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a) 192=(20-1)2=202-2.20.1+12=400-40+1=361;
282=(30-2)2=302-2.30.2+22=900-120+4=784;
812=(80+1)2=802+2.80.1+12=6400+160+1=6561;
912=(90+1)2=902+2.90.1+12=8100+180+1=8281;
b) 19.21=(20-1)(20+1)=202-1=400-1=399;
29.31=(30-1)(30+1)=302-1=900-1=899;
39.41=(40-1)(40+1)=402-1=1600-1=1599
c) 292-82=(29-8)(29+8)=21.37=37(20+1)=740+37=777
562-462=(56-46)(56+46)=10.100=1000
672-562=(67-56)(67+56)=11.123=123(10+1)=1230+123=1353
Ta có: a + b + c = 0
<=> a2 + b2 + c2 + 2(ab + bc + ac) = 0
<=> a2 + b2 + c2 = -2(ab + bc + ac)
<=> a4 + b4 + c4 + 2(a2b2 + b2c2 + a2c2 = 4[a2b2 + b2c2 + a2c2 + 2abc(a + b + c)] (vì a + b + c= 0)
<=> a4 + b4 + c4 + 2(a2b2 + b2c2 + a2c2) = 4(a2b2 + b2c2 + a2c2)
<=> a4 + b4 + c4 = 2(a2b2 + b2c2 + a2c2) (đpcm)
b) Từ a4 + b4 + c4 = 2(a2b2 + b2c2 + a2c2)
<=> (a4 + b4 + c4)/2 = a2b2 + b2c2 + a2c2 + 2abc(a + b + c) (vì a + b + c) = 0
<=> (a4 + b4 + c4)/2 = (ab + bc + ac)2
<=> a4 + b4 + c4 = 2(ab + bc + ac)2 (đpcm)
c) Từ a4 + b4 + c4 = 2(a2b2 + b2c2 + a2c2)
<=> 2(a4 + b4 + c4) = a4+ b4 + c4 + 2(a2b2 + b2c2 + a2c2)
<=> 2(a4 + b4 + c4) = (a2 + b2 + c2)2
<=> a4 + b4 + c4 = (a2 + b2 + c2)2/2 (đpcm)
Bài 209 : đăng tách ra cho mn cùng làm nhé
a,sửa đề : \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)
b, \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
\(2B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2B=3^{64}-1\Rightarrow B=\frac{3^{64}-1}{2}\)
c, \(C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2=2\left[\left(a-b+c\right)^2-\left(b-c\right)^2\right]\)
\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)=2a\left(a-2b+2c\right)\)
Trả lời:
Bài 1. Tính:
a) ( x + 2y )2 = x2 + 2.x.2y + ( 2y )2 = x2 + 4xy + 4y2
b) ( x - 3y ) ( x + 3y ) = x2 - ( 3y )2 = x2 - 9y2
c) ( 5 - x )2 = 52 - 2.5.x + x2 = 25 - 10x + x2
d) ( x - 1 )2 = x2 + 2x + 1
e) ( 3 - y )2 = 32 - 2.3.y + y2 = 9 - 6y + y2
Trả lời:
Bài 2. Viết các biểu thức sau dưới dạng bình phương của một tổng:
a) x2 + 6x + 9 = x2 + 2.x.3 + 32 = ( x + 3 )2
b) lỗi đề
c) 2xy2 + x2y4 + 1 = ( xy2 )2 + 2.xy2 + 1 = ( xy2 + 1 )2
Bài 3. Rút gọn biểu thức:
a) (x + y)2 + (x - y)2 = [ ( x + y ) - ( x - y ) ] [ ( x + y ) + ( x - y ) ] = ( x + y - x + y ) ( x + y + x - y ) = 2y.2x = 4xy
b) 2 ( x - y ) ( x + y ) + ( x - y )2 + ( x + y )2 = ( x - y )2 + 2 (x - y ) ( x + y ) + ( x + y )2 = ( x - y + x + y )2 = ( 2x )2 = 4x2
c) ( x - y + z )2 + ( z - y )2 + 2 ( x - y + z )( y - z ) = ( x - y + z )2 + 2 ( x - y + z )( y - z ) + ( y - z )2 = ( x - y + z + y - z )2 = x2
\(\left(x-1\right)-\left(x-2\right)\left(x+2\right)\)
\(=\left(x-1\right)-\left(x^2-2^2\right)\)
\(=\left(x-1\right)-x^2+2^2\)
\(=x-1-x^2+2^2\)
\(=x-x^2+\left(2-1\right)\left(2+1\right)\)
\(=x-x^2+3\)
a/ (x-1)2-(x-2)(x+2)
=(x-1)-(x2-22)
=(x-1)-x2-22
=x-x2 +(2-1)(2+1)
=x-x2+3
Trả lời:
a, \(-xy.\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+3xy\)
b, \(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y\)
\(=12x^6y^5:6x^2y^2-3x^3y^4:6x^2y+4x^2y+6x^2y\)
\(=2x^4y^3-\frac{1}{2}xy^3+\frac{2}{3}\)
a.\(\left(-xy\right)\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+6xy\)
b.\(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y=2x^4y^4-\frac{1}{2}xy^3+\frac{2}{3}\)
192=(20−1)2=202−2.20.1+12=400−40+1=361
a)
19^2=(20−1)^2=20^2−2.20.1+1^2=400−40+1=361
28^2=(30−2)^2=30^2−2.30.2+2^2=900−120+4=784
81^2=(80+1)^2=80^2+2.80.1+1^2=6400+160+1=6561
91^2=(90+1)^2=90^2+2.90.1+1^2=8100+180+1=8281