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18 tháng 10 2020

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

29 tháng 9 2019

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 \(x\in\left\{\frac{1}{4};2017\right\}\)

29 tháng 9 2019

Bài 1 

a) \(8x\left(x-2017\right)-2x+4034=0\)

\(\Rightarrow8x\left(x-2017\right)-2\left(x-2017\right)=0\)

\(\Rightarrow\left(x-2017\right)\left(4x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}\)

làm nhiều rồi 

hehe

hihi

30 tháng 8 2019

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 :

\(a,16x^2-\left(4x-5\right)^2=15\)

\(16x^2-16x^2+40x-25-15=0\)

\(40x-40=0\)

\(40x=40\)

\(x=1\)

\(b,\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(4x^2+12x+9-4x^2+4=49\)

\(12x=36\)

\(x=3\)

\(c,\left(2x+1\right)\left(2x-1\right)+\left(1-2x\right)^2=18\)

\(4x^2-1+1-4x+4x^2=18\)

\(8x^2-4x-18=0\)

\(2\left(4x^2-2x-9\right)=0\)

\(x=\frac{1-\sqrt{37}}{4}\)

\(d,2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(2x^2+4x+2-x^2+9-x^2+8x-16=0\)

\(12x=4\)

\(x=\frac{1}{3}\)

28 tháng 9 2019

ko ai thèm trả lời đâu cu

28 tháng 9 2019

a) \(4x^2-6x=2x\left(2x-3\right)\)

b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)

c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(5x+3\right)\left(x-y\right)\)

d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)

e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)

\(=5\left(1-3x\right)\left(x+3y\right)\)

f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)

\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)

10 tháng 7 2019

\(\left(x+5\right)^2-3\left(x+5\right)\)

\(=\left(x+5\right)\left(x+5-3\right)\)

\(=\left(x+5\right)\left(x+2\right)\)

10 tháng 7 2019

\(2x\left(x-3\right)-\left(x-3\right)^2\)

\(=\left(x-3\right)\left(2x-x+3\right)\)

\(=\left(x-3\right)\left(x+3\right)\)

mệt rời o 

thông cảm 

hihi

Bài 7 

\(a,A=x^2-2x+5\)

\(=\left(x^2-2x+1\right)+4\)

\(=\left(x-1\right)^2+4\ge4\forall x\)

GTNN \(A=4\) khi \(\left(x-1\right)^2=0\Rightarrow x=1\)

\(b,B=x^2-x+1\)

\(=\left(x^2-2\cdot\frac{1}{2}x+\frac{1}{4}\right)+\frac{3}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

\(c,C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

Đặt \(x^2+5x=t\)

\(\Rightarrow C=\left(t-6\right)\left(t+6\right)\)

\(=t^2-36\)

\(\left(x^2+5x\right)^2-36\ge36\forall x\)

\(d,D=x^2+5y^2-2xy+4y-3\)

\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)-4\)

\(=\left(x-y\right)^2+\left(2y+1\right)^2-4\ge-4\)

đợi mink tí

30 tháng 8 2019

\(a,\left(x+2\right)^2=x^2+4x+4\)

\(b,\left(x-1\right)^2=x^2-2x+1\)

\(c,\left(x^2+y^2\right)^2=x^4+2x^2y^2+y^4\)

\(d,\left(x^3+2y^2\right)^2=x^6+4x^3y^2+4y^4\)

6 tháng 11 2019

\(C1:=3+1-3y\)

\(=4-3y\)

\(C2:\)

\(a.=3x\left(2y-1\right)\)

\(b.=\left(x-y\right)\left(x+y\right)+4\left(x+y\right)\)

\(=\left(x-y+4\right)\left(x+y\right)\)

\(C3:\)

\(a.6x^2+2x+12x-6x^2=7\)

\(14x=7\)

\(x=\frac{1}{2}\)

\(b.\frac{1}{5}x-2x^2+2x^2+5x=-\frac{13}{2}\)

\(\frac{26}{5}x=-\frac{13}{2}\)

\(x=-\frac{13}{2}\times\frac{5}{26}\)

\(x=-\frac{5}{4}\)

3 tháng 7 2020

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, \(\frac{1}{2}x\left(\frac{2}{5}-4x\right)+\left(2x+5\right)x=-6\frac{1}{2}\)

\(< =>\frac{x}{2}.\frac{2}{5}-\frac{x}{2}.4x+2x^2+5x=-\frac{13}{2}\)

\(< =>\frac{x}{5}-2x^2+2x^2+5x=-\frac{13}{2}\)

\(< =>\frac{26x}{5}=\frac{-13}{2}\)

\(< =>26x.2=\left(-13\right).5\)

\(< =>52x=-65< =>x=-\frac{65}{52}=-\frac{5}{4}\)