Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(C1:=3+1-3y\)
\(=4-3y\)
\(C2:\)
\(a.=3x\left(2y-1\right)\)
\(b.=\left(x-y\right)\left(x+y\right)+4\left(x+y\right)\)
\(=\left(x-y+4\right)\left(x+y\right)\)
\(C3:\)
\(a.6x^2+2x+12x-6x^2=7\)
\(14x=7\)
\(x=\frac{1}{2}\)
\(b.\frac{1}{5}x-2x^2+2x^2+5x=-\frac{13}{2}\)
\(\frac{26}{5}x=-\frac{13}{2}\)
\(x=-\frac{13}{2}\times\frac{5}{26}\)
\(x=-\frac{5}{4}\)
Bạn Moon làm kiểu gì vậy ?
1) \(\left(3x^2y^2+x^2y^2\right):\left(x^2y^2\right)-3y\)
\(=\left[\left(x^2y^2\right)\left(3+1\right)\right]:\left(x^2y^2\right)-3y\)
\(=4-3y\)
2) a, \(6xy-3x=\left(3x\right)\left(2y-1\right)\)
b, \(x^2-y^2+4x+4y=\left(x+y\right)\left(x-y\right)+4\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+4\right)\)
3) a, \(2x\left(3x+1\right)+\left(4-2x\right)3x=7\)
\(< =>6x^2+2x+12x-6x^2=7\)
\(< =>14x=7< =>x=\frac{7}{14}\)
b, \(\frac{1}{2}x\left(\frac{2}{5}-4x\right)+\left(2x+5\right)x=-6\frac{1}{2}\)
\(< =>\frac{x}{2}.\frac{2}{5}-\frac{x}{2}.4x+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{x}{5}-2x^2+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{26x}{5}=\frac{-13}{2}\)
\(< =>26x.2=\left(-13\right).5\)
\(< =>52x=-65< =>x=-\frac{65}{52}=-\frac{5}{4}\)
3/
a/ \(A=\left(x-y\right)^2+\left(x+y\right)^2.\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)
\(A=x^2-2xy+y^2+x^2+2xy+y^2\)
\(A=2x^2+2y^2\)
b/ \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(B=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)
\(B=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(B=8ab\)
c/ \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(C=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(C=x^2+2xy+y^2-x^2+2xy-y^2\)
\(C=4xy\)
d/ \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(D=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(D=4x^2-4x+1-8x^2+24x-18+4\)
\(D=-4x^2+20x-13\)
Bài 6
\(\left(a-b\right)^2=a^2-2ab+b^2\)
\(=\left(a^2+2ab+b^2\right)-4ab\)
\(=\left(a+b\right)^2-4ab\)
Bài 5 :
\(a,16x^2-\left(4x-5\right)^2=15\)
\(16x^2-16x^2+40x-25-15=0\)
\(40x-40=0\)
\(40x=40\)
\(x=1\)
\(b,\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)
\(4x^2+12x+9-4x^2+4=49\)
\(12x=36\)
\(x=3\)
\(c,\left(2x+1\right)\left(2x-1\right)+\left(1-2x\right)^2=18\)
\(4x^2-1+1-4x+4x^2=18\)
\(8x^2-4x-18=0\)
\(2\left(4x^2-2x-9\right)=0\)
\(x=\frac{1-\sqrt{37}}{4}\)
\(d,2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)
\(2x^2+4x+2-x^2+9-x^2+8x-16=0\)
\(12x=4\)
\(x=\frac{1}{3}\)
\(\left(x^2+\frac{2}{5}y\right)\left(x^2+\frac{2}{5}y\right)=x^4+\frac{4}{5}x^2y+\frac{4}{25}y\)
b\(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)=\left(x^2-\frac{1}{3}\right)^3-x^2-\frac{1}{3}\)
\(a\frac{x^2-49}{x+5}:\left(x-7\right)\)
\(=\frac{\left(x-7\right)\left(x+7\right)}{x+5}.\frac{1}{\left(x-7\right)}\)
\(=\frac{x+7}{x+5}\)
\(b,\frac{2x+7}{x+2}-\frac{x+8}{2x+4}\)
\(=\frac{2\left(2x+7\right)}{2\left(x+2\right)}-\frac{x+8}{2\left(x+2\right)}=\frac{4x+14-x+8}{2\left(x+2\right)}\)
\(=\frac{3x+22}{2\left(x+2\right)}\)
a) \(\frac{x^2-49}{x+5}\div\left(x-7\right)=\frac{\left(x-7\right)\left(x+7\right)}{x+5}.\frac{1}{x-7}=\frac{x+7}{x+5}\)
b) \(\frac{2x+7}{x+2}-\frac{x+8}{2x+4}=\frac{2\left(2x+7\right)}{2\left(x+2\right)}-\frac{x+8}{2\left(x+2\right)}=\frac{\left(4x+14\right)-\left(x+8\right)}{2\left(x+2\right)}\)
\(=\frac{4x+14-x-8}{2\left(x+2\right)}=\frac{3x+6}{2\left(x+2\right)}=\frac{3\left(x+2\right)}{2\left(x+2\right)}=\frac{3}{2}\)
a. \(8x\left(x-2017\right)-2x+4034=0\)
\(8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\left(8x-2\right)\left(x-2017\right)=0\)
\(\Rightarrow TH1:8x-2=0\)
\(8x=2\)
\(x=\frac{1}{4}\)
\(TH2:x-2017=0\)
\(x=2017\)
Vậy \(x\in\left\{\frac{1}{4};2017\right\}\)
Bài 1
a) \(8x\left(x-2017\right)-2x+4034=0\)
\(\Rightarrow8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}\)
a) ( 3 - x )( x2 + 2x - 7 ) + ( x - 3 )( x2 + x - 5 )
= ( 3 - x )( x2 + 2x - 7 ) - ( 3 - x )( x2 + x - 5 )
= ( 3 - x )( x2 + 2x - 7 - x2 - x + 5 )
= ( 3 - x )( x - 2 )
b) ( x - 5 )2 + 3( 5 - x )
= ( x - 5 )2 - 3( x - 5 )
= ( x - 5 )( x - 5 - 3 ) = ( x - 5 )( x - 8 )
c) 2x( x - 1 )2 - ( 1 - x )3
= 2x( 1 - x )2 - ( 1 - x )3
= ( 1 - x )2( 2x - 1 + x ) = ( 1 - x )2( 3x - 1 )
d) x2 + 8x + 16 = ( x + 4 )2
e) x2 - 4xy + 4y2 = ( x - 2y )2
g) 4x2 - 25y2 = ( 2x )2 - ( 5y )2 = ( 2x - 5y )( 2x + 5y )
h) 25( x + 1 )2 - 4( x - 3 )2
= 52( x + 1 )2 - 22( x - 3 )2
= ( 5x + 5 )2 - ( 2x - 6 )2
= ( 5x + 5 - 2x + 6 )( 5x + 5 + 2x - 6 )
= ( 3x + 11 )( 7x - 1 )
i) x3 + 27 = ( x + 3 )( x2 - 3x + 9 )
k) 8x3 - 125 = ( 2x )3 - 53 = ( 2x - 5 )( 4x2 + 10x + 25 )
l) x3 + 6x2 + 12x + 8 = ( x + 2 )3
m) -x3 + 9x2 - 27x + 27 = -( x3 - 9x2 + 27x - 27 ) = -( x - 3 )3
\(\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
\(=\frac{2x+4-2x+4}{x^2-4}.\frac{\left(x+2\right)^2}{8}\)
\(=\frac{8}{x^2-4}.\frac{\left(x+2\right)^2}{8}\)
\(=\frac{x+2}{x-2}\)
Ta có:
\(\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
\(=\left(\frac{2\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right).\frac{x^2+4x+4}{8}\)
\(=\left(\frac{2x+4}{x^2-4}-\frac{2x-4}{x^2-4}\right).\frac{x^2+4x+4}{8}\)
\(=\frac{0}{x^2-4}.\frac{x^2+4x+4}{8}\)
\(=0.\frac{x^2+4x+4}{8}\)
\(=0\)
Nhân phá ra thôi bn
\(\left(x+1\right)\left(x^6-x^5+x^4-x^3+x^2-x+1\right)\)
\(=x^7+x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)
\(=x^7+1\)
\(\left(x+1\right)\left(x^6-x^5+x^4-x^3+x^2-x+1\right)\)
\(=\)\(x^7-x^6+x^5-x^4+x^3-x^2+x+x^6-x^5+x^4-x^3+x^2-x+1\)
\(=\)\(x^7+1\)