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5: \(=4b^2-2b+\dfrac{1}{4}-\dfrac{1}{4}+a-a^2\)
\(=\left(2b\right)^2-2\cdot2b\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(a^2-a+\dfrac{1}{4}\right)\)
\(=\left(2b-\dfrac{1}{2}\right)^2-\left(a-\dfrac{1}{2}\right)^2\)
\(=\left(2b-\dfrac{1}{2}-a+\dfrac{1}{2}\right)\left(2b-\dfrac{1}{2}+a-\dfrac{1}{2}\right)\)
\(=\left(2b-a\right)\left(2b+a-1\right)\)
6:
\(=b^2-4b+4-9c^2\)
\(=\left(b-2\right)^2-9c^2\)
\(=\left(b-2-3c\right)\left(b-2+3c\right)\)
Bài 5:
1) \(\left(5+7\right)\left(7-5\right)=7^2-5^2\)
2) \(\left(x+y\right)\left(y-x\right)=y^2-x^2\)
3) \(\left(x-y\right)\left(-x-y\right)=-\left(x+y\right)\left(x-y\right)=-\left(x^2-y^2\right)=y^2-x^2\)
6) \(\left(2+3x^2\right)\left(3x^2-2\right)=9x^4-4\)
7) \(\left(\dfrac{1}{2}+x\right)\left(-x+\dfrac{1}{2}\right)=\left(\dfrac{1}{2}+x\right)\left(\dfrac{1}{2}-x\right)=\dfrac{1}{4}-x^2\)
8) \(\left(4m-5n\right)\left(5n+4m\right)=\left(4m-5n\right)\left(4m+5n\right)=16m^2-25n^2\)
9) \(\left(7a+1\right)\left(1-7a\right)=\left(1+7a\right)\left(1-7a\right)=1-49a^2\)
10) \(\left(1+9\right)\left(1-9\right)=1-9^2\)
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:
+, \(62.58=\left(60+2\right)\left(60-2\right)=60^2-2^2=3600-4=3596\)
+, \(199^2=\left(200-1\right)^2=200^2-2.200.1+1^2=40000-400+1=39601\)
+, \(499^2=\left(500-1\right)^2=500^2-2.500.1+1^2=250000-1000+1=249001\)
+, \(299.301=\left(300-1\right)\left(300+1\right)=300^2-1=90000-1=89999\)
a: =x^2+6x+9+x^2-6x+9+2x^2-32
=4x^2-14
b: =(x+3-10+x)^2=(2x-7)^2=4x^2-28x+49
c: =(x-3-x+5)^2=2^2=4
e: =x^2+10x+25-x^2+10x-25=20x
d: A=(5-1)(5+1)(5^2+1)(5^4+1)/4
=(5^2-1)(5^2+1)(5^4+1)/4
=(5^4-1)(5^4+1)/4
=(5^8-1)/4
g: =x^2-9-x^2-4x+5
=-4x-4
=\(\left(2-1\right)\left(2+1\right)\left(2^2-1\right)....\left(2^{20}-1\right)\) +1
=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{20}+1\right)+1\)
=\(\left(2^4-1\right)\left(2^4+1\right)....\left(2^{20}+1\right)+1\)
=.....
=\(\left(2^{20}-1\right)\left(2^{20}+1\right)+1\)
=\(2^{40}-1+1\)
=\(2^{40}\)
Chuc ban hoc tot
Sai rồi, nếu mũ là 32 thì bài này làm thế đc chứ mũ 20 thì ko làm như này được
\(x^2+20x+100=\left(x+10\right)^2\)
\(16x^2+24xy+\left(3y\right)^2=\left(4x+3y\right)^2\)
\(y^2-14y+49=\left(y-7\right)^2\)
\(a,\) \(x^2+20x+100=\left(x+10\right)^2\)
\(b,\) \(16x^2+24x+9=\left(4x+3\right)^2\)
\(c,\) \(y^2-14x+49=\left(y-7\right)^2\)
\(10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
Chúc bạn học tốt và nhớ click cho mình với nhá!
= (5x-25) + (5x - x2)
= 5(x-5) + x(5-x)
= 5(x-5) - x(x-5)
= (5 - x)(x - 5)
\(499^2+499+500\)
\(=499^2+499+\left(499+1\right)\)
\(=499^2+2.499+1\)
\(=\left(499+1\right)^2\)
\(=500^2\)
\(=2500\)