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b: \(x^4+x^2+1\)

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

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

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

c: \(x^7+x^5+1\)

\(=x^7+x^6+x^5-x^6-x^5-x^4+x^5+x^4+x^3-x^3+1\)

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

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

8 tháng 7 2017

ĐỀ BÀI ???

a: \(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)

\(\Leftrightarrow12x^2+15x^2-18=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-1\right)=0\)

=>x=-6 hoặc x=1

b: \(x^4+3x^2-4=0\)

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

=>x=1 hoặc x=-1

 

 

 

4 tháng 7 2017

a, \(4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)

\(\Leftrightarrow4x^2-20x-7x^2+28x+3x^2=12\)

\(\Leftrightarrow8x=12\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy...

b, \(-3x\left(x-5\right)+5\left(x-1\right)+3x^2=4-x\)

\(\Leftrightarrow-3x^2+15x+5x-5+3x^2=4-x\)

\(\Leftrightarrow21x=9\)

\(\Leftrightarrow x=\dfrac{3}{7}\)

Vậy...

c, \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)

\(\Leftrightarrow x^2-9x+20-x^2+x+2=7\)

\(\Leftrightarrow-8x=-15\Leftrightarrow x=\dfrac{15}{8}\)

Vậy...

d, \(-\left(x+3\right)\left(x-4\right)+\left(x-1\right)\left(x+1\right)=10\)

\(\Leftrightarrow-x^2+x+12+x^2-1=10\)

\(\Leftrightarrow x=-1\)

Vậy...

e, \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)

\(\Leftrightarrow x^3-27+5x-x^3=6x\)

\(\Leftrightarrow x=-27\)

Vậy...

4 tháng 7 2017

a) \(4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)

\(4x^2-20x-7x^2+28x+3x^2-12=0\)

\(8x-12=0\)

\(4\left(2x-3\right)=0\)

\(2x-3=0\Rightarrow x=\dfrac{3}{2}\)

b) \(-3x\left(x-5\right)+5\left(x-1\right)+3x^2=4-x\)

\(-3x^2+15x+5x-5+3x^2-4+x=0\)

\(21x-9=0\)

\(3\left(7x-3\right)=0\)

\(\Rightarrow7x-3=0\Rightarrow x=\dfrac{3}{7}\)

c) \(\left(x-5\right)\left(x-4\right)-\left(x-1\right)\left(x-2\right)=7\)

\(x^2-4x-5x+20-x^2+2x+x-2-7=0\)

\(-6x+11=0\Rightarrow x=\dfrac{11}{6}\)

d) \(-\left(x-3\right)\left(x-4\right)+\left(x-1\right)\left(x+1\right)=10\)

\(-x^2+4x+3x-12+x^2-1-10=0\)

\(7x-23=0\)

\(x=\dfrac{23}{7}\)

e) \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)

\(x^3-27+5x-x^3-6x=0\)

\(-x-27=0\Rightarrow x=-27\)

b, \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)

\(\Rightarrow x^2-9x+20-x^2+x+2=7\)

\(\Rightarrow-8x+22=7\)

\(\Rightarrow-8x=-15\)

\(\Rightarrow x=\frac{15}{8}\)

c, \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)

\(\Rightarrow3x^2-10x+8=3x^2-27x-3\)

\(\Rightarrow3x^2-10x-3x^2+27x=\left(-3\right)+\left(-8\right)\)

\(\Rightarrow17x=-11\)

\(\Rightarrow x=-\frac{11}{17}\)

d, \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)

\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27+5x-x^3=6x\)

\(\Rightarrow6x=-27\)

\(\Rightarrow x=-\frac{27}{6}\)

\(\Rightarrow x=-\frac{9}{2}\)

e, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)

\(\Rightarrow3x^2-2x-5-3x^2-2x+1=x-4\)

\(\Rightarrow-4=x-4\)

\(\Rightarrow x=0\)

9 tháng 7 2019

b)    (x - 5)(x - 4) - (x + 1)(x - 2) = 7
<=> x2 - 9x + 20 - x2 + x + 2 - 7 = 0
<=> 8x - 15 = 0 <=> x = 15/8

c)    (3x - 4)(x - 2) = 3x(x - 9) - 3
<=> 3x2 - 10x + 8 = 3x2 - 27x - 3
<=> 17x = -11 <=> x = -11/17

d)    (x - 3)(x2 + 3x + 9) + x(5 - x2) = 6x
<=> x3 - 27 - x3 + 5x - 6x = 0
<=> x = -27

e)    (3x - 5)(x + 1) - (3x - 1)(x + 1) = x - 4
<=> (x + 1)(3x - 5 - 3x + 1) - x + 4 = 0
<=> -4x - 4 - x + 4 = 0 <=> x = 0

b: \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5-6x}{4-x^2}\)

\(=\dfrac{4x-8+2x+4+6x-5}{\left(x-2\right)\left(x+2\right)}=\dfrac{12x-9}{\left(x-2\right)\left(x+2\right)}\)

c: \(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\)

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

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

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

e: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)

\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)

\(=\dfrac{-x^2+7x+78}{x\left(x+6\right)}\)

\(=\dfrac{-x^2+13x-6x+78}{x\left(x+6\right)}\)

\(=\dfrac{-x\left(x-13\right)-6\left(x-13\right)}{x\left(x+6\right)}\)

\(=\dfrac{\left(13-x\right)\left(x+6\right)}{x\left(x+6\right)}=\dfrac{13-x}{x}\)

13 tháng 7 2021

bài vách ngọc ngà và bài cà phê ko đường