=>10y=36,8

hay y=3,68

y = 3,68

=> \(y\times\left(7,1+2,9\right)=36,8\)
=> \(y\times10=36,8\)
=> \(y=36,8:10\)
=> \(y=3,68\)

a, 3.2^x+1-2=94

B,  (3x-1)³=125

C,  2^x+2^x+1+...........+2^x+99=2¹⁰⁰-1

. là dấu nhân 

MIK CẦN GẤP

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!

\(\hept{\begin{cases}x-\left(y+z\right)=\frac{-1}{12}\\y-\left(x+z\right)=\frac{-1}{2}\\z-\left(x+y\right)=\frac{-7}{12}\end{cases}}\)

\(\Leftrightarrow-\left(x+y+z\right)=\frac{-7}{6}\)

\(\Leftrightarrow\hept{\begin{cases}-\left(x+y\right)=z-\frac{7}{6}\\-\left(x+z\right)=y-\frac{7}{6}\\-\left(y+z\right)=x-\frac{7}{6}\end{cases}}\)

Thay vô tinh tiếp, đc chứ??

Ta suy ra :x - (y + z) + y - (x + z) + z - (x+y) = \(\frac{-1}{12}\)+\(\frac{-1}{2}\)+\(\frac{-7}{12}\)= \(\frac{-7}{6}\)

=> x - y -z + y - x - z + z - x - y = -(x + y + z)= \(\frac{-7}{6}\)

Ta có x - (y + z) = x + (-(y +z)) = -(y + z) - x + 2x = -(x + y + z) + 2x = \(\frac{-7}{6}\)+ 2x = \(\frac{-1}{12}\)

=> 2x = \(\frac{13}{12}\)

x =\(\frac{13}{12}\) : 2 = \(\frac{13}{24}\)

Tương tự : y - (z + x) = - (x + y + z) + 2y = \(\frac{-7}{6}\)+ 2y = \(\frac{-1}{2}\)

=> 2y = \(\frac{4}{6}\)=\(\frac{2}{3}\)

=> y = \(\frac{2}{3}\): 2 = \(\frac{2}{6}\)=\(\frac{1}{3}\)

Vậy : - (x + y + z) = -(\(\frac{13}{24}\)+ \(\frac{1}{3}\)+ z) = -(\(\frac{7}{8}\)+ z) = -\(\frac{7}{6}\)

=> z = \(\frac{7}{6}\)- \(\frac{7}{8}\)= \(\frac{7}{24}\)

(x-3)(2y+5)=0

Th1 X-3=0

       x=3

Th2 2y+5=0 

         y=-5/2

1/ Vì x,y,z tỉ lệ với 3,5,7 nên \(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}=\frac{x-z}{3-7}=\frac{12}{-4}=-3\)

=> x/3 = -3 => x = -9

y/5 = -3 => y = -15

z/7 = -3 => z = -21

Vậy x=-9,y=-15,z=-21

2/ 

Ta có: 3x = 4y => x/4 = y/3

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x}{4}=\frac{y}{3}=\frac{x+y}{4+3}=\frac{21}{7}=3\)

=> x/4 = 3 => x = 12

y/3 = 3 => y = 9

Vậy x=12,y=9

3/ 

Ta có: 5a = 2b => a/2 = b/5 => 3a/6 = 2b/10

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{3a}{6}=\frac{2b}{10}=\frac{3a+2b}{6+10}=\frac{32}{16}=2\)

=> a/2 = 2 => a = 4

b/5 = 2 => b = 10

Vậy a=4,b=10

Đặt x/3 = y/2 = z/5= k

=> x= 3k 

y = 2k 

z= 5k

=> xyz = 22 * 5

3k * 2k * 5k = 110

30 * k^3  = 110

k^3 = 11/3

k= ....                    (đề sao mà ra kq nhiều số qá)

Tìm đc k rồi thì thay vào chỗ x= 3k; y= 2k gì đó, vậy là ra x;y;z.

\(\frac{x}{3}=\frac{y}{2}=\frac{z}{5}=\frac{x\times y\times z}{3\times2\times5}=\frac{22,5}{30}=0,75\)

\(\Rightarrow\frac{x}{3}=0,75\Rightarrow x=2,25\)

\(\Rightarrow\frac{y}{2}=0,75\Rightarrow y=1,5\)

\(\Rightarrow\frac{z}{5}=0,75\Rightarrow z=3,75\)

\(\frac{x}{3}=\frac{y}{2}=\frac{z}{5}=k\)

=> \(\hept{\begin{cases}x=3k\\y=2k\\z=5k\end{cases}}\)

=> 3k . 2k . 5k = 22,5

=> 30 k³ = 22,5

=> k³ = \(\frac{3}{4}\)

=> k \(\sqrt[3]{\frac{3}{4}}\)

=> \(\hept{\begin{cases}x=3.\sqrt[3]{\frac{3}{4}}\approx2,7256\\y=2.\sqrt[3]{\frac{3}{4}}\approx1,8171\\z=5.\sqrt[3]{\frac{3}{4}}\approx4,5428\end{cases}}\)

tuy minh la ban nha

\(a,\left(y^{54}\right)^2=y\)\(\Rightarrow y^{108}=y\)\(\Rightarrow y=\pm1\)

\(b,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)

\(\Rightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)

\(\Rightarrow x\in\left\{0;1;2\right\}\)

\(c,x\left(6-x\right)^{2019}=\left(6-x\right)^{2019}\)

\(\Rightarrow\left(6-x\right)^{2019}\left(x-1\right)=0\)

\(\Rightarrow x\in\left\{1;6\right\}\)

\(\left(y^{54}\right)^2=y\)

\(\Rightarrow y^{108}=y\)

\(\Rightarrow y^{108}-y=0\)

\(\Rightarrow y\cdot\left(y^{107}-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}y=0\\y^{107}-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}y=0\\y^{107}=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}\)

a) \(\left(y^{54}\right)^2=y\)

\(\Leftrightarrow y^{108}=y\)

\(\Leftrightarrow y^{108}-y=0\)

\(\Leftrightarrow y.\left(y^{107}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=0\\y^{107}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}}\)

Vậy \(y\in\left\{0;1\right\}\)

ai giúp mk đi