giúp mình với

C

\(\dfrac{3^8\cdot20^5-3^9\cdot5^5\cdot2^9}{6^8\cdot10^4-3^8\cdot2^9\cdot5^4}=\dfrac{3^8\cdot2^{10}\cdot5^5-3^9\cdot5^5\cdot2^9}{2^8\cdot3^8\cdot2^4\cdot5^4-3^8\cdot2^9\cdot5^4}\\ =\dfrac{3^8\cdot2^9\cdot5^5\left(2-3\right)}{2^9\cdot3^8\cdot5^4\left(2^3-1\right)}=\dfrac{-5}{2^3-1}=\dfrac{-5}{7}\)

Áp dụng t/c dtsbn

\(\dfrac{x+1}{2}=\dfrac{y+2}{3}=\dfrac{z+2}{4}=\dfrac{3x+3-2y-4+z+2}{6-6+4}=\dfrac{-105+1}{4}=\dfrac{-104}{4}=-26\\ \Rightarrow\left\{{}\begin{matrix}x+1=-52\\y+2=-78\\z+2=-104\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-54\\y=-80\\z=-106\end{matrix}\right.\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x+1}{2}=\dfrac{y+2}{3}=\dfrac{z+2}{4}=\dfrac{3x-2y+z+3-4+2}{6-6+4}=\dfrac{-105+1}{4}=-26\)

Do đó: \(\left\{{}\begin{matrix}x+1=-52\\y+2=-78\\z+2=-104\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-53\\y=-80\\z=-106\end{matrix}\right.\)

a, 3 - 2 | 5x - 4 | = -11

2|5x - 4| = 14

|5x - 4| = 7

Th1: 5x -4 =7

5x = 11

x= 11/5

Th2:

5x -4 =-7

5x = -3

x= -3/5

a) => 2/5x-4/=14

   =>   /5x-4/=7

  => 5x-4=7 hoac 5x-4=-7

       x=11/5         x=-3/5

a: =>4/3x=7/9-4/9=1/3

=>x=1/4

b: =>5/2-x=9/14:(-4/7)=-9/8

=>x=5/2+9/8=29/8

c: =>3x+3/4=8/3

=>3x=23/12

hay x=23/36

d: =>-5/6-x=7/12-4/12=3/12=1/4

=>x=-5/6-1/4=-10/12-3/12=-13/12

em moi lop 4 mà

B

B

\(x=\dfrac{1}{y}\Rightarrow\dfrac{1}{y}-y=4\\ \Rightarrow y^2+4y-1=0\\ \Leftrightarrow\left[{}\begin{matrix}y=-2-\sqrt{5}\Rightarrow x=2-\sqrt{5}\\y=-2+\sqrt{5}\Rightarrow x=2+\sqrt{5}\end{matrix}\right.\)

Với \(x=2-\sqrt{5};y=-2-\sqrt{5}\)

\(A=x^2+y^2=18\\ B=x^3-y^3=76\\ C=x^4+y^2=322\)

Với \(x=2+\sqrt{5};y=-2+\sqrt{5}\)

\(A=x^2+y^2=18\\ B=x^3-y^3=76\\ C=x^4+y^4=322\)

A=x^2+y^2

=(x-y)^2+2xy

=4^2+2=18

B=(x-y)^3+3xy(x-y)

=4^3+3*1*4

=64+12=76

C=(x^2+y^2)^2-2x^2y^2

=18^2-2

=322

Bài 9: 

a: \(A=-0.5x^2yz\cdot\left(-3\right)xy^3z=1.5x^3y^4z^2\)

b: Hệ số là 1,5

Bậc là 9