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a, 25^2 - 15^2 = ( 25 - 15 )( 25 + 15) = 10 . 40 = 400
b, 87^2 + 73^2 - 27^2 - 13^2
= 87^2 - 27^2 + 73^2 - 13^2
= ( 87 - 27)( 87 + 27) + (73 - 13 )(73+ 13)
= 60 . 114 + 60 . 86
= 60( 114 + 86)
= 60 .200
= 12000
c, x^3 + 27 + 9 x^2 + 27x
= x^3 + 27x + 9x^2 + 27
=(x + 3)^3
thay x =97 ta có
= (97 + 3)^3
= 100^3
=1000000
d, 1,6^2 + 4.0,8.3,4 + 3,4^2 ( nè 3,4^2 chứ không phải 3,42)
= 1,6^2 + 2.2.0,8.3,4 + 3,4^2
=1,6^2 + 2.1,6.3,4 + 3,4^2
= (1,6 + 3,4)^2
= 5^2
= 25
e, x = 11 => 12 =x + 1 thay vào ta có
x^4 - ( x+ 1)x^3 + (x+1)x^2 -(x+1)x + 11
= x^4 - x^4 - x^13 + x^3 + x^2 - x^2 - x + 11
= -x + 11
= -11 + 11
= 0
ĐÚng ch o tui nha
\(100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+....+\left(2-1\right)\left(2+1\right)\)
\(=1.199+1.195+...+1.3\)
\(=199+195+....+3\)
\(=\left[\left(\dfrac{199-3}{4}\right)+1\right]:2.\left(199+3\right)=5050\)
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)
\(3\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{64}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right).....\left(2^{64}+1\right)\)
\(=\left(2^8-1\right)......\left(2^{64}+1\right)=2^{128}-1\)
như thế này chứ:
A=1002-992+982-972+...+22-12
B=12-22+32-42+...-20082-20092
C=3.(22+1)(24+1)(28+1)(216+1)-232
Bài 1:
a) \(100^2-99^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+...+2+1\)
=> tự làm tiếp :))
b) tương tự
Bài 2 :
a) \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(\left(2-1\right)A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^8-1\right)\left(2^8+1\right)\)
\(A=2^{16}-1< 2^6=B\)
b) Phân tích \(2004\cdot2006=\left(2005-1\right)\left(2005+1\right)=\left(2005^2-1\right)\)rồi áp dụng hđt thứ 3 tự làm tiếp như câu a)
Bài 3:
a) Cứ khai triển hết ra
b) \(a^2+b^2+c^2=ab+bc+ac\)
\(a^2+b^2+c^2-ab-bc-ac=0\)
Nhân 2 vào cả 2 vế được :
\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+c^2\right)=0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
mà mũ 2 luôn lớn hơn hoặc bằng 0
\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow}a=b=c\left(đpcm\right)}\)
P.s: toàn bài nâng cao làm hơi ẩu tí ^^
a) \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+....+\left(2-1\right)\left(2+1\right)\)
\(=199+195+....+3\)
\(=\frac{\left(199+3\right)\left[\left(199-3\right):4+1\right]}{2}\)
\(=5050\)
Bài 12:
a: \(=\left(xy+1+x+y\right)\left(xy+1-x-y\right)\)
\(=\left[x\left(y+1\right)+\left(y+1\right)\right]\left[x\left(y-1\right)-\left(y-1\right)\right]\)
\(=\left(x+1\right)\left(x-1\right)\left(y+1\right)\left(y-1\right)\)
b: \(=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\cdot\left(3x^2+y^2\right)\)
c: \(=3y^2\left(x^4+x^3+x+1\right)\)
\(=3y^2\left[x^3\left(x+1\right)+\left(x+1\right)\right]\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)