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\(\text{a) }301^2\\ =\left(300+1\right)^2\\ =300^2+2\cdot300\cdot1+1^2\\ =90000+600+1\\ =90601\\ \)
\(\text{b) }499^2\\ =\left(500-1\right)^2\\ =500^2-2\cdot500\cdot1+1^2\\ =250000-1000+1\\ =249001\\ \)
\(\text{c) }68\cdot72\\ =\left(70-2\right)\left(70+2\right)\\ =70^2-2^2\\=4900-4\\ =4896\\ \)
a) \(301^2\)
\(=\left(300+1\right)^2\)
\(=300^2+600+1\)
\(=90601\)
b) \(499^2\)
\(=\left(500-1\right)^2\)
\(=500^2-1000+1\)
\(=249001\)
c) \(68\cdot72\)
\(=\left(70-2\right)\left(70+2\right)\)
\(=70^2-2^2\)
\(=4896\)
a, \(\left(301\right)^2=\left(300+1\right)^2=300^2+2.300.1+1^2\)
=90000+6000+1=90601
b,\(499^2=\left(500-1\right)^2=500^2-2.500.1+1^2\)
=10000-10000+1=1
a) 3012 = ( 300 + 1 )2 = 3002 + 2.300.1 + 12 = 90601
b) 4992 = ( 500 - 1 )2 = 5002 - 2.500.1 + 12 = 249001
c) 68.72 = ( 70 - 2). ( 70 + 2) = 702 - 42 = 4900 - 16 = 4884
Bài1:
\(\left(3+xy^2\right)^2=81+6xy^2+x^2y^4\)
Các câu sau tương tự
Bài2:
\(a,\left(4x^2+4xy+y^2\right)\)
=\(\left(2x+y\right)^2\)
b)\(9m^2+n^2-6mn=\left(3m-n\right)^2\)
c)\(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)
d)\(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
Bài3:
\(a,301^2=\left(300+1\right)^2=900+600+1=1501\)
b/\(499^2=\left(500-1\right)^2=2500-1000+1=1501\)
c/\(68.72=\left(70-2\right)\left(70+2\right)=70^2-2^2=4900-4=4896\)
62 . 58 = (60 + 2)(60 - 2) = 60\(^2\) - 2\(^2\) = 3600 - 4 = 3596
199\(^2\) = (200 -1)\(^2\) = 200\(^2\) - 2.200.1 + 1\(^2\) = 40 000 - 400 + 1 = 39601
499\(^2\) = (500 - 1)\(^2\) = 500\(^2\) - 2.500.1 + 1\(^2\) = 250 000 - 1000 + 1 = 249 001
299 . 301 = (300 - 1)(300 + 1) = 300\(^2\) - 1\(^2\) = 90 000 - 1 = 89 999
Học tốt
Đúng thì k cho mk nhé
Trả lời:
+, \(62.58=\left(60+2\right)\left(60-2\right)=60^2-2^2=3600-4=3596\)
+, \(199^2=\left(200-1\right)^2=200^2-2.200.1+1^2=40000-400+1=39601\)
+, \(499^2=\left(500-1\right)^2=500^2-2.500.1+1^2=250000-1000+1=249001\)
+, \(299.301=\left(300-1\right)\left(300+1\right)=300^2-1=90000-1=89999\)
bai 1 : ta có a+b+c=0=>(a+b+c)^2=0
=>a^2+b^2+c^2+2ab+2ac+2bc=0
=>1+2(ab+bc+ac)=0(vì a^2+b^2+c^2=1)
=>ab+bc+cd=-1/2
=>(ab+bc+cd)^2=1/4
=>a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2=1/4
=>a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c)=1/4
=>a^2b^2 +a^2c^2+b^2c^2=1/4(vì a+b+c=0)*
mặt khác a^2+b^2+c^2=1(gt)
=>(a^2+b^2+c^2)^2=1
=>a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1
=>a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)=1
=>a^4+b^4+c^4+2.1/4=1(theo *)
=>a^4+b^4+c^4=1- 1/2=1/2(dpcm)
mk chi giai dc nhu v thoi
a. 34^2 + 66^2 + 68 x 66
= 34 x 34 + 66 x 66 + 68 x 66
= 34 x 34 + 66 x (66 + 68)
= 34 x 34 + 66 x 134
= 34 x 34 + 66 x 34 + 66 x 100
= 34 x (34 + 66) + 66 x 100
= 34 x 100 + 66 x 100
= (34 + 66) x 100
= 100 x 100
= 10000
\(34^2+66^2+68.66\)
\(=34.34+66.66+68.66\)
\(=34.34+66.\left(68+66\right)\)
\(=34.34+66.134\)
\(=34.34+66.\left(100+34\right)\)
\(=34.34+66.100+66.34\)
\(=34.\left(66+34\right)+66.100\)
\(=34.100+66.100\)
\(=\left(34+66\right).100\)
\(=100^2\)
\(=10000\)
( 34^2+66^2+68.66
= 34^2 + 66^2 + 2. 34.66
= ( 34+66)^2
= 100^2 = 10 000
b) 74^2 + 24^2 - 48.74
= 74^2 + 24^2 - 2. 74 . 24
= (74-24)^2 = 50^2 = 2500
a) 34^2 + 66^2 + 68 . 66 = 5580,66
b) 74^2 + 24^2 – 48 . 74 = 6003.26
Trả lời:
\(51^2=\left(50+1\right)^2\)
\(=50^2+2\times50\times1+1^2\)
\(=2500+100+1\)
\(=2601\)
\(301^2=\left(300+1\right)^2\)
\(=300^2+2\times300\times1+1^2\)
\(=90000+600+1\)
\(=90601\)
\(99^2=\left(100-1\right)^2\)
\(=100^2-2\times100+1\)
\(=10000-200+1\)
\(=9801\)
Học tốt
ĐẶT \(\frac{1}{1357}=a;\frac{1}{301}=b\)
\(\Leftrightarrow M=a.\left(5+b\right)-\left(2+1-a\right).2b-3ab+6b\)
\(\Leftrightarrow M=5a+ab-4b-2b+2ab-3ab+6b\)
\(\Leftrightarrow M=5a\)
thay vào ta được
\(M=5.\frac{1}{1357}=\frac{5}{1357}\)
Giải:
a) \(301^2\)
\(=\left(300+1\right)^2\)
\(=300^2+2.300.1+1^2\)
\(=90000+600+1\)
\(=90601\)
b) \(499^2\)
\(=\left(500-1\right)^2\)
\(=500^2-2.500.1+1^2\)
\(=250000-1000+1\)
\(=249001\)
c) \(68.72\)
\(=\left(70-2\right).\left(70+2\right)\)
\(=70^2-2^2\)
\(=4900-4\)
\(=4896\)
Chúc bạn học tốt!!!