Câu hỏi của Thaodethuong

Question

1/ Viết các biểu thức sau dưới dạng tổng:a. (x+y+z).(x-y-z)b. (x-y+z).(x+y+z)c. -16+(x-3)$^2$

Lê Tuấn Kiệt · Accepted Answer

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.Câu trả lời: 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.

Aikawa Maiya · Answer

$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$Câu trả lời: $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$

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