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Bài 1 :
a ) Ta có :
\(3^4.5^4-\left(15^2+1\right)\left(15^2-1\right)\)
\(=15^4-\left(15^4-1\right)\)
\(=15^4-15^4+1\)
\(=1\)
b ) Ta có :
\(x=11\Rightarrow x+1=12\)
Thay \(x+1=12\) vào biểu thức ta được :
\(x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+111\)
\(=x^4-x^4-x^3+x^3-x^2+x^2-x+111\)
\(=111-x\)
Thay \(x=11\) vào biểu thức vừa rút gọn ta được :
\(111-11=100\)
\(a,3^4.5^4-\left(15^2+1\right)\left(15^2-1\right)\)
\(=\left(3.5\right)^4-\left(15^4-1\right)\)
\(=15^4-15^4+1\)
\(=1\)
Bài 2:
\(a,\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)^2-2.\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
\(=2^2=4\)
\(b,3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
Thay 12 = x + 1 vào biểu thức trên, ta có:
x4 - (x + 1)x3 + (x + 1)x2 - (x + 1)x + 111
= x4 - x4 - x3 + x3 + x2 - x2 - x + 111
= 111 - x (*)
Thay x = 11 vào (*), ta có:
111 - 11
= 100
Vậy giá trị của biểu thức trên là 100 tại x = 11
(x + y + z)3 - x3 - y3 - z3
= x3 + y3 + z3 + 3(x + y)(x + z)(y + z) - x3 - y3 - z3
= 3(x + y)(x + z)(y + z)
A = 2x2 + 10x - 1
\(=2\left(x^2+5x+\frac{25}{4}-\frac{25}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(x+\frac{5}{2}\right)^2-\frac{27}{4}\right]\)
\(=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge-\frac{27}{2}\)
\(MinA=-\frac{27}{2}\Leftrightarrow x=-\frac{5}{2}\)
2.
a) . -x3 + 3x2 - 3x + 1
=13-3.12x+3.1.x2-x3
=(1-x)3
b)8- 12x + 6x2 - x3
=23-3.22.x+3.2.x2-x3
=(2-x)3
3.
a) x3 + 12x2 + 48x + 64 tại x = 6
=x3+3.x2.4+3x4+432
=(x+4)3thay x=6 ta được :
(6+4)3=103=1000
b) x3 - 6x2 + 12x - 8 tại x= 22
=x3-3.x2.2+3.x.22 -23
=(x-2)3 thay x=22 ta đc:
=(22-2)3=203=8000
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
Bài 1:
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=36x^2+72x+1+36x^2-72x+1-2\left(36x^2-1\right)\)
\(=36x^2+72x+1+36x^2-72x+1-72x^2+2\)
\(=4\)
b) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
c) \(x\left(2x^3-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^4-3x-5x^3-x^2+x^2\)
\(=2x^4-5x^3-3x\)
d) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
a)\(x^3y^3+x^2y^2+4\)
\(=x^3y^3-x^2y^2+2xy+2x^2y^2-2xy+4\)
\(=xy\left(x^2y^2-xy+2\right)+2\left(x^2y^2-xy+2\right)\)
\(=\left(xy+2\right)\left(x^2y^2-xy+2\right)\)
b)\(x^4+x^3+6x^2+5x+5\)
\(=x^4+x^2+x^2+5x^2+5x+5\)
\(=x^2\left(x^2+x+1\right)+5\left(x^2+x+1\right)\)
\(=\left(x^2+5\right)\left(x^2+x+1\right)\)
c)\(x^4-2x^3-12x^2+12x+36\)
\(=x^4-2x^3-6x^2-6x^2+12x+36\)
\(=x^2\left(x^2-2x-6\right)-6\left(x^2-2x-6\right)\)
\(=\left(x^2-6\right)\left(x^2-2x-6\right)\)
d)\(x^8y^8+x^4y^4+1\)
\(=x^8y^8+2x^4y^4+1-x^4y^4\)
\(=\left(x^4y^4+1\right)^2-\left(x^2y^2\right)^2\)
\(=\left(x^4y^4+1+x^2y^2\right)\left(x^4y^4+1-x^2y^2\right)\)
\(=\left(x^4y^4+2x^2y^2+1-x^2y^2\right)\left(x^4y^4+1-x^2y^2\right)\)
\(=\left(\left(x^2y^2+1\right)^2-\left(xy\right)^2\right)\left(x^4y^4+1-x^2y^2\right)\)
\(=\left(x^2y^2+1-xy\right)\left(x^2y^2+1+xy\right)\left(x^4y^4+1-x^2y^2\right)\)
a/ 34.54-(152+1)(152-1)
=154-(154-152+152-1)
=154-154+1=1
b/ x4-12x3+12x2-12x+111
=x4-x3-11x3+11x2+x2-x-11x+11+100
=x3(x-1)-11x2(x-1)+x(x-1)-11(x-1)+100
=(x3-11x2+x-11)(x-11)+100
Thay x=11 vào ta được:
=(113-11.112+11-11)(11-11)+100
=0.10+100=100
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