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\(31,8^2-2.31,8.21,8+21,8^2\)
\(=\left(31,8-21,8\right)^2=10^2=100\)
\(58,2^2+2.58,2.41,8+41,8^2\)
\(=\left(58,2+41,8\right)^2=100^2=10000\)
Bài 1:
a) \(\left(n+2\right)^2-\left(n-2\right)^2=n^2+4n+4-\left(n^2-4n+4\right)=8n\) \(⋮\)\(8\) (đpcm)
b) \(\left(n+7\right)^2-\left(n-5\right)^2=n^2+14n+49-\left(n^2-10n+25\right)=24n-24\)\(⋮\)\(24\) (đpcm)
Bài 2:
mk biến đổi về pt tích sau đó bạn giải nốt nhé
a) \(\left(x-4\right)^2-36=0\)
<=> \(\left(x-4-6\right)\left(x-4+6\right)=0\)
<=> \(\left(x-10\right)\left(x+2\right)=0\)
................
b) \(4x^2-12x=-9\)
<=> \(4x^2-12x+9=0\)
<=> \(\left(2x-3\right)^2=0\)
..............
c) \(\left(x+8\right)^2=121\)
<=> \(\left(x+8\right)^2-121=0\)
<=> \(\left(x+8+11\right)\left(x+8-11\right)=0\)
<=> \(\left(x+19\right)\left(x-3\right)=0\)
...................
a) \(1001^2=\left(1000+1\right)^2=1000^2+2.1000.1+1^2=1002001\)
b) \(29,9\times30,1=\left(30-0,1\right).\left(30+0,1\right)=30^2-\left(0,1\right)^2=899,99\)
c) \(\left(31,8\right)^2-2.31,8.21,8+\left(21,8\right)^2=\left(31,8-21,8\right)^2=10^2=100\)
bai 3
a) 36-12x+x2
= x2-12x+36
=(x-6)2
b) 4x2+12x+9
=(2x+3)2
c) -25x6-y8+10x3y4
=-(25x6+y8-10x3y4)
=-(5x3)2+(y4)2-10x3y4
=-(5x3)2-2.5x3y4+(y4)
=-(5x3-y4)2
d) \(\dfrac{1}{4}\) x2-5xy+25y2
=(\(\dfrac{1}{2}\) x-5y)2
Bài3:
Bạn =kia làm r nhé
Bài4:
\(a,75^2-25^2\\ =\left(75-25\right)\left(75+25\right)\\ =50.100=5000\\ b,53^2-47^2\\ =\left(53-47\right)\left(53+47\right)\\ =6.100=600\\ c,31,8^2-2.31,8.21,8+21,8^2\\ =\left(31,8-21,8\right)^2\\ =10^2=100\\ d,58,2+2.58,2.41,8+41,8^2\\ =\left(58,2+41,8\right)^2\\ =100^2=1000\)
Cái đó áp dụng HDT binh phương của 1 hiệu =>(31,8-21,8)2=102=100
\(c,=\left(31,8-21,8\right)^2=10^2=100\\ 12,\\ a,\left(n+2\right)^2-\left(n-2\right)^2\\ =\left(n+2-n+2\right)\left(n+2+n-2\right)\\ =4\cdot2n=8n⋮8\\ b,\left(n+7\right)^2-\left(n-5\right)^2\\ =\left(n+7-n+5\right)\left(n+7+n-5\right)\\ =12\left(2n+2\right)=24\left(n+1\right)⋮24\)
B1:
a) \(1001^2=\left(1000+1\right)^2\)
\(=1000^2+2.1000+1=1000000+2000+1\)
= \(1002001\)
b) \(29,9.30,1\)
= \(\left(30-0,1\right)\left(30+0,1\right)\)
= \(30^2-0,1^2=900-0,01=899,99\)
c) \(31,8^2-2.31,8.21,8+21,8^2\)
= \(\left(31,8-21,8\right)^2=10^2=100\)
B2:
a) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
b) \(a^6-b^3=\left(a^2\right)^3-b^3\)
= \(\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
c) \(8y^3-125=\left(2y\right)^3-5^3\)
= \(\left(2y-5\right)\left(4y^2+10y+25\right)\)
d) \(8z^3+27=\left(2z\right)^3+3^3\)
= \(\left(2z+3\right)\left(4z^2-6z+9\right)\)
B3:
a) A = \(x^2-20x+101\)
= \(x^2-20x+100+1\)
= \(\left(x-10\right)^2+1\ge1\) với mọi x
MinA = 1 khi và chỉ khi x = 10
b) B = \(4a^2+4a+2\)
= \(4a^2+4a+1+1\)
= \(\left(2a+1\right)^2+1\ge1\) với mọi x
MinB = 1 khi và chỉ khi a = \(-\dfrac{1}{2}\)
Giải:
a) \(\left(31,8\right)^2-2.31,8.21,8+\left(21,8\right)^2\)
\(=\left(31,8-21,8\right)^2\)
\(=10^2=100\)
Vậy ...
b) \(\left(7,5\right)^2+2.7,5.2,5+\left(2,5\right)^2\)
\(=\left(7,5+2,5\right)^2\)
\(=10^2=100\)
Vậy ...
HĐT: \(\left(A\pm B\right)^2=A^2\pm2.A.B+B^2\)
a, (31,8)2 - 2.31,8.21,8+(21,8)2
=(31,8+21,8)2 =102 =100
b,(7,5)2 + 2.7,5.2,5 +(2,5)2
=(7,5 +2,5)2 =102 =100