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\(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)
\(\Leftrightarrow\)\(x^2+y^2+z^2+3-2x-2y-2z\ge0\)
\(\Leftrightarrow\)\(\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(z^2-2z+1\right)\ge0\)
\(\Leftrightarrow\)\(\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2\ge0\)
Dáu "=" xảy ra \(\Leftrightarrow\) \(x=y=z=1\)
a,b,c,d > 0 ta có:
- a < b nên a.c < b.c
- c < d nên c.b < d.b
Áp dụng tính chất bắc cầu ta được: a.c < b.c < b.d hay a.c < b.d (đpcm)
a) \(=x^2+2xy+y^2+x^2-2xy+y^2=2\left(x^2+y^2\right)\)
b) \(=2\left(x^2-y^2\right)+2\left(x^2+y^2\right)=2x^2+2x^2+2y^2-2y^2=4x^2\)( cái này áp dụng luôn kết quả câu trên nha)
c) \(\left(x-y+z\right)^2++2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2=\left(x-y+z+y-z\right)^2=x^2\)
tớ cũng giống Nguyễn Thị Bích Hậu
tích cho nha 1 cái thôi cũng được .
a ) \(\left(x+y\right)^2+\left(x-y\right)^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2\)
\(=2x^2+2y^2\)
b ) \(2.\left(x-y\right).\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=\left[\left(x-y\right)+\left(x+y\right)\right]\)
\(=2x\)
c tương tự
=a, (x-3)(x+3)-(x-7)(x+7)= x2 - 9 - x2 + 7
= -2
b, (4x-5)2+(3x-2)2-2(4x+5)(3x-2)= (4x-5)2 - 2(4x+5)(3x-2) + (3x-2)2
= ( 4x - 5 - 3x + 2 )2
= ( x - 3 )2
c, 2(3x-y)(3x+y)+(3x-y)2+(3x+y)2= 2(3x-y)(3x+y)+(3x-y)2+(3x+y)2
= (3x-y)2+ 2(3x-y)(3x+y)+ (3x+y)2
= ( 3x - y + 3x + y )2
= ( 6x )2
= 36x2
d, (x-y+z)2+(z-y)2+2(x-y+z+2(x-y+z)(y-z-y+z)(y-z)
1, rút gọn
a, (x-3)(x+3)-(x-7)(x+7)
= x^2 - 9 - (x^2 - 49)
= x^2 - 9 - x^2 + 49
= 40
b, (4x-5)2+(3x-2)2-2(4x+5)(3x-2)
= 16x^2 - 40x + 25 + 9x^2 - 12x + 4 - 2(12x^2 - 8x + 15x - 10)
= 25x^2 - 52x + 29 - 24x^2 + 16x - 30x + 20
= x^2 - 66x + 49
c, 2(3x-y)(3x+y)+(3x-y)2+(3x+y)2
= 2(9x^2 - y^2) + 9x^2 - 6xy + y^2 + 9x^2 + 6xy + y^2
= 18x^2 - 2y^2 + 18x^2 + 2y^2
= 36x^2
d, (x-y+z)2+(z-y)2+2(x-y+z+2(x-y+z)(y-z-y+z)(y-z)
= dài vl
nguyễn hoàng mai
MÌnh không ghi đề bài đâu .
\(a,=x^2+2xy+y^2+x^2-2xy+x^2=2\left(x^2+y^2\right)\)2)
\(b,=2\left(x^2-y^2\right)+2\left(x^2+y^2\right)=2x^2+2x^2+2y^2-2y^2=4x^2\)( áp dụng kết quả câu trên )
\(c,\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2=\left(x-y+z+y-z\right)^2=x^2\)
a) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(24+x^3\right)\)
\(=x^3+2^3-24-x^3\)
\(=\left(x^3-x^3\right)+\left(8-24\right)\)
\(=-16\)
phần c hình như sai đầu bài !
a ) \(\left(x+y+z\right)^2=x^2+y^2+z^{2^{ }}+2xy+2yz+2zx\)
Biến đổi vế trái ta được :
\(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)\)
\(=x^2+xy+xz+xy+y^2+yz+zx+zy+z^2\)
\(=x^2+y^2+z^{2^{ }}+2xy+2yz+2zx\)
Vậy \(\left(x+y+z\right)^2=x^2+y^2+z^{2^{ }}+2xy+2yz+2zx\)
Câu 1: Rút gọn
a. (x+y)2 + (x-y)2
=x2+2xy+y2+x2-2xy+y2=2x2+2y2
b. 2.(x-y) . (x+y) + (x+y)2 + (x-y)2
=2.(x2-y2)+2x2+2y2=4x2
c. (x-y+z)2 + (z-y)2 +2.(x-y+z) . (z-y)
=x2+y2+z2-2xy-2yz+2zx+z2-2yz+y2+2.(xz-xy-yz+y2+z2-zy)
=x2+2y2+2z2-2xy+2zx-4yz+2xz-2xy-4yz+2y2+2z2
=x2+4y2+4z2-4xy-8yz+4xz
Câu 2: Chứng minh
(ac+bd)2 + (ad-bc)2=a2c2+2abcd+b2d2+a2d2-2abcd+b2c2= a2c2+b2d2+a2d2+b2c2 =(a2+b2) . (c2+d2)
Câu 1:
a. \(\left(x+y\right)^2+\left(x-y\right)^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2\)
\(=2\left(x^2+y^2\right)\)
b. \(2\left(x-y\right)\left(x+y\right)+\left(x+y^2\right)+\left(x-y\right)^2\)
\(=\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)^2\)
\(=\left(2x\right)^2\)
\(=4x^2\)
a,\(x^2+2xy+y^2+x^2-2xy+y^2=2\left(x^2+y^2\right)\)
b,\(=[\left(x+y\right)+\left(x-y\right)]^2=\left(2x\right)^2=4x^2\)
c,\(=[\left(x-y+z\right)+\left(z-y\right)]^2=\left(x+2z-2y\right)^2\)
bài 2
biến đổi vế trái
\(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
\(=\left(ac+bd\right)^2-2abcd+a^2d^2+b^2c^2\)
\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)(đc c/m)
Câu 1: a) (x+y)2 + (x-y)2 = x2+2xy+y2+x2-2xy+y2 = 2x2 + 2y2 = 2.(x2+y2)
b) 2.(x-y).(x+y)+(x+y)2+(x-y)2 = 2.(x2-y2)+x2+2xy+y2+x2-2xy+y2
= 2x2-2y2+2x2+2y2
= 4x2
c) (x-y+z)2+(z-y)2+2.(x-y+z).(z-y) = (x-y+z+z-y)2 = (x-2y+2z)2
Câu 2: VP = (ac+bd)2+(ad-bc)2 = a2c2+2acbd+b2d2+a2d2-2abcd+b2c2
= a2c2+b2d2+a2d2+b2c2
= (a2c2+b2c2)+(b2d2+a2d2)
= c2.(a2+b2)+ d2.(a2+b20
= (a2+b2).(c2+d2) = VT (đpcm)
Câu 1:
a/ \(\left(x+y\right)^2+\left(x-y\right)^2\)
= \(x^2+2xy+y^2+x^2-2xy+y^2\)
= \(2x^2+2y^2\)
= \(2\left(x^2+y^2\right)\)
b/ \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
= \(2\left(x^2-y^2\right)+x^2+2xy+y^2+x^2-2xy+y^2\)
= \(2x^2-2y^2+2x^2+2y^2\)
= \(4x^2\)
c/ \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(z-y\right)\)
= \(x^2+y^2+z^2-2xy-2yz+2xz+z^2-2yz+y^2+2\left(xz-yz+z^2-xy+y^2-yz\right)\)
= \(x^2+2y^2+2z^2-2xy-4yz+2xz+2xz-2yz+2z^2-2xy+2y^2-2yz\)
= \(x^2+4y^2+4z^2-4xy-8yz+4xz\)
= \(\left(x-2y+2z\right)^2\)
Câu 2:
Ta có \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)= \(\left(ac\right)^2+2abcd+\left(bd\right)^2+\left(ad\right)^2-2abcd+\left(bc\right)^2\)
= \(a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
= \(a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)
= \(\left(c^2+d^2\right)\left(a^2+b^2\right)\)(đpcm)