2x-y=1 ; 2y-z=2 và 2z-x=3
Hãy tính tổng của x+y+z
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NH
0
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 4 2021
a. \(y'=\dfrac{-1}{\left(x-1\right)}\)
b. \(y'=\dfrac{5}{\left(1-3x\right)^2}\)
c. \(y=\dfrac{\left(x+1\right)^2+1}{x+1}=x+1+\dfrac{1}{x+1}\Rightarrow y'=1-\dfrac{1}{\left(x+1\right)^2}=\dfrac{x^2+2x}{\left(x+1\right)^2}\)
d. \(y'=\dfrac{4x\left(x^2-2x-3\right)-2x^2\left(2x-2\right)}{\left(x^2-2x-3\right)^2}=\dfrac{-4x^2-12x}{\left(x^2-2x-3\right)^2}\)
e. \(y'=1+\dfrac{2}{\left(x-1\right)^2}=\dfrac{x^2-2x+3}{\left(x-1\right)^2}\)
g. \(y'=\dfrac{\left(4x-4\right)\left(2x+1\right)-2\left(2x^2-4x+5\right)}{\left(2x+1\right)^2}=\dfrac{4x^2+4x-14}{\left(2x+1\right)^2}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 4 2021
2.
a. \(y'=4\left(x^2+x+1\right)^3.\left(x^2+x+1\right)'=4\left(x^2+x+1\right)^3\left(2x+1\right)\)
b. \(y'=5\left(1-2x^2\right)^4.\left(1-2x^2\right)'=-20x\left(1-2x^2\right)^4\)
c. \(y'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{2x+1}{x-1}\right)'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{-3}{\left(x-1\right)^2}\right)=\dfrac{-9\left(2x+1\right)^2}{\left(x-1\right)^4}\)
d. \(y'=\dfrac{2\left(x+1\right)\left(x-1\right)^3-3\left(x-1\right)^2\left(x+1\right)^2}{\left(x-1\right)^6}=\dfrac{-x^2-6x-5}{\left(x-1\right)^4}\)
e. \(y'=-\dfrac{\left[\left(x^2-2x+5\right)^2\right]'}{\left(x^2-2x+5\right)^4}=-\dfrac{2\left(x^2-2x+5\right)\left(2x-2\right)}{\left(x^2-2x+5\right)^4}=-\dfrac{4\left(x-1\right)}{\left(x^2-2x+5\right)^3}\)
f. \(y'=4\left(3-2x^2\right)^3.\left(3-2x^2\right)'=-16x\left(3-2x^2\right)^3\)
12 tháng 12 2020
a, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x+1}+1\)
\(=\frac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^3-x^2-2x+x-1-x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^3-2x^2-x-2}{\left(x-1\right)\left(x+1\right)}\)
K
1
12 tháng 12 2020
a) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}+\dfrac{1}{1+x+1}\) \(=\dfrac{x^2.\left(x-1\right)\left(x+2\right)}{\left(x+1\right).\left(x-1\right)\left(x+2\right)}+\dfrac{2x.\left(x+2\right)}{\left(x-1\right).\left(x+1\right).\left(x+2\right)}+\dfrac{\left(x-1\right).\left(x+1\right)}{\left(x-1\right)\left(x+1\right).\left(x+2\right)}\)
\(=\dfrac{x^2.\left(x-1\right).\left(x+2\right)+2x.\left(x+2\right)+\left(x-1\right)\left(x+1\right)}{\left(x+1\right).\left(x-1\right).\left(x+2\right)}\)
\(=\dfrac{x^4+x^3-2x^2+2x^2+4x+x^2-1}{\left(x-1\right)\left(x+1\right).\left(x+2\right)}\)
\(=\dfrac{x^4+x^3+x^2+4x-1}{\left(x^2-1\right).\left(x+2\right)}\)
\(=\dfrac{x^4+x^3+x^2+4x-1}{x^3+2x^2-x-2}\)
TP
2
4 tháng 9 2021
a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)
\(=6x^2-2x-6x^2-2x+18x+6\)
=14x+6
b: Ta có: \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)+3\left(2x-3\right)\)
\(=4x^2-12x+9-4x^2+1+6x-9\)
\(=-6x+1\)
c: Ta có: \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y-1-x-y\right)^2\)
=1
4 tháng 9 2021
a) \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)=6x^2-2x-6x^2-2x+18x+6=14x+6\)
b) \(\left(2x-3\right)^2-\left(1+2x\right)\left(2x-1\right)+3\left(2x-3\right)=4x^2-12x+9-4x^2+1+6x-9=-6x+1\)
c) \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-1-x-y\right)^2=\left(-1\right)^2=1\)
LL
1
7 tháng 7 2017
a. \(\left(2x^2+y\right)\left(2x^2-y\right)-4x^2+y^2=\left(2x^2\right)^2-y^2-4x^2+y^2\)
\(=4x^4-4x^2\)
b. \(\left(2x^2+y\right)^2-\left(2x^2-y^2\right)=4x^4+4x^2y+y^2-2x^2+y^2\)
\(=4x^4+4x^2y-2x^2+2y^2\)
c. \(\left(2x+1\right)\left(2x-1\right)-4x^2=\left(2x\right)^2-1^2-4x^2=4x^2-1-4x^2=-1\)
d. \(\left(2x^{3y}+y\right)^2-\left(y-2x^{3y}\right)^2\)
\(=\left(2x^{3y}+y+y-2x^{3y}\right)\left(2x^{3y}+y-y+2x^{3y}\right)\)
\(=2y.2.2x^{3y}=4y.2x^{3y}\)
2x-y=1 ; 2y-z=2 ; 2z-x=3
=> 1+y=2x ; 2+z=2y ; 3+x=2z
=> 2x+2y+2z = 1+y+2+z+3+x
=> 2.(x+y+z)=x+y+z+(1+2+3)
=> 2.(x+y+z)=x+y+z+6
=> x+y+z =6
Vậy x+y+z=6
yamate cư đa sai