Cho mk hỏi con này ra bao nhiu z: \(A=\frac{\left(\sqrt{x}+2\right)\cdot\left(x-1\right)-\left(\sqrt{x}-2\right)\cdot\left(x+1\right)\cdot\left(x+1\right)}{\left(x+1\right)^2\cdot\left(x-1\right)}\)
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15 tháng 6 2018
1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
8 tháng 8 2019
\(S=\frac{yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left(x-z\right)+xy\left(z+1\right)\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
+ \(yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left(x-z\right)+xy\left(z+1\right)\left(x-y\right)\)
\(=yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left[\left(y-z\right)+\left(x-y\right)\right]\)
\(+xy\left(z+1\right)\left(x-y\right)\)
\(=\left(y-z\right)\left[yz\left(x+1\right)-zx\left(y+1\right)\right]+\left(x-y\right)\left[xy\left(z+1\right)-zx\left(y+1\right)\right]\)
\(=\left(y-z\right)\left[z\left(y-x\right)\right]+\left(x-y\right)\cdot x\cdot\left(y-z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
\(\Rightarrow S=\frac{1}{xyz}\)
2 tháng 10 2020
MTC: (x+y)(x+1)(1-y)
\(=\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}=\frac{\left(x+y\right)\left(1+x\right)\left(1-y\right)\left(x-y+xy\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=x-y+xy\)
Với \(x\ne-1;x\ne-y;y\ne1\)thì giá trị biểu thức được xác định
7 tháng 6 2016
a
(x+1)-(x-1)-3(x+1)(x-1)
=(x+1)-(x-1)-3x+1.(x-1)
=(x+1)-(x-1)-3x+x-1
=x+1-x+1-3x+x-1
=x-x-3x+x+1+1-1
=-2x
b,
5(x+2)(x-2)-1/2(6-8x)^2+17
=5x+10(x-2)-1/2(36-64x2)+17
=5x+10x-20-18+32x2+17
=5x+10x-20-18+17+32x2
=15x-21+32x2
7 tháng 6 2016
a
(x+1)-(x-1)-3(x+1)(x-1)
=(x+1)-(x-1)-3x+1.(x-1)
=(x+1)-(x-1)-3x+x-1
=x+1-x+1-3x+x-1
=x-x-3x+x+1+1-1
=-2x
b,
5(x+2)(x-2)-1/2(6-8x)^2+17
=5x+10(x-2)-1/2(36-64x2)+17
=5x+10x-20-18+32x2+17
=5x+10x-20-18+17+32x2
=15x-21+32x2
31 tháng 3 2017
\(\Rightarrow\frac{3}{4}x+5-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}+3\)+3
\(\Rightarrow\left(\frac{3}{4}x-\frac{2}{3}x-\frac{1}{6}x\right)+\left(5+4-1\right)=\frac{1}{3}x+\left(4-\frac{1}{3}+3\right)\)
=>\(\frac{-1}{12}x+8=\frac{1}{3}x+\frac{20}{3}\)\(\Rightarrow\frac{-1}{12}x+8-\frac{1}{3}x=\frac{20}{3}\)
\(\Rightarrow\left(\frac{-1}{12}-\frac{1}{3}\right)x+8=\frac{20}{3}\)
\(\Rightarrow\frac{-5}{12}x+8=\frac{20}{3}\Rightarrow\frac{-5}{12}x=\frac{20}{3}-8\)
\(\Rightarrow\frac{-5}{12}x=\frac{-4}{3}\Rightarrow x=\frac{-4}{3}:\frac{-5}{12}=\frac{16}{5}\)
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