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a. 1619 = (24)19 = 276
825 = (23)25 = 275
Vì 276 > 275 nên 1619 > 825
b. 3111 < 3211 = (25)11 = 255
1714 > 1614 = (24)14 = 256
Vì 255 < 256 nên 3111 < 1714
a^3+b^3+c^3=( a+b+c )(a^2+b^2+c^2-ab-bc-ac) +3abc
suy ra kq laf 33
Theo đề ta có:
a+b+c=0 => c=-(a+b) (1)
Thay (1) vao a^3+b^3+c^3 ta có:
a^3+b^3+[-(a+b)]^3=3ab[-(a+b)]
<=>a^3+b^3-(a+b)=-3ab(a+b)
<=> a3+ b3- a3 -3a2b- 3ab2- b3= -3a2b- 3ab2
<=> 0= 0
vậy ta có đpcm.
a)
Ta có :
\(x+y=3\)
\(x^2+y^2=5\Leftrightarrow\left(x+y\right)^2-2xy=5\Leftrightarrow9-2xy=5\Leftrightarrow2xy=4\Rightarrow xy=2\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3.\left(5-2\right)=9\)
b)
Ta có :
\(x-y=5\)
\(x^2+y^2=15\Leftrightarrow\left(x-y\right)^2+2xy=15\Leftrightarrow25+2xy=15\Rightarrow xy=-5\)
=> \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=\left(5\right)\left(15+-5\right)=50\)
11.\(8a^3-12a^2+6a-1=\left(3a-1\right)^3\)
12.\(x^3+12x^2+48x+64=\left(x+4\right)^3\)
13.\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
14.\(a^3-b^3+c^3+3abc\)
\(=\left(a-b\right)^3+3ab\left(a-b\right)+c^3+3abc\)
\(=\left(a-b+c\right)\left[\left(a-b\right)^2+\left(a-b\right)c+c^2\right]+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2-2ab+b^2+ac-bc+c^2\right)+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2+b^2+c^2+ab-bc+ac\right)\)
15.\(4x^4+1=4x^4+4x^2+1-4x^2\)
\(=\left(2x^2+1\right)^2-4x^2\)
\(=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
16.\(4x^4+y^4=4x^4+4x^2y^2+y^4-4x^2y^2\)
\(=\left(2x^2+y^2\right)^2-4x^2y^2\)
\(=\left(2x^2-2xy+y^2\right)\left(2x^2+2xy+y^2\right)\)
17.\(x^4+324=x^4+36x^2+18^2-36^2\)
\(=\left(x^2+18\right)^2-36x^2\)
\(=\left(x^2+6x+18\right)\left(x^2-6x+18\right)\)
18.\(x^5+x+1=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
19.\(x^{11}+x+1=x^{11}-x^8+x^8-x^5+x^5-x^2+x^2+x+1\)
\(=x^8\left(x^3-1\right)+x^5\left(x^3-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^8\left(x-1\right)\left(x^2+x+1\right)+x^5\left(x-1\right)\left(x^2+x+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^5+x^3-x^2+1\right)\)
20.\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)
\(=\left[\left(x-3\right)\left(x-10\right)\right]\left[\left(x-5\right)\left(x-6\right)\right]-24x^2\)
\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)
Đặt \(t=x^2-11x+30\) thay vào phương trình ta được:
\(\left(t-2x\right).t-24x^2\)
\(=t^2-2tx-24x^2\)
\(=t^2+4tx-6tx-24x^2\)
\(=t\left(t+4x\right)-6x\left(t+4x\right)\)
\(=\left(t+4x\right)\left(t-6x\right)\)
\(=\left(x^2-11x+30+4x\right)\left(x^2-11x+30-6x\right)\)
\(=\left(x^2-7x+30\right)\left(x^2-17x+30\right)\)
đừng có chép câu TL của tui nhá cu cÒng
Điều đó là không tốt đâu thằng đệ à
Hahahaha!!!
a) Ta có:
1619= (24)19=276
825=(23)25=275
Vì 276>275=> 169>825
b)Ta có:
3111<3211 = (25)11=255
1714>1614 =(24)14=256
Vì 255<256=> 3211 <1614
=>3111<1714
c)Ta có:
2711=(33)11=333
818=(34)8=332
Vì 333 >332=> 2711>818
e)Ta có:
32n =(32)n=9n
23n=(23)n=8n
Vì 9n>8n=>32n>23n
d)Ta có:
536= (53)12=12512
1124=(112)12 =12112
Vì 12512>12112
=>536>1124