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\(A=\left[\left(a+b\right)+\left(c+d\right)\right]^2+\left[\left(a+b\right)-\left(c+d\right)\right]^2+\left[\left(a-b\right)+\left(c-d\right)\right]^2+\left[\left(a-b\right)-\left(c-d\right)\right]^2\)
Ta có
\(\left[\left(a+b\right)+\left(c+d\right)\right]^2=\left(a+b\right)^2+2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)
\(\left[\left(a+b\right)-\left(c+d\right)\right]^2=\left(a+b\right)^2-2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)
\(\left[\left(a-b\right)+\left(c-d\right)\right]^2=\left(a-b\right)^2+2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)
\(\left[\left(a-b\right)-\left(c-d\right)\right]^2=\left(a-b\right)^2-2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)
\(A=2\left(a+b\right)^2+2\left(a-b\right)^2+2\left(c+d\right)^2+2\left(c-d\right)^2\)
\(A=2\left(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2\right)\)
\(A=4\left(a^2+b^2+c^2+d^2\right)\)
a: \(=a^2+2a\left(b-c\right)+\left(b-c\right)^2+a^2-2a\left(b-c\right)+\left(b-c\right)^2-2\left(b-c\right)^2\)
\(=2a^2+2\left(b-c\right)^2-2\left(b-c\right)^2=2a^2\)
b: \(=a^2+2a\left(b+c\right)+\left(b+c\right)^2+a^2-2a\left(b+c\right)+\left(b+c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2\)
\(=2a^2+2\left(b+c\right)^2+\left(a-b+c\right)^2+\left(a+b-c\right)^2\)
\(=2a^2+2\left(b+c\right)^2+a^2-2a\left(b-c\right)+\left(b-c\right)^2+a^2+2a\left(b-c\right)+\left(b-c\right)^2\)
\(=2a^2+2\left(b+c\right)^2+2a^2+2\left(b-c\right)^2\)
\(=4a^2+2\left(b^2+2bc+c^2+b^2-2bc+c^2\right)\)
\(=4a^2+4b^2+4c^2\)
Cách khác cho bài 1, 2 nha! Akai Haruma em tháy nó nhanh hơn!
1/Đặt \(a=x;b-c=y\)
biểu thức trở thành \(\left(x+y\right)^2+\left(x-y\right)^2-2y^2=2\left(x^2+y^2\right)-2y^2=2x^2=2a^2\)
2/ Đặt \(a-b-c=x;b-c-a=y;c-a-b=z\Rightarrow\left(a+b+c\right)^2=\left(-\left(a+b+c\right)\right)^2=\left(x+y+z\right)^2\)
Khi đó \(B=\left(x+y+z\right)^2+x^2+y^2+z^2\)
\(=2\left(x^2+y^2+z^2+xy+yz+zx\right)\)
\(=\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\)
\(=4\left(a^2+b^2+c^2\right)\)(thay x, y, z bởi các biến đã đặt rồi rút gọn thôi:))
Lời giải:
1.
\((a+b-c)^2+(a-b+c)^2-2(b-c)^2\)
\(=a^2+b^2+c^2+2ab-2ac-2bc+a^2+b^2+c^2-2ab+2ac-2bc-2(b^2-2bc+c^2)\)
\(=2(a^2+b^2+c^2)-4bc-2(b^2+c^2)+4bc\)
\(=2a^2\)
2.
\((a+b+c)^2+(a-b-c)^2+(b-c-a)^2+(c-a-b)^2\)
\(=(a+b+c)^2+a^2+(b+c)^2-2a(b+c)+b^2+(a+c)^2-2b(a+c)+c^2+(a+b)^2-2c(a+b)\)
\(=(a+b+c)^2+a^2+b^2+c^2+[(a+b)^2+(b+c)^2+(c+a)^2]-4(ab+bc+ac)\)
\(=a^2+b^2+c^2+2(ab+bc+ac)+a^2+b^2+c^2+(2a^2+2b^2+2c^2+2ab+2bc+2ac)-4(ab+bc+ac)\)
\(=4(a^2+b^2+c^2)\)
3.
\((a+b+c+d)^2+(a+b-c-d)^2+(a+c-b-d)^2+(a+d-b-c)^2\)
\(=(a+b)^2+(c+d)^2+2(a+b)(c+d)+(a+b)^2+(c+d)^2-2(a+b)(c+d)+(a-b)^2+(c-d)^2+2(a-b)(c-d)+(a-b)^2+(d-c)^2+2(a-b)(d-c)\)
\(=2(a+b)^2+2(c+d)^2+2(a-b)^2+2(c-d)^2\)
\(=2[(a+b)^2+(a-b)^2+(c+d)^2+(c-d)^2]\)
\(=2(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2)\)
\(=2(2a^2+2b^2+2c^2+2d^2)=4(a^2+b^2+c^2+d^2)\)
1.
a) \((a + b + c)^2 + (a - b - c)^2 +( b - c - a) ^2 + (c - a - b)^2 \)
\(= (a + b + c)^2 + (a + b - c)^2 + (a - b - c)^2 + (a - b + c)^2 \)
\(= (a + b)^2 + 2c(a + b) + c^2 + (a + b)^2 - 2c(a + b) + c^2 + (a - b)^2 - 2c(a - b) + c^2 + (a - b)^2 + 2c(a - b) +c^2 \)
\(= 2(a + b)^2 + 2c^2 + 2(a - b)^2 + 2c^2 \)
\(= 2[(a + b)^2 + (a - b)^2] + 4c^2 \)
\(=2(2a^2 + 2b^2) + 4c^2 \)
\(= 4(a^2 + b^2 + c^2)\)
b) Đặt: \(x=a+b; y=c+d; z=a-b; t=c-d \)
Ta được:
\((x+y)^2+(x-y)^2+(z+t)^2+(z-t)^2 \)
\(= (x^2+2xy+y^2)+(x^2-2xy+y^2)+(z^2+2zt+t^2)+(z^2-2zt+t^2) \)
\(= 2x^2+2y^2+2z^2+2t^2 \)
\(= 2(x^2+y^2+z^2+t^2) \)
\(=2.\left[(a+b)^2+(c+d)^2+(a-b)^2+(c-d)^2 \right]\)
\(= 2(a^2+2ab+b^2+c^2+2cd+d^2+a^2-2ab+b^2+c^2-2cd+d^2) \)
\(= 2(2a^2+2b^2+2c^2+2d^2) \)
\(= 4(a^2+b^2+c^2+d^2)\)
làm bừa thui,ai tích mình mình tích lại
số dư lớn nhất bé hơn 175 là 174
số nhỏ nhất có 4 chữ số là 1000
Mà 1000:175=5( dư 125)
số đó là:
a) (a + b + c + d)(a - b - c - d)
= a(a + b + c + d) - b(a + b + c + d) - c(a + b + c + d) - d(a + b + c + d)
= (aa + ab + ac + ad) - (ba + bb + bc + bd) - (ca + cb + cc + cd) - (da + db + dc + dd)
= aa - bb - cc - dd
a) \(A=\left(x-y\right)^2+\left(x+y\right)^2\)
\(=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)
\(=x^2-2xy+y^2+x^2+2xy+y^2\)
\(=\left(x^2+x^2\right)-\left(2xy-2xy\right)+\left(y^2+y^2\right)\)
\(=2x^2+2y^2\)
\(=2.\left(x^2+y^2\right)\)
b) \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)
\(=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(=\left(4a^2-4a^2\right)+\left(4ab+4ab\right)+\left(b^2-b^2\right)\)
\(=8ab\)\
c) \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(=x^2+2xy+y^2-x^2+2xy-y^2\)
\(=\left(x^2-x^2\right)+\left(2xy+2xy\right)+\left(y^2-y^2\right)\)
\(=4xy\)
d) \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+1-8x^2+24x-18+4\)
\(=\left(4x^2-8x^2\right)-\left(4x-24x\right)+\left(1-18+4\right)\)
\(=-4x^2+20x-13\)
\(=-4x^2+20x-25+12\)
\(=-\left(4x^2-20x+25\right)-8\)
\(=-\left[\left(2x\right)^2-2.4x.5+5^2\right]-8\)
\(=-\left(2x-5\right)^2-8\)
2(a+b)2+2(c+d)2+2(a−b)2+2(d−c)2=2(2a2+2b2+2d2+2c2=4(∑a2)⇔2(a+b)2+2(c+d)2+2(a−b)2+2(d−c)2=2(2a2+2b2+2d2+2c2=4(∑a2)