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1. 20182 - 2017.2019 = 20182 - ( 2018 - 1 )( 2018 + 1 ) = 20182 - ( 20182 - 12 ) = 20182 - 20182 + 12 = 1
2. 20012 = ( 2000 + 1 )2 = 20002 + 2.2000.1 + 12 = 4 000 000 + 4000 + 1 = 4 004 001
3. 19992 = ( 2000 - 1 )2 = 20002 - 2.2000.1 + 12 = 4 000 000 - 4000 + 1 = 3 996 001
4. 113 = ( 10 + 1 )3 = 103 + 3.102.1 + 3.10.12 + 13 = 1000 + 300 + 30 + 1 = 1331
5. 193 = ( 20 - 1 )3 = 203 - 3.202.1 + 3.20.1 - 13 = 8000 - 1200 + 60 - 1 = 6859
6. 57.63 = ( 60 - 3 )( 60 + 3 ) = 602 - 32 = 3600 - 9 = 3591
7. 953 + 3.952.5 + 3.95.52 + 53 = ( 95 + 5 )3 = 1003 = 1 000 000
1,\(2x\left(x-5\right)-\left(x-2\right)^2-\left(x+3\right)\left(x-3\right)\)
\(=2x^2-10x-x^2+4x-4-x^2+9\)
\(=\left(2x^2-x^2-x^2\right)+\left(-10x+4x\right)+\left(-4+9\right)\)
\(=-6x+5\)
2,\(\left(x+1\right)^2-3\left(x-5\right)\left(x+5\right)-\left(2x-1\right)^2\)
\(=x^2+2x+1-3\left(x^2-25\right)-\left(4x^2-4x+1\right)\)
\(=x^2+2x+1-3x^2+75-4x^2+4x-1\)
\(=-6x^2+6x+75\)
3,\(\left(x-1\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-1\right)^3-\left(x^3-27\right)\)
\(=x^3-3x^2+3x-1-x^3+27\)
\(=-3x^2+3x+26\)
4,\(\left(x+5\right)\left(x^2-5x+25\right)-\left(x+2\right)^3\)
\(=\left(x^3+125\right)-\left(x^3+6x^2+12x+8\right)\)
\(=x^3+125-x^3-6x^2-12x-8\)
\(=-6x^2-12x+117\)
5,\(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)^2+\left(x+1\right)^2\)
\(=2x^2-14x-\left(x+3\right)\left(x^2-4x+4\right)+x^2+2x+1\)
=\(2x^2-14x-x^3+4x^2-4x-3x^2+12x-12+x^2+2x+1\)
\(=-x^3+4x^2-4x+1\)
6,\(\left(2x+5\right)\left(x-3\right)-\left(x+5\right)\left(x-1\right)-\left(x-4\right)^2\)
\(=2x^2-6x+5x-15-x^2+x-5x+5-x^2+8x-16\)
\(=3x-26\)
7,\(\left(x+5\right)\left(x-5\right)\left(x+2\right)-\left(x+2\right)^3\)
=\(\left(x^2-25\right)\left(x+2\right)-x^3-6x^2-12x-8\)
\(=x^3+2x^2-25x-50-x^3-6x^2-12x-8\)
\(=-4x^2-27x-58\)
Nếu đúng thì tick cho mk nha ^_^
Lời giải:
Khai triển:
\(\text{VT}=5(x^5+y^5+z^5)+5\underbrace{[x^3(y^2+z^2)+y^3(x^2+z^2)+z^3(x^2+y^2)]}_{M}\)
Xét riêng $M$ kết hợp với điều kiện $x+y+z=0$ ta có
\(M=x^2y^2(x+y)+y^2z^2(y+z)+z^2x^2(x+z)=-(x^2y^2z+y^2z^2x+z^2x^2y)\)
\(\Leftrightarrow M=-xyz(xy+yz+xz)=\frac{-1}{2}xyz[(x+y+z)^2-(x^2+y^2+z^2)]=\frac{1}{2}xyz(x^2+y^2+z^2)\)
Ta biết đến một hằng thức rất quen thuộc: Nếu $x+y+z=0$ thì \(x^3+y^3+z^3=3xyz\)
Cách chứng minh: \(x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=0-3(-x)(-y)(-z)=3xyz\)
Do đó \(M=\frac{1}{6}(x^3+y^3+z^3)(x^2+y^2+z^2)=\frac{\text{VT}}{30}\)
\(\Rightarrow \text{VT}=5(x^5+y^5+z^5)+5M=5(x^5+y^5+z^5)+\frac{\text{VT}}{6}\)
\(\Rightarrow \text{VT}=6(x^5+y^5+z^5)\) (đpcm)
b) Theo phần a)
\(\left\{\begin{matrix} M=\frac{1}{2}xyz(x^2+y^2+z^2)\\ M=\frac{5(x^2+y^2+z^2)(x^3+y^3+z^3)}{30}\end{matrix}\right.\Rightarrow \frac{5(x^2+y^2+z^2)(x^3+y^3+z^3)}{30}=\frac{xyz(x^2+y^2+z^2)}{2}\)
Mà \(5(x^2+y^2+z^2)(x^3+y^3+z^3)=6(x^5+y^5+z^5)\Rightarrow \frac{6(x^5+y^5+z^5)}{30}=\frac{xyz(x^2+y^2+z^2)}{2}\)
\(\Leftrightarrow 2(x^5+y^5+z^5)=5xyz(x^2+y^2+z^2)\) (đpcm)
b)Vì x+y+z=0
=>x+y=-z =>(x+y)^5=-z^5
hay x^5+y^5+5(x^4y+xy^4+2x³y²+2x²y³+)=-z^5
<=>x^5+y^5+z^5+5xy(x³+y³+2x²y+2x²y)=0
<=>x5+y^5+z^5+5xy(x+y)(x²-xy+y²+2xy)=0
<=>x^5+y^5+z^5-5xyz(x²+xy+y²)=0
<=>x^5+y^5+z^5=5xyz(x²+xy+y²)
<=>2(x^5+y^5+z^5)=5xyz(2x²+2xy+2y²)
<=>2(x^5+y^5+z^5)=5xyz[x²+y²+(x+y)²]
<=>2(x^5+y^5+z^5)=5xyz(x³+y²+z²)
1: \(\Leftrightarrow5x^2+4x-1-2x^2+12x-18=3x^2+5x-2-x^2-8x-16+x^2-x\)
\(\Leftrightarrow3x^2+16x-19=3x^2-4x-18\)
=>20x=1
hay x=1/20
2: \(\Leftrightarrow5x^2-20x-41=x^2-10x+25+4x^2+4x+1-\left(x^2-2x\right)+\left(x-1\right)^2\)
\(\Leftrightarrow5x^2-20x-41=4x^2-4x+26+x^2-2x+1\)
\(\Leftrightarrow-20x-41=-6x+27\)
=>-14x=68
hay x=-34/7
a) (x3 + 8y3) : (2y + x)
= (x + 2y)(x2 - 2xy + 4y2) : (2y + x)
= x2 - 2xy + 4y2
b) (x3 + 3x2y + 3xy2 + y3) : (2x + 2y)
= (x + y)3 : 2(x + y)
= \(\dfrac{\left(x+y\right)^2}{2}\)
c) (6x5y2 - 9x4y3 + 15x3y4) : 3x3y2
= 3x3y2(2x2 - 3xy + 5y2) : 3x3y2
= 2x2 - 3xy + 5y2
\(\left(-3\right)^5\div\left(-3\right)^2\)
\(=\left(-3\right)^{5-2}\)
\(=\left(-3\right)^3\)
(-3)5 : (-3)2 = (-3)5-2 = (-3)3 = -27
-HT-