Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=\left(892+108\right)^2=1000^2=1000000\)
b: \(=\left(36-26\right)^2=10^2=100\)
c: \(=9.8\cdot10+10.2\cdot10=20\cdot10=200\)
a) 20042 - 16
= 20042 - 42
= (2004 - 4)(2004 + 4)
= 2000.2008
= 4016000
b) 8922 + 892.216 + 1082
= 8922 + 2.892.108 + 1082
= (892 + 108)2
= 10002
= 1000000
c) 10,2.9,8 - 9,8.0,2 + 10,2.0,2
= 9,8(10,2 - 0,2) + 2,04
= 9,8.10 + 2,04
= 98 + 2,04
= 100,04
d) 362 + 262 - 52.36
= 362 - 2.36.26 + 262
= (36 - 26)2
= 102
= 100
8922 + 892.216 + 1082
= 8922 + 2.892.108 + 1082
= ( 892 + 108 )2
= 10002 = 1 000 000
362 + 262 - 52.36
= 362 - 2.36.26 + 262
= ( 36 - 26 )2
= 102 = 100
\(A=2004^2-16=\left(2004-4\right)\left(2004+4\right)=2000\cdot2008=4016000\)
\(B=892^2+892\cdot216+108^2=892^2+2\cdot892\cdot108+108^2=\left(892+108\right)^2=1000^2=1000000\)
\(C=10,2\cdot9,8-9,8\cdot0,2+10,2^2-10,2\cdot0,2\)
\(=9,8\left(10,2-0,2\right)+10,2\left(10,2-0,2\right)\)
\(=10\cdot\left(9,8+10,2\right)=10\cdot20=200\)
\(D=36^2+26^2-52\cdot36=\left(36-26\right)^2=10^2=100\)
\(E=105^2-25=\left(105-5\right)\left(105+5\right)=100\cdot110=11000\)
\(F=73^2-27^2=\left(73-27\right)\left(73+27\right)=46\cdot100=4600\)
A. 20042-16=20042-42=(2004-4)(2004+4)=2000.2008=2016000
B. 8922+892.216+1082= 8922+2.892.108+1082=(892+108)2=10002=1000000
C.Khó nhìn quá
D.362+262-52.36=362-2.36.26+262=(36-16)2=202=400
E.1052-25=(105-5)(105+5)=100.110=11000
F. 732-272=(73-27)(73+27)=46.100=4600
bài 1.
a.\(\left(x+4\right)\left(x^2-4x+16\right)=x^3-4^3=x^3-64\)
b.\(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)=\left(x^2\right)^3-\left(\frac{1}{3}\right)^3=x^6-\frac{1}{27}\)
bài 2.
a.\(892^2+892.216+108^2=892^2+2.892.108+108^2\)
\(=\left(892+108\right)^2=1000^2=1_{ }000_{ }000\)
b.\(36^2+26^2-52.36=36^2+26^2-2.26.36=\left(36-26\right)^2=10^2=100\)
Bài 4:
\(x^3-2x^2+x=x\left(x-1\right)^2\)
\(5\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)
\(x^2-12x+36=\left(x-6\right)^2\)
Phân tích đa thức thành nhân tử:
\(6xy+5x-5y-3x^2-3y^2\)
\(=-3x^2+6xy-3y^2+5x-5y\)
\(=-3\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)
\(=-3\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left[-3\left(x-y\right)+5\right]\)
\(=\left(x-y\right)\left(-3x+3y+5\right)\)
Thực hiện phép tính:
a)\(\left(x^2+x-3\right)\left(x^2-x+3\right)\)
\(=\left[x^2+\left(x-3\right)\right]\left[x^2-\left(x-3\right)\right]\)
\(=\left(x^2\right)^2-\left(x-3\right)^2\)
\(=x^4-\left(x^2-6x+9\right)\)
\(=x^4-x^2+6x-9\)
b)\(\left(5x-1\right)\left(x+3\right)-\left(x-2\right)\left(5x-4\right)\)
\(=\left(5x^2+15x-x-3\right)-\left(5x^2-4x-10x+8\right)\)
\(=5x^2+15x-x-3-5x^2+4x+10x-8\)
\(=28x-11\)
1
a,\(\left(2x+1\right)\left(3x+1\right)-\left(6x-1\right)\left(x+1\right)\)
=\(6x^2+2x+3x+1-\left(6x^2+6x-x-1\right)\)
\(=6x^2+5x+1-6x^2-6x+x+1\)
\(=2\)
c,\(\left(a+1\right)\left(a^2-a+1\right)+\left(a+1\right)\left(a-1\right)\)
\(=\left(a^3+1\right)+\left(a^2-1\right)\)
\(=a^3+1+a^2-1\)
\(=a^3+a^2\)
2,
a,\(4ab+a^2-3a-12b\)
\(=\left(4ab-12b\right)+\left(a^2-3a\right)\)
\(=4b\left(a-3\right)+a\left(a-3\right)\)
\(=\left(4b+a\right)\left(a-3\right)\)
b,\(x^3+3x^2+3x+1-27y^3\)
\(=\left(x+1\right)^3-\left(3y\right)^3\)
\(=\left(x+1-3y\right)\left[\left(x+1\right)^2+\left(x+1\right).3y+\left(3y\right)^2\right]\)
\(=\left(x+1-3y\right)\left(x^2+2x+1+3xy+3y+9y^2\right)\)
4
a,\(2004^2-16\)
\(=2004^2-4^2\)
\(=\left(2004-4\right)\left(2004+4\right)\)
\(=2000.2008\)
\(=4016000\)
b,\(892^2+892.216+108^2\)
\(=\left(892+108\right)^2\)
\(=1000^2=1000000\)
c,\(10,2.9,8-9,8.0,2+10,2^2-10,2.0,2\)
\(=9,8\left(10,2-0,2\right)+10,2\left(10,2-0,2\right)\)
\(=9,8.10+10,2.10\)
\(=98+102\)
\(=200\)
d,\(36^2+26^2-52.36\)
=\(\left(36-26\right)^2\)
\(=10^2=100\)
3)\(A=-x^2+2x-3\)
\(\Leftrightarrow A=-x^2+2x-1-2\)
\(\Leftrightarrow A=-\left(x^2-2x+1\right)-2\)
\(\Leftrightarrow A=-\left(x-1\right)^2-2\)
Vậy GTLN của A=-2 khi x=1