a, \(=-91x-y+5z\)

b, \(=4x^2+x^2y-5y^2-\dfrac{5}{3}x^3+6xy^2+x^2y\)

\(=4x^2+2x^2y-5y^2-\dfrac{5}{3}x^3+6xy^2\)

Ta có: \(\left(3x-1\right)^2=64\)    \(\Rightarrow\left(3x-1\right)^2=8^2\)    \(\Rightarrow3x-1=8\Rightarrow3x=8+1=9\)\(\Rightarrow x=9\div3\Rightarrow x=3\)

Tick nha

\(64-x(3x-27)\div2=40\)

\(\Rightarrow x(3x-27)\div2=64-40\)

\(\Rightarrow x(3x-27)\div2=24\)

\(\Rightarrow x(3x-27)=24\div2\)

\(\Rightarrow x(3x-27)=12\)

\(\Rightarrow4x+27x=12\)

\(\Rightarrow x(27+4)=12\)

\(\Rightarrow31x=12\)

\(\Rightarrow x\in\varnothing\)

=>10x=-100

hay x=-10

⇒3x+7x=-64-36

⇒10x=-100

⇒x=-10

\(\frac{2^{7x}}{2^{3x}}=64\)

\(\Rightarrow\frac{\left(2^7\right)^x}{\left(2^3\right)^x}=2^6\)

\(\Rightarrow\left(\frac{128}{8}\right)^x=2^6\)

\(\Rightarrow16^x=2^6\)

\(\Rightarrow2^{4x}=2^6\)

\(\Rightarrow4x=6\)

\(\Rightarrow x=\frac{6}{4}=\frac{3}{2}\)

Ta co          2^7x /  2^3x = 64

     ===>    2^7x-3x = 64

     ===> 2^4x = 64 = 2⁶

      ^{=====> 4x = 6}

     ===> x = 1,5

1,5 đó bạn

(3x - 2)³ = 64

<=> (3x - 2)³ = 4³

<=>3x-2 = 4

=> 3x = 4+2

=> 3x = 6

=> x = 6 : 2

=> x = 3

bn ơi! mk nhầm

phải là 4:2 chứ ko phải 6:2

mong bn thông cm

bn giải lại đk ko, mik cũng làm như v nhưng tính lại kq sai

Hic

Ta có:64=8\(^2\)

\(\Rightarrow\)3x-1=8

3x=8+1

3x=9

x=9:3

x=3

Vậy x=3.

\(\left(3x-1\right)^2=64\)

\(\Rightarrow\left(3x-1\right)^2=8^2\)

\(\Rightarrow3x-1=8\)

\(\Rightarrow3x=9\)

\(\Rightarrow x=3\)

Vậy x = 3

Ta có:

\(\left(3x-1\right)^2=64\)

\(\Rightarrow\)\(\left(3x-1\right)^2=8^2\)

\(\Rightarrow\)\(3x-1=8\)

\(3x=8+1\)

\(3x=9\)

\(x=9:3\)

\(x=3\)

Vậy x = 3

         ♥♥♥  chúc bạn học tốt ♥♥♥

\(b,\left(5x-1\right)^2:2=8\\ \Leftrightarrow\left(5x-1\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=4\\5x-1=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{5}\end{matrix}\right.\\ c,\left(1-3x\right)^3=-64\\ \Leftrightarrow1-3x=-4\\ \Leftrightarrow3x=5\\ \Leftrightarrow x=\dfrac{5}{3}\)

\(3x+36=-7x-64\)

\(10x=-100\)

\(x=-10\)

\(-5x-178=14x+145\)

\(-19x=323\)

\(x=-17\)

\(\left(3x-4\right).2^3=64\)

\(\Rightarrow3x-4=64:8=8\)

\(\Rightarrow3x=4+8=12\)

\(\Rightarrow x=4\)

(3x-4).2^3 = 64

=> (3x-4).8 = 64

=) 3x-4 = 8

=> 3x = 12

=> x = 12:3

=> x = 4 (tmđk)

Vậy, x = 4

\(\left(3x-4\right).2^3=64\)

\(3x-4=\frac{64}{2^3}\)

\(3x-4=\frac{2^6}{2^3}\)

\(3x-4=2^{6-3}\)

\(3x-4=2^3\)

\(3x=8+4\)

\(3x=12\)

\(x=4\)

vậy \(x=4\)