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a)Ta có \(101^2\)=\(\left(100+1\right)^2\)=10000+200+1
=10201
b)\(199^2\)=\(\left(200-1\right)^2=40000-400+1\)=39601
c)47.53=\(\left(50-3\right)\left(50+3\right)=50^2-3^2\)=2500-9=2491
a) 1012=(100+1)2=1002+2.50.2+12=10000+200+1=10201
b)1992=(200-1)2=2002 -2.200.1+12=40000-400+1=39601
c) 47.53=(50-3)(50+3)=502-32=2500-9=2491
a) 1012 =(100+1)2 =10000+1=10001
b) 1992 =(199+1)2 =2002 =40000
c) 47.53=(40+7 .50+3)=20000+10=20010
Ta có : B = 202 - 192 + 182 - 172 + ..... + 22 - 12
=> B = (20 - 19)(20 + 19) + (18 - 17)(18 + 17) + ..... + (2 - 1)(2 + 1)
=> B = 39 + 35 + 31 + ..... + 3
Số số hạng của dãy trên là :
(39 - 3) : 4 + 1 = 10 (số)
Tổng B là :
(39 + 3) x 10 : 2 = 210
Vậy B = 210
Ta có : \(C=\left(15^4-1\right)\left(15^4+1\right)-3^8.5^8\)
\(\Rightarrow C=\left(15^4\right)^2-1-15^8\)
\(\Rightarrow C=15^8-1-15^8\)
=> C = -1
Vậy C = - 1
a) Đặt \(A=\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=2.\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^4-1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(...\)
\(2A=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2A=3^{64}-1\)
\(A=\frac{3^{64}-1}{2}\)
Câu a :
\(x^2+6x+9=\left(x+3\right)^2\)
Câu b :
\(10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
Câu c :
\(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\) \(=\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+\dfrac{1}{2}.2x+\left(\dfrac{1}{2}\right)^2\right]\)
Cau d :
\(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}\right)^2-\left(8y\right)^2=\left(\dfrac{1}{5}+8\right)\left(\dfrac{1}{5}-8\right)\)
a: \(=\dfrac{2\cdot5^5-4\cdot5^3+5^4}{5^3}=2\cdot5^2-4+5=50+1=51\)
b: \(=\dfrac{3^8-3^6+3^6\cdot2^3}{3^5}=3^3-3+3\cdot2^3=24+24=48\)
c: \(=\dfrac{7^6\cdot2^3-7^3}{7^3}=14^3-1\)
d: \(=28^4-28^4+1=1\)
Bài 2
\( a)4{\left( {x + 1} \right)^2} + {\left( {2x - 1} \right)^2} - 8\left( {x - 1} \right)\left( {x + 1} \right) = 11\\ \Leftrightarrow 4\left( {{x^2} + 2x + 1} \right) + 4{x^2} - 4x + 1 - 8\left( {{x^2} - 1} \right) = 11\\ \Leftrightarrow 4{x^2} + 8x + 4 + 4{x^2} - 4x + 1 - 8{x^2} + 8 = 11\\ \Leftrightarrow 4x + 13 = 11\\ \Leftrightarrow 4x = 11 - 13\\ \Leftrightarrow 4x = - 2\\ \Leftrightarrow x = - \dfrac{1}{2} \)
Bài 2:
\( b)\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + x\left( {x + 2} \right)\left( {2 - x} \right) = 1\\ \Leftrightarrow {x^3} - 27 + x\left( {2 + x} \right)\left( {2 - x} \right) = 1\\ \Leftrightarrow {x^3} - 27 + x\left( {4 - {x^2}} \right) = 1\\ \Leftrightarrow {x^3} - 27 + 4x - {x^3} = 1\\ \Leftrightarrow 4x = 1 + 27\\ \Leftrightarrow 4x = 28\\ \Leftrightarrow x = 7 \)
B1:a)(3x-5)2-(3x+1)2=8
[(3x-5)+(3x+1)].[(3x-5)-(3x+1)]=8
(3x-5+3x+1)(3x-5-3x-1)=8
9x2-15x-9x2-3x-15x+25+15x+5+9x2-15x-9x2-3x+3x-5-3x-1=8
-36x+24=8
-36x=8-24=16
x=16:(-36)=\(\dfrac{-4}{9}\)
Bài 5:
a: \(=\left(xy-u^2v^3\right)\left(xy+u^2v^3\right)\)
b: \(=\left(2xy^2-3xy^2+1\right)\left(2xy^2+3xy^2-1\right)\)
\(=\left(1-xy^2\right)\left(5xy^2-1\right)\)
Bài 6:
a: \(\left(a+b+c-d\right)\left(a+b-c+d\right)\)
\(=\left(a+b\right)^2+\left(c-d\right)^2\)
\(=a^2+2ab+b^2+c^2-2cd+d^2\)
b: \(\left(a+b-c-d\right)\left(a-b+c-d\right)\)
\(=\left(a-d\right)^2-\left(b-c\right)^2\)
\(=a^2-2ad+d^2-b^2+2bc-c^2\)
d )
\(B=5\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^{64}-1\right)\left(2^{64}+1\right)\)
\(\Rightarrow B=\frac{5}{3}\left(2^{128}-1\right)\)
Sửa lại dấu \(\Rightarrow\)dòng 3 :
\(B=\frac{5}{3}\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)