\(^{1,9^2}\)

b) \(9,9^2\)

c)...">

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Khách

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a: \(A=-\left(x^2-4x-3\right)\)

\(=-\left(x^2-4x+4-7\right)\)

\(=-\left(x-2\right)^2+7< =7\)

Dấu '=' xảy ra khi x=2

b: \(B=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\)

Dấu '=' xảy ra khi x=1/2

c: \(C=-2\left(x^2-x+\dfrac{5}{2}\right)\)

\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)

\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)

Dấu '=' xảy ra khi x=1/2

e: \(E=-\left(x^2+6x+9+1\right)=-\left(x+3\right)^2-1< =-1\)

Dấu = xảy ra khi x=-3

Bài 1: Phân tích đa thức thành nhân tử

a) Ta có: \(16x^2-y^2+6y-9\)

\(=16x^2-\left(y^2-6y+9\right)\)

\(=\left(4x\right)^2-\left(y-3\right)^2\)

\(=\left[4x-\left(y-3\right)\right]\left[4x+\left(y-3\right)\right]\)

\(=\left(4x-y+3\right)\left(4x+y-3\right)\)

b) Ta có: \(a^2-16a^2b^2+b^2+2ab\)

\(=\left(a^2+2ab+b^2\right)-\left(4ab\right)^2\)

\(=\left(a+b\right)^2-\left(4ab\right)^2\)

\(=\left(a+b-4ab\right)\left(a+b+4ab\right)\)

c) Ta có: \(x^3-6x^2-9x\)

\(=x\left(x^2-6x-9\right)\)

d) Ta có: \(mx^2+my^2-nx^2-ny^2\)

\(=m\left(x^2+y^2\right)-n\left(x^2+y^2\right)\)

\(=\left(x^2+y^2\right)\left(m-n\right)\)

e) Ta có: \(a^3+b^3+a^2c+b^2c-abc\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)\)

\(=\left(a^2-ab+b^2\right)\left(a+b+c\right)\)

f) Ta có: \(4x^2-y^2-4x+1\)

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

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

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

g) Ta có: \(\left(2x+3\right)^2+5\cdot\left(2x+3\right)\)

\(=\left(2x+3\right)\left(2x+3+5\right)\)

\(=\left(2x+3\right)\left(2x+8\right)\)

\(=2\left(2x+3\right)\left(x+4\right)\)

h) Ta có: \(3x^2-10x-8\)

\(=3x^2-12x+2x-8\)

\(=3x\left(x-4\right)+2\left(x-4\right)\)

\(=\left(x-4\right)\left(3x+2\right)\)

18 tháng 7 2020

cảm ơn nhiều ạ :)))

24 tháng 9 2017

Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@

B1: Phân tích thành nhân tử:

a) \(6x^2+9x=3x\left(2x+3\right)\)

b) \(4x^2+8x=4x\left(x+2\right)\)

c) \(5x^2+10x=5x\left(x+2\right)\)

d) \(2x^2-8x=2x\left(x-4\right)\)

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

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

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

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

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

h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)

i) \(9x^2-18x+9=\left(3x-3\right)^2\)

k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)

l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)

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

\(=-\left(2x-y\right)^2\)

n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)

\(=\left(x-31\right)\left(x+1\right)\)

o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)

\(=\left(2+x\right)\left(8+x\right)\)

p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)

\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)

\(=\left(5x-5\right)\left(9x-3\right)\)

24 tháng 9 2017

Bài 1 :

a ) \(6x^2+9x=3x\left(x+3\right)\)

b ) \(4x^2+8x=4x\left(x+2\right)\)

c ) \(5x^2+10x=5x\left(x+2\right)\)

d ) \(2x^2-8x=2x\left(x-4\right)\)

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

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

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

h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)

i ) \(9x^2-18x+9=\left(3x-3\right)^2\)

k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)

l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)

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

n ) \(\left(x-15\right)^2=x^2-30x+15^2\)

o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)

p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)

Bài 2 :

a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)

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

c ) \(2x+x^2-2y-2xy=......................\)

d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)

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

f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)

30 tháng 9 2019

a) \(x^2-3x+xy-3y\)

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

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

b) \(x^2+y^2-2xy-25\)

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

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

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

30 tháng 9 2019

m) \(81-x^2+2xy-y^2\)

\(=9^2-\left(x-y\right)^2\)

\(=\left(9-x+y\right)\left(9+x-y\right)\)

k) \(x^2-xy-x+y\)

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

\(=\left(x-1\right)\left(x-y\right)\)

b: \(x^2\left(x-2x^3\right)=x^3-2x^5\)

b: \(\left(x^2+1\right)\left(5-x\right)\)

\(=5x^2-x^3+5-x\)

c: \(\left(x-2\right)\left(x^2+3x-4\right)\)

\(=x^3+3x^2-4x-2x^2-6x+8\)

\(=x^3+x^2-10x+8\)

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

27 tháng 7 2018

a) \(A=x^2-2x-6\)

\(A=\left(x^2-2x+1\right)-7\)

\(A=\left(x-1\right)^2-7\)

\(\left(x-1\right)^2\) luôn \(\ge\)\(0\) => GTNN của biểu thức là -7 với \(\left(x-1\right)^2=0\) tức x=1

a: \(=x^2-2x+1-7=\left(x-1\right)^2-7>=-7\)

Dấu '=' xảy ra khi x=1

b: \(=4x^2-4x+1+6=\left(2x-1\right)^2+6>=6\)

Dấu '=' xảy ra khi x=1/2

c: \(=9x^2-6x+1-1=\left(3x-1\right)^2-1>=-1\)

Dấu '=' xảy ra khi x=1/3

d: \(=x^2+12x+36-36=\left(x+6\right)^2-36>=-36\)

Dấu '=' xảy ra khi x=-6

e: \(=x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}>=-\dfrac{9}{4}\)

Dấu '=' xảy ra khi x=3/2

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

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

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

d) \(\dfrac{1}{9}x^2-\dfrac{2}{3}x+1=\left(\dfrac{1}{3}x-1\right)^2\)

e) \(\left(4x^2+1\right)^2-16x^2=\left(4x^2+1+4x^2\right)\left(4x^2+1-4x^2\right)=8x^2+1\)

f) \(16x^2-\left(x^2+4\right)^2=\left(4x^2+x^2+4\right)\left(4x^2-x^2-4\right)=\left(5x^2+4\right)\left(3x^2-4\right)\)

g) \(x^2+6x^2+12x+8=\left(x+2\right)^3\)

h) \(27x^3-54x^2+36x-8=\left(3x-2\right)^3\)

i) \(x^3-\dfrac{3}{2}x^2+\dfrac{3}{4}x-\dfrac{1}{8}=\left(x-\dfrac{1}{2}\right)^3\)

k) \(0,125x^3-0,75x^2+1,5x-1=\left(0,5-1\right)^3\)

13 tháng 9 2018

thanks nha

22 tháng 8 2018

a) 2x2 + 4x + xy + 2y

= (2x2 + xy) + (4x + 2y)

= x(2x + y) + 2(2x + y)

= (x + 2)(2x + y)

22 tháng 8 2018

b) x2 + xy - 7x - 7y

= x(x + y) - 7(x + y)

= (x - y)(x + y)

27 tháng 6 2017

Bài 1:

a) -16 +(x-3)2

<=> (x-3)2-16

<=> (x-3)2 -42

<=> (x-3-4)(x-3+4)

<=> (x-7)(x+1)

b) 64+16y+y2

<=> y2 + 2.8.y + 82

<=> (y+8)2

c) \(\dfrac{1}{8}-8x^3\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^3-\left(2x\right)^3\)

\(\Leftrightarrow\left(\dfrac{1}{2}-2x\right)\left(\dfrac{1}{4}+x+4x^2\right)\)

d)\(x^2-x+\dfrac{1}{4}\)

\(\Leftrightarrow x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2\)

e) x4 + 4x2 + 4

<=> (x2)2 + 2.2.x2 +22

<=> (x2 + 2)2

g)\(8x^3+60x^2y+150xy^2+125y^3\)

\(\Leftrightarrow\left(2x+5y\right)^3\)

28 tháng 6 2017

Ban giup minh bai 2 luon voi nha Hậu Trần Công