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\(a.x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
\(b.3x\left(x-2\right)-x+2=0\)
\(\Leftrightarrow3x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=2\end{matrix}\right.\)
\(c.x\left(x-4\right)-2x+8=0\)
\(\Leftrightarrow x\left(x-4\right)-2\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
\(d.3x\left(x+5\right)-3x-15=0\)
\(\Leftrightarrow3x\left(x+5\right)-3\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-3=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
\(d'.x\left(2x-3\right)-3\left(3-2x\right)=0\)
\(\Leftrightarrow x\left(2x-3\right)+3\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{3}{2}\end{matrix}\right.\)
a: =>(x+5)(x-5-1)=0
=>(x+5)(x-6)=0
=>x=-5 hoặc x=6
b: =>(x-2)(3x-1)=0
=>x=2 hoặc x=1/3
c: =>(x-4)(x-2)=0
=>x=4 hoặc x=2
d: =>(x+5)(3x-5)=0
=>x=5/3 hoặc x=-5
e: =>x(2x-3)+3(2x-3)=0
=>(2x-3)(x+3)=0
=>x=3/2 hoặc x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
d.
$(\frac{x}{2}-y)^3=(\frac{x}{2})^3-3(\frac{x}{2})^2y+3.\frac{x}{2}y^2-y^3$
$=\frac{x^3}{8}-\frac{3x^2y}{4}+\frac{3xy^2}{2}-y^3$
e.
$(\frac{x}{2}+\frac{y}{3})^3=(\frac{x}{2})^3+3(\frac{x}{2})^2\frac{y}{3}+3.\frac{x}{2}(\frac{y}{3})^2+(\frac{y}{3})^3$
$=\frac{x^3}{8}+\frac{x^2y}{4}+\frac{xy^2}{6}+\frac{y^3}{27}$
f.
$(\frac{2x}{3}-2y)^3=(\frac{2x}{3})^3-3(\frac{2x}{3})^2.2y+3.\frac{2x}{3}(2y)^2-(2y)^3$
$=\frac{8x^3}{27}-\frac{8x^2y}{3}+8xy^2-8y^3$
g.
$(x+y)^3+(x-y)^3=(x^3+3x^2y+3xy^2+y^3)+(x^3-3x^2y+3xy^2-y^3)$
$=2x^3+6xy^2$
Lời giải:
a.
$(3-y)^3=3^3-3.3^2y+3.3y^2-Y63=27-27y+9y^2-y^3$
b.
$(3x+2y^2)^3=(3x)^3+3.(3x)^2(2y^2)+3.3x(2y^2)^2+(2y^2)^3$
$=8y^6+24xy^4+24x^2y^2+8x^3$
c.
$(x-3y^2)^3=x^3-3x^2(3y^2)+3x(3y^2)^2-(3y^2)^3$
$=x^3-9x^2y^2+27xy^4-27y^6$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,-x^3+3x^2-3x+1=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(b,8-12x+6x^2-x^3=2^3-3.2^2x+3.2x^2-x^3=\left(2-x\right)^3\)
\(c,x^3-6x^2y+12xy^2-8y^3=x^3-3.2y.x^2+2.\left(2y\right)^2x-\left(2y\right)^3=\left(x-2y\right)^3\)
\(d,8x^3+12x^2+6x+1\\ =\left(2x\right)^3+3.1\left(2x\right)^2+3.2x.1^2+1^3=\left(2x+1\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(c,x^3+x^2+\dfrac{x}{3}+\dfrac{1}{27}=x^3+3.x^2.\dfrac{1}{3}+3.x.\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3=\left(x+\dfrac{1}{3}\right)^3\\ f,x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=x^3+3.x^2.\dfrac{1}{2}=3.x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3=\left(x+\dfrac{1}{2}\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x - 2)³ - x(x + 1)(x - 1) + 6x(x + 3)
= x³ - 6x² + 12x - 8 - x(x² - 1) + 6x² + 18x
= x³ + 30x - 8 - x³ + 8
= 30x
b) (x - 2)(x² - 2x + 4)(x + 2)(x² + 2x + 4)
= [(x - 2)(x² + 2x + 4)][(x + 2)(x² - 2x + 4)]
= (x³ - 2³)(x³ + 2³)
= x⁶ - 2⁶
= x⁶ - 64
c) (2x + y)(4x² - 2xy + y²) - (2x - y)(4x² + 2xy + y²)
= [(2x)³ + y³] - [(2x)³ - y³]
= 8x³ + y³ - 8x³ + y³
= 2y³
d) (x + y)³ - (x - y)³ - 2y³
= x³ + 3x²y + 3xy² + y³ - x³ + 3x²y - 3xy² + y³ - 2y³
= 6x²y
e) (x + y + z)² - 2(x + y + z)(x + y) + (x + y)
= x² + y² + z² + 2xy + 2xz + 2yz - 2x² - 2xy - 2xy - 2y² - 2xz - 2yz + x + y
= -x² - y² + z² + x + y
a: =x^3-6x^2+12x-8+6x^2-18x-x(x^2-1)
=x^3-6x-8-x^3+x
=-5x-8
b: =(x-2)(x^2+2x+4)(x+2)(x^2-2x+4)
=(x^3-8)(x^3+8)=x^6-64
c: =8x^3+y^3-8x^3+y^3=2y^3
d: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
e: =(x+y+z)(x+y+z-2x-2y)+(x+y)
=(x+y+z)(-x-y+z)+(x+y)
=z^2-(x+y)^2+(x+y)
=z^2-x^2-2xy-y^2+x+y
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3\left(2a-1\right)+5\left(3-a\right)\)
\(=6a-3+15-5a\)
\(=a-12\)
Thay \(a=\dfrac{-3}{2}\) vào biểu thức trên
\(a-12\)
\(=\dfrac{-3}{2}-12\)
\(=\dfrac{-27}{2}\)
\(b,25x-4\left(3x-1\right)+7\left(5-2x\right)\)
\(=25x-12x+4+35-14x\)
\(=-1x+39\)
Thay \(x=2,1\) vào biểu thức trên
\(-1x+39\)
\(=-1.2,1+39\)
\(=-2,1+39\)
\(=36,9\)
\(c,4a-2\left(10a-1\right)+8a-2\)
\(=4a-20a+2+8a-2\)
\(=-8a\)
Thay \(a=-0,2\) vào biểu thức trên
\(-8a\)
\(=-8.\left(-0,2\right)\)
\(=1,6\)
\(d,12\left(2-3b\right)+35b-9\left(b+1\right)\)
\(=24-36b+35b-9b-9\)
\(=-10b-15\)
Thay \(b=\dfrac{1}{2}\) vào biểu thức trên
\(-10b-15\)
\(=-10.\dfrac{1}{2}-15\)
\(=-20\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =6y^3-3y^2-y^2+2y-y+y^2-y^3
=5y^3-3y^2+y
b: =2x^2a-a-2x^2a-a-x^2-ax
=-x^2-ax-2a
c: =2p^3-p^3+1+2p^3+6p^2-3p^5
=3p^3+6p^2-3p^5+1
d: =-3a^3+5a^2+4a^3-4a^2=a^3+a^2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x^2-25\right)^2-\left(x-5\right)^2\)
\(=\left(x-5\right)^2\left(x+5\right)^2-\left(x-5\right)^2\)
\(=\left(x-5\right)^2\left[\left(x+5\right)^2-1\right]\)
\(=\left(x-5\right)^2\left(x^2+10x+25-1\right)\)
\(=\left(x-5\right)^2\left(x^2+10x+24\right)\)
b) \(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)
\(=\left(2x-5\right)^2\left(2x+5\right)^2-9\left(2x-5\right)^2\)
\(=\left(2x-5\right)^2\left[\left(2x+5\right)^2-9\right]\)
\(=\left(2x-5\right)^2\left(4x^2+20x+25-9\right)\)
\(=\left(2x-5\right)^2\left(4x^2+20x+16\right)\)
c) \(4\left(2x-3\right)^2-9\left(4x^2-9\right)^2\)
\(=4\left(2x-3\right)^2-9\left(2x-3\right)^2\left(2x+3\right)^2\)
\(=\left(2x-3\right)^2\left[4-9\left(2x+3\right)^2\right]\)
\(=\left(2x-3\right)^2\left[4-9\left(4x^2+12x+9\right)\right]\)
\(=\left(2x-3\right)^2\left(4-36x^2-108x-81\right)\)
\(=(2x-3)^2(-36x^2-108x-77)\)
d) \(a^6-a^4+2a^3+2a^2\)
\(=\left(a^6-a^4\right)+\left(2a^3+2a^2\right)\)
\(=a^4\left(a^2-1\right)+2a^2\left(a-1\right)\)
\(=a^4\left(a+1\right)\left(a-1\right)+2a^2\left(a-1\right)\)
\(=\left(a-1\right)\left[a^4\left(a+1\right)+2a^2\right]\)
\(=a^2\left(a-1\right)\left[a^2\left(a+1\right)+2\right]\)
\(=a^2\left(a-1\right)\left(a^3+a^2+2\right)\)
e) \(\left(3x^2+3x+2\right)^2-\left(3x^2+2x-2\right)^2\)
\(=\left[\left(3x^2+3x+2\right)-\left(3x^2+3x-2\right)\right]\left[\left(3x^2+3x+2\right)+\left(3x^2+3x-2\right)\right]\)
\(=\left(3x^2+3x+2-3x^2-3x+2\right)\left(3x^2+3x+2+3x^2+3x-2\right)\)
\(=4\left(6x^2+6x\right)\)
\(=4\cdot6x\left(x+1\right)\)
\(=24x\left(x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)
\(=\left[\left(3x+1\right)-2\left(x-2\right)\right]\left[\left(3x+1\right)+2\left(x-2\right)\right]\)
\(=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\)
\(=\left(x+5\right)\left(5x-3\right)\)
c) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)
\(=\left[3\left(2x+3\right)-2\left(x+1\right)\right]\left[3\left(2x+3\right)+2\left(x+1\right)\right]\)
\(=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\)
\(=\left(4x-7\right)\left(8x+11\right)\)
f) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left[2bc-\left(b^2+c^2-a^2\right)\right]\left[2bc+\left(b^2+c^2-a^2\right)\right]\)
\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)
g: =(ax+by-ay-bx)(ax+by+ay+bx)
=[a(x-y)-b(x-y)]*[a(x+y)+b(x+y)]
=(x-y)(x+y)(a-b)(a+b)
h: =(a^2+b^2-5-2ab-4)(a^2+b^2-5+2ab+4)
=[(a-b)^2-9][(a+b)^2-1]
=(a-b-3)(a-b+3)(a+b-1)(a+b+1)
i: =(4x^2-3x-18-4x^2-3x)(4x^2-3x-18+4x^2+3x)
=(-6x-18)(8x^2-18)
=-12(x+3)(4x^2-9)
=-12(x+3)(2x-3)(2x+3)
k: =(3x+3y-3)^2-(4x+6y+2)^2
=(3x+3y-3-4x-6y-2)(3x+3y-3+4x+6y+2)
=(-x-3y-5)(7x+9y-1)
i: =25-(2x-3y)^2
=(5-2x+3y)(5+2x-3y)
m: =(x-y)^2-(2m-n)^2
=(x-y-2m+n)(x-y+2m-n)
e: =>(3x-1-x-5)(3x-1+x+5)=0
=>(2x-6)(4x+4)=0
=>x=3 hoặc x=-1
f: =>(2x-1-x+3)(2x-1+x-3)=0
=>(x+2)(3x-4)=0
=>x=4/3 hoặc x=-2
g: =>4x^2-4x+1-4x^2+1=0
=>-4x+2=0
=>x=1/2
g: =>(x^2+4)(x^2-1)=0
=>x^2-1=0
=>x=1 hoặc x=-1
m: =>(x+3)(2-x)=0
=>x=2 hoặc x=-3
l: =>(x-2)(x^2+2x+4-x+12)=0
=>(x-2)(x^2+x+16)=0
=>x-2=0
=>x=2
k: =>(2x-5)(2x+5-2x-7)=0
=>2x-5=0
=>x=5/2