\(R=5x-x^2\)

2. Cho \(x+y=...">

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Câu 1: 

\(R=-\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{25}{4}=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}< =\dfrac{25}{4}\)

Dấu '=' xảy ra khi x=5/2

Câu 3:

\(\Leftrightarrow x^2+4x+4+y^2-2y+1=0\)

=>(x+2)^2+(y-1)^2=0

=>y=1 và x=-2

24 tháng 2 2020

a)\(\left(4x^3-xy^2+y^3\right)\left(x^2y+2xy^2-2y^3\right)\)

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

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

\(=4x^5y-x^3y^3+x^2y^4+8x^4y^2-2x^2y^4+2xy^5\)

\(-8x^3y^3+2xy^5-2y^6\)

\(=-2y^6+4x^5y+\left(2xy^5+2xy^5\right)+8x^4y^2+\left(x^2y^4-2x^2y^4\right)\)

\(-\left(x^3y^3+8x^3y^3\right)\)

\(=-2y^6+4x^5y+4xy^5+8x^4y^2-x^2y^4-9x^3y^3\)

24 tháng 2 2020

b) 

(!)  \(2\left(x+y\right)^2-7\left(x+y\right)+5\)

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

\(=2\left(x+y\right)\left(x+y-1\right)-5\left(x+y-1\right)\)

\(=\left(2x+2y-5\right)\left(x+y-1\right)\)

(!!) \(\left(x+y+z\right)^2-x^2-y^2-z^2\)

\(=\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-x^2-y^2-z^2\)

\(=2\left(xy+yz+zx\right)\)

14 tháng 11 2016

1.

a) \(2x\left(x-4\right)+\left(x-1\right)\left(x+2\right)=2x^2-8x+x^2+x-2=x^2-7x-2\)

b) \(\left(x-3\right)^2-\left(x-2\right)\left(x^2+2x+4\right)=x^2-6x+9-x^3+8=-x^3+x^2-6x+17\)

2.

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

b) \(x^3+2x^2y+xy^2-16x=x\left(x^2+2xy+y^2-16\right)=x\left[\left(x+y\right)^2-16\right]=\)làm tiếp chắc dễ

3. 

\(\frac{x^4?2x^3+4x^2+2x+3}{x^2+1}\) Giữa x^4 và 2x^3 (vị trí dấu ? là dấu + hay -)

4) \(A=x^2-3x+4=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\)

\(A\ge\frac{7}{4}\)

Vậy GTNN của A là 7/4

2 tháng 9 2018

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

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

\(=3x^2-7x-2\)

hk tốt

25 tháng 6 2018

Nguyễn Thanh Hằng giúp vs !!! khocroi

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

a: \(A=2x^2-2xy-y^2+2xy=2x^2-y^2\)

\(=2\cdot\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)

b: \(B=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)

\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)

=1/5-1=-4/5

\(C=x^3+6x^2+12x+8=\left(x+2\right)^3=\left(-9\right)^3=-729\)

d: \(D=20x^3-10x^2+5x-20x^2+10x+4\)

\(=20x^3-30x^2+15x+4\)

\(=20\cdot5^3-30\cdot5^2+15\cdot2+4=1784\)

AH
Akai Haruma
Giáo viên
24 tháng 11 2018

a)

\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)

b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)

c)

\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)

d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)

AH
Akai Haruma
Giáo viên
24 tháng 11 2018

e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)

f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)

g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)

\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)