Rút gọn biểu thức:P=12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
CM đẳng thức: (a+b+c)^3= a^3+b^3+c^3+3(a+b)(b+c)(c+a)
1. Biết số tự nhiên a chia cho 5 dư 4. Chứng minh rằng \(a^2\) chia cho 5 dư 1
2. Rút gọn biểu thức : \(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
3. Chứng minh hằng đẳng thức: \(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(\frac{1}{2}\left(5^{32}+1\right)=\frac{5^{32}+1}{2}\)
a)
Ta có
a chia 5 dư 4
=> a=5k+4 ( k là số tự nhiên )
\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16\)
Vì 25k^2 chia hết cho 5
40k chia hết cho 5
16 chia 5 dư 1
=> đpcm
2) Ta có
\(12=\frac{5^2-1}{2}\)
Thay vào biểu thức ta có
\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)}{2}\)
\(\Rightarrow P=\frac{\left[\left(5^2\right)^2-1^2\right]\left[\left(5^2\right)^2+1^2\right]\left(5^8+1\right)}{2}\)
\(\Rightarrow P=\frac{\left[\left(5^4\right)^2-1^2\right]\left[\left(5^4\right)^2+1^2\right]}{2}\)
\(\Rightarrow P=\frac{5^{16}-1}{2}\)
3)
\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+b^3+c^2+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Rút gọn biểu thức
A= 12(52 +1)(54 +1)(58 +1)(516 +1)
Chứng minh
(a+b+c)3 = a3+b3+c3+3(a+b)(b+c)(c+a)
Bài 1:
A = \(12.\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> \(\left(5^2-1\right)A\) = \(12\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> 24A = \(12\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> A = \(\dfrac{12}{24}.\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> A = \(\dfrac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
=> A = \(\dfrac{1}{2}\left(5^{32}-1\right)\)
Bài 2:
Ta có: \(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3\)
= \(\left(a+b\right)^3+c^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)
= \(a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)\left(ac+bc+c^2\right)\)
= \(a^3+b^3+c^3+3\left(a+b\right)\left(ab+bc+ca+c^2\right)\)
= \(a^3+b^3+c^3+3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
= \(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\) => đpcm
Rút gọn biểu thức:
a) P= (5x-1)+2x(1-5x)x(4+5x)+(5x+4)^2
b) Q= (x-y)^3+(y+x)^3+(y-x)^3-3xy(x+y)
c) 12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
Rút gọn biểu thức:
a) P= (5x-1)+2x(1-5x)x(4+5x)+(5x+4)^2
b) Q= (x-y)^3+(y+x)^3+(y-x)^3-3xy(x+y)
c) 12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
a: \(P=\left(5x-1-5x-4\right)^2=\left(-3\right)^2=9\)
b: \(Q=\left(x+y\right)^3-3xy\left(x+y\right)=x^3+y^3\)
c: \(=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{5^{32}-1}{2}\)
Rút gọn các biểu thức:
a, (3x+1)^2-2(3x+1)(3x+5)+(3x+5)^2
b,(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
c,(a+b-c)^2+(a-b+c)^2-2(b-c)^2
d,(a+b+c)^2+(a-b-c)^2+(b-c-a)^2+(c-a-b)^2
e,(a+b+c+d)^2+(a+b-c-d)^2+(a+c-b-d)^2+(a+d-b-c)^2
Rút gọn các biểu thức sau a)√27-✓12+✓48-5✓3 b)5✓18-✓5+✓20+✓1 2 C)✓25:✓16=✓36:✓9 D)✓12+✓27-5✓3 E)2✓3-✓75+2✓12
a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)
Rút gọn các biểu thức sau:
a) 2x(2x-1)^2 - 3x(x+3)(x-3) - 4x(x+1)^2
b) (a-b+c)^2 - (b-c)^2 + 2ab-2ac
c) (3x+1)^2 - 2(3x+1)(3x+5) + (3x+5)^2
d) (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
e) (a+b-c)^2 + (a-b+c)^2 - 2(b-c)^2
g) (a+b+c)^2 + (a-b-c)^2 + (b-c-a)^2 + (c-a-b)^2
h) (a+b+c+d)^2 + (a+b-c-d)^2 + (a+c-b-d)^2 + (a+d-b-c)^2
Rút gọn các biểu thức sau:
a) 2x(2x-1)^2 - 3x(x+3)(x-3) - 4x(x+1)^2
b) (a-b+c)^2 - (b-c)^2 + 2ab-2ac
c) (3x+1)^2 - 2(3x+1)(3x+5) + (3x+5)^2
d) (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
e) (a+b-c)^2 + (a-b+c)^2 - 2(b-c)^2
g) (a+b+c)^2 + (a-b-c)^2 + (b-c-a)^2 + (c-a-b)^2
h) (a+b+c+d)^2 + (a+b-c-d)^2 + (a+c-b-d)^2 + (a+d-b-c)^2
Tìm giá trị nhỏ nhất: M=x2+y2-x+6y+10
Rút gọn biểu thức:P:(5x-1)+2(1-5x)(4+5x)+(5x+4)
Q=(x+y)3+(y+x)3+(y-x)3-3xy(x+y)
H=12(52+1)(54+1)(58+1)(516+1)
C/m hằng đẳng thức:(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
Phân tích thành nhân tử: x3+y3+z3-3xyz
xy(x+y)+yz(y+z)+xz(x+z)+2xyz
Tính nhanh giá trị của đa thức : 3(x-3)(x+7)+(x-4)2+48 tại x= 0,5
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Ta có :
\(x^3+y^3+z^3-3xyz\)
\(\Rightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(\Rightarrow\left(x+y+z\right)\left[\left(x+y^2\right)-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
P/s tham khảo nha
hok tốt