1.Bỏ ngoặc rồi tính :
a) (x-1)-(x-3)
b) (2+y)-(y+1)
c) -x-(y-x-z)
d) x- z-(x-y+z)
K
Khách
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NL
1 tháng 7 2017
a.\(\left(258-79\right)-\left(111-90\right)\)
\(=258-79-111+90\)
\(=258+\left(-79\right)+\left(-111\right)+90\)
\(=258+\left[\left(-79\right)+\left(-111\right)\right]+90\)
\(=258+\left(-190\right)+90\)
\(=258+\left(-100\right)\)
\(=158\)
b.\(\left(617-304\right)-\left(617+254\right)\)
\(=617-304-617-254\)
\(=617+\left(-304\right)+\left(-617\right)+\left(-254\right)\)
\(=\left[617+\left(-617\right)\right]+\left[\left(-304\right)+\left(-254\right)\right]\)
\(=0+\left(-558\right)\)
\(=-558\)
c.\(\left(217-372\right)-\left(217-465\right)-\left(465+372\right)\)
\(=217-372-217+465-465-372\)
\(=217+\left(-372\right)+\left(-217\right)+465+\left(-465\right)+\left(-372\right)\)
\(=\left[217+\left(-217\right)\right]+\left[\left(-372\right)+\left(-372\right)\right]+\left[465+\left(-465\right)\right]\)
\(=0+\left(-744\right)+0\)
\(=-744\)
d.\(\left(x+y+z+t\right)-\left(x+y+z-t\right)-\left(y+z+t-x\right)-\left(z+t+x-y\right)-\left(t+x+y-z\right)\)
\(=x+y+z+t-x-y-z+t-y-z-t+x-z-t-x+y-t-x-y+z\)viết hoài mỏi tay nên khúc sau tự làm nhé ^_^
G
7 tháng 7 2019
ta có: \(x+y+z=a\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=a^2\)
\(\Rightarrow b+2\left(xy+yz+xz\right)=a^2\Rightarrow xy+yz+xz=\frac{a^2-b}{2}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{c}\Rightarrow c\left(xy+yz+xz\right)=xyz\)
Ta có:\(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)
\(=a\left(b-\frac{a^2-b}{2}\right)+\frac{3c\left(a^2-b\right)}{2}\)
S
0
AH
Akai Haruma
Giáo viên
14 tháng 10 2021
Lời giải:
$xy+yz+xz=\frac{1}{2}[(x+y+z)^2-(x^2+y^2+z^2)]=\frac{1}{2}(a^2-b^2)$
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}$
$\Rightarrow xyz=c(xy+yz+xz)=\frac{1}{2}c(a^2-b^2)$
Khi đó:
$P=(x+y+z)^3-3(x+y)(y+z)(x+z)$
$=(x+y+z)^3-3[(x+y+z)(xy+yz+xz)-xyz]=(x+y+z)^3-3(xy+yz+xz)(x+y+z)+3xyz$
$=a^3-\frac{3}{2}a(a^2-b^2)+\frac{3}{2}c(a^2-b^2)$
1. Bỏ ngoặc rồi tính:
a, ( x - 1)-(x - 3)
= x - 1 - x + 3
= -1 + 3 = 2
b, (2 + y) - ( y + 1)
= 2 + y - y - 1
= 2 - 1 = 1
c,- x -( y - x - z)
= -x - y + x + z
= -y + z
d, x- z - ( x - y + z)
= x - z - x + y - z
= -2z + y
Ticks mình nhé!
a, \(\left(x-1\right)-\left(x-3\right)=x-1-x+3=2\)
b, \(\left(2+y\right)-\left(y+1\right)=2+y-y-1=1\)
c, \(-x-\left(y-x-z\right)=-x-y+x+z=-y+z\)
d, \(x-z-\left(x-y+z\right)=x-z-x+y-z=2z+y\)