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a, \(\left(x-y+z\right)\left(x-y-z\right)\)
\(=\left(x-y\right)^2-z^2\)(hằng đẳng thức số 3)
b, Sửa đề:\(\left(\dfrac{1}{2}x+y-z\right)\left(\dfrac{1}{2}x+y+z\right)\)
\(=\left(\dfrac{1}{2}x+y\right)^2-z^2\)(hằng đẳng thức số 3)
Chúc bạn học tốt!!!
Bài 1:
a) -16 +(x-3)2
<=> (x-3)2-16
<=> (x-3)2 -42
<=> (x-3-4)(x-3+4)
<=> (x-7)(x+1)
b) 64+16y+y2
<=> y2 + 2.8.y + 82
<=> (y+8)2
c) \(\dfrac{1}{8}-8x^3\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^3-\left(2x\right)^3\)
\(\Leftrightarrow\left(\dfrac{1}{2}-2x\right)\left(\dfrac{1}{4}+x+4x^2\right)\)
d)\(x^2-x+\dfrac{1}{4}\)
\(\Leftrightarrow x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2\)
e) x4 + 4x2 + 4
<=> (x2)2 + 2.2.x2 +22
<=> (x2 + 2)2
g)\(8x^3+60x^2y+150xy^2+125y^3\)
\(\Leftrightarrow\left(2x+5y\right)^3\)
\(a.\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2xz\\ b.\left(x-y+z\right)^2=x^2+y^2+z^2-2xy-2yz+2xz\\ c.\left(x-y-z\right)^2=x^2+y^2+z^2-2xy+2yz-2xz\)
a, \(\left(x+y+z\right)^2=x^2+y^2+c^2+2xy+2yz+2zx\)
b, \(\left(x-y-z\right)^2=x^2+y^2+z^2-2xy+2yz-2zx\)
c, \(\left(x-y+z\right)^2=x^2+y^2+z^2-2xy-2yz+2xz\)
f: \(x^2y^2+2xy+1=\left(xy+1\right)^2\)
g: \(\left(3x-2y\right)^2+2\left(3x-2y\right)+1=\left(3x-2y+1\right)^2\)
h: \(\left(x-3y\right)^2-8\left(x-3y\right)+16=\left(x-3y-4\right)^2\)
i: \(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)^2=4x^2\)
Ngu như bò đực lặt.
Bài này mà làm ko ra.......................................a
a,16x2-9
=42x2-32
=(4x-3)(4x+3) HĐT thứ 3
b,9a2-25b2
=32a2-52b2
=(3a-5b)(3a+5b) HĐT thứ 3
c,81-y4
=32.32-y2.y2
=(32-y2)
=(3-y)(3+y) HĐT thứ 3
d,(2x+y)2-1
=(2x+y-1)(2x+y-1) HĐT thứ 3
e,(x+y+z)2-(x-y-z)2
cái này là HĐT thứ 8 mở rộng bạn lên mạng tìm nha
a) \(16x^2-9=\left(4x\right)^2-3^2=\left(4x-3\right).\left(4x+3\right)\)
b) \(9a^2-25b^4=\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right).\left(3a+5b^2\right)\)
c) \(81-y^4=9^2-\left(y^2\right)^2=\left(9-y^2\right).\left(9+y^2\right)\)
d)\(\left(2x+y\right)^2-1=\left(2x+y\right)^2-1^2=\left(2x+y-1\right).\left(2x+y+1\right)\)
a, (x + y + z)(x - y - z)
= x^2 - xy - xz + xy - y^2 - zy + zx - zy - z^2
= x^2 + y^2 + z^2 + (xy - xy) + (xz - xz) - (zy + zy)
= x^2 + y^2 + z^2 - 2zy
b, (x - y + z)(x + y + z)
= x^2 + xy + xz - xy - y^2 - zy + zx + zy + z^2
= x^2 + y^2 + z^2 + (xy - xy) + xz + xz + (zy - zy)
= x^2 + y^2 + z^2 + 2zx
a) Ta có:
(x+y+z)(x-y-z) = x^2 -xy -xz +yx- y^2 -yz+zx -zy -z^2
=x^2 - y^2 - 2yz - z^2.
b) Ta có: (x-y+z)(x+y+z) = x^2 +xy+xz -yx-y^2 -yz +zx+zy +z^2
=x^2 +2xz- y^2 +z^2.
c) Ta có: -16 + (x-3)^2 = -16 + ( x^2-6x+9)
= -16 + x^2 - 6x + 9
= x^2 - 6x - 7.
\(a,\left(x+y+z\right)\left(x-y-z\right)\)
\(=x\left(x-y-z\right)+y\left(x-y-z\right)+z\left(x-y-z\right)\)
\(=x^2-xy-xz+xy-y^2-yz+xz-yz-z^2\)
\(=x^2-y^2-2yz-z^2\)
\(b,\left(x-y+z\right)\left(x+y+z\right)\)
\(=x\left(x+y+z\right)-y\left(x+y+z\right)+z\left(x+y+z\right)\)
\(=x^2+xy+xz-xy-y^2-yz+xz+yz+z^2\)
\(=x^2+2xz-y^2+z^2\)
\(c,-16+\left(x-3\right)^2\)
\(=-16+x^2-6x+9\)
\(=x^2-6x-7\)