= 3x2+ 21x - 2x - 14
= 3x2 +19x - 14
c) (2x- 5 )(x2-7x + 4 )
= 2xx2 - 2x7x + 2x4 - 5x2 +5 . 7x - 5.4
= 2x3- 14x2+8x - 5x 2+ 35x - 20
= 2x3+ ( - 14x2- 5x2 )+(8x+35x) - 20
= 2x3 - 9x2 + 43x - 20
Câu 2 :
a) (5x - 3x)2
= (5x)2 - 2.5x3y + (3y)2
= 25x2 - 30xy +9y
b) (2x - 5)(2x + 5)
= (2x)2 - 52
= 4x2 - 25
c) (3x - 7y)3
= ( 3x)3 - 3(3x)27y + 3.3x(7y)2 - (7y)2
= 27x3 - 189x2y + 441xy2- 343y3
d) (x+5)(x2-5x+25)
= (x+5)(x2-5x+52)
= x3+53
= x3+125
Câu 3 :
a) 25x-30x2
= 5.5x-5.6xx
= 5x(5-6x)
b) 8x(x+4y)+16(x+4y)
= (x+4y)(8x+16)
c) 24x8y7 -15x6y5+18x6y3
= 3.8x6x2y3y4-3.5x6y3y2+3.6x6y3
= 3x6y3(8x2y4-5y2+6)
Câu 3:
a: \(25x-30x^2=5x\left(5-6x\right)\)
b: \(8x\left(x+4y\right)+16\left(x+4y\right)=8\left(x+4y\right)\left(x+2\right)\)
\(a,=\dfrac{12x^2y\left(x+y\right)^2:3x\left(x+y\right)}{3x\left(x+y\right):3x\left(x+y\right)}=4xy\left(x+y\right)\\ b,=\dfrac{x^2+6x+9+24-x^2+4x-3}{4\left(x-3\right)\left(x+3\right)}=\dfrac{10\left(x+3\right)}{4\left(x-3\right)\left(x+3\right)}=\dfrac{4}{2\left(x-3\right)}\)