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bài 5:

1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)

2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)

\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)

3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)

\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)

\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)

4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)

\(=\dfrac{1}{6\left(x^2+x+1\right)}\)

5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)

\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)

\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)

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

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

Bài 3:

1: \(9x^3-xy^2\)

\(=x\cdot9x^2-x\cdot y^2\)

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

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

2: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

\(=x\left(x-3y\right)-6\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-6\right)\)

3: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

\(=x\left(x-3y\right)-6\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-6\right)\)

4: \(6xy-x^2+36-9y^2\)

\(=36-\left(x^2-6xy+9y^2\right)\)

\(=36-\left(x-3y\right)^2\)

\(=\left(6-x+3y\right)\left(6+x-3y\right)\)

5: \(x^4-6x^2+5\)

\(=x^4-x^2-5x^2+5\)

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

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

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

6: \(9x^2-6x-y^2+2y\)

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

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

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

10 tháng 11 2018

câu a đề có sai số mũ ko vậy

b) \(\dfrac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)

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

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

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

c) \(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)

\(=\dfrac{\left(x^2+3x\right)^2-1}{x^4+6x^3+9x^2-2x^2-6x+1}\)

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

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

31 tháng 7 2023

1) \(8x^3-12x^2+6x-1=0\)

\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)

\(\Leftrightarrow\left(2x-1\right)^3=0\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

2) \(x^3-6x^2+12x-8=27\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)

\(\Leftrightarrow\left(x-2\right)^3=27\)

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

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=3+2\)

\(\Leftrightarrow x=5\)

3) \(x^2-8x+16=5\left(4-x\right)^3\)

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

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

\(\Leftrightarrow5\left(4-x\right)=1\)

\(\Leftrightarrow4-x=\dfrac{1}{5}\)

\(\Leftrightarrow x=4-\dfrac{1}{5}\)

\(\Leftrightarrow x=\dfrac{19}{5}\)

4) \(\left(2-x\right)^3=6x\left(x-2\right)\)

\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)

\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)

\(\Leftrightarrow8-x^3=0\)

\(\Leftrightarrow x^3=8\)

\(\Leftrightarrow x^3=2^3\)

\(\Leftrightarrow x=2\)

5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)

\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-10+4\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=\dfrac{-6}{12}\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)

\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)

\(\Leftrightarrow-54x-2x^3=36x^2-54x\)

\(\Leftrightarrow-2x^3=36x^2\)

\(\Leftrightarrow-2x^3-36x^2=0\)

\(\Leftrightarrow-2x^2\left(x+18\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)

30 tháng 10 2018

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

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

\(=x^2-16-x^2+9\)

\(=-7\)

\(2,\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)

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

\(=2^2=4\)

a: Ta có: \(x^2-4-\left(x+2\right)^2\)

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

=-4x-8

b: Ta có: \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)

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

=2x-1

c: ta có: \(\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)\)

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

\(=-3x+6\)

d: Ta có: \(\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)

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

=4

e: ta có: \(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)

\(=21a^2-35a+8a^2+2a-12a-3-\left(36a^2-24a+4\right)\)

\(=29a^2-45a-3-36a^2+24a-4\)

\(=-7a^2-21a-7\)

g: ta có: \(\left(5y-3\right)\left(5y+3\right)-\left(5y-4\right)^2\)

\(=25y^2-9-25y^2+40y-16\)

=40y-25

h: Ta có: \(\left(3x+1\right)^3-\left(1-2x\right)^3\)

\(=27x^3+27x^2+9x+1-1+6x-12x^2+8x^3\)

\(=35x^3+15x^2+15x\)

i: Ta có: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

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

\(=16x^2\)

25 tháng 6 2017

a,\(A=\left(x^4-3x^2+9\right)\left(x^2+3\right)+\left(3-x^2\right)^2\)

\(A=x^6-3x^4+9x^2+3x^4-9x^2+27+9-6x^2+x^4\)

\(A=x^6+x^4-6x^2+36\)

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

\(M=5\left(x^2+4xy+4y^2\right)-\left(9y^2+12xy+4x^2\right)+\left(16x^2-8xy+y^2\right)+3\left(x^2-4y^2\right)\)

\(M=5x^2+20xy+20y^2-9y^2-12xy-4x^2+16x^2-8xy+y^2+3x^2-12y^2\)

\(M=20x^2\)

Các câu còn lại làm tương tự! Chúc bạn học tốt!!!

25 tháng 6 2017

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

\(\Leftrightarrow\left(6x+1\right)^2-2\left(1+6x\right)\left(6x-1\right)+\left(6x-1\right)^2\)

\(\Leftrightarrow\left[\left(6x+1\right)-\left(6x-1\right)\right]^2\)

\(\Leftrightarrow\left(6x+1-6x+1\right)^2=2^2=4\)

28 tháng 9 2021

\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)

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

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

\(=4\)

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

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

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

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

\(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=\left(3x^2+5x^2-8x^2\right)-\left(6x+5x\right)+24\)

\(=-11x+24\)