a) \(\left(2x-3\right)^4=5^4\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}=4\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\)

b) \(\left(2x-3\right)^3=-64\)

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

\(\Rightarrow2x-3=-4\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=-\dfrac{1}{2}\)

a/

\(\left(2x-3\right)^4=5^4\\ \Leftrightarrow\left|2x-3\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

Vậy...

b/ 2x-3=-64

<=>2x = -61

<=> x = -61/2

\(4^x=64\\ 4^x=4^3\\ x=3\)

4^x = 64

4^x = 4³

x  = 3

a: \(27^{2-x}< =9\)

=>\(\left(3^3\right)^{2-x}< =3^2\)

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

=>6-3x<=2

=>-3x<=-4

=>\(x>=\dfrac{4}{3}\)

b: \(7^{3-x}< 49\)

=>\(7^{3-x}< 7^2\)

=>3-x<2

=>-x<2-3=-1

=>x>1

c: \(27^{3-x}>9\)

=>\(\left(3^3\right)^{3-x}>3^2\)

=>\(3^{9-3x}>3^2\)

=>9-3x>2

=>-3x>-7

=>\(x< \dfrac{7}{3}\)

d: \(2^{3-x}< 2^3\)

=>3-x<3

=>-x<0

=>x>0

e: \(27^{3-x^2}< 27^{x+1}\)

=>\(3-x^2< x+1\)

=>\(-x^2-x+2< 0\)

=>\(x^2+x-2>0\)

=>(x+2)(x-1)>0

=>\(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)

a. x × 3 = 27

x = 27 : 3

x = 9

b. 4 × x = 20

x = 20 : 4

x = 5

c. 10 + x : 2 = 20

x : 2 = 20 – 10

x : 2 = 10

x = 10 × 2

x = 20

d. x × 3 = 27 + 3

x × 3 = 30

x = 30 : 3

x = 10

e. 27 : x = 789 – 780

27 : x = 9

x = 27 : 9

x = 3

Phương pháp giải:

- Muốn tìm một số hạng ta lấy tổng trừ đi số hạng kia.

- Muốn tìm một thừa số ta lấy tích chia cho thừa số kia.

Lời giải chi tiết:

a)

● x + 2 = 8

    x = 8 − 2

    x = 6

● x × 2 = 8

    x = 8 : 2

    x = 4

b)

● x + 3 = 12

    x = 12 − 3

    x = 9

● x × 3 = 12

    x = 12 : 3

    x = 4

c)

● 3 + x = 27

    x = 27 − 3

    x = 24

● 3 × x = 27

    x = 27 : 3

    x = 9

a) x + 2 = 8

x = 8 -2 

x = 6

 x × 2 = 8

x = 8 :2

x = 4

b) x + 3 = 12

x = 12 - 3

x = 9

x × 3 = 12

x = 12 : 3

x = 4

c) 3 + x = 27

x = 27 - 3

x = 24

3 × x = 27

x = 27 :3

x = 9

2(x-5)+3=-7

=>2x-10=-7+3=-4

=>2x=-4+10=6

=>x=6:2=3

4x+(27+2^3)=1

=>4x+(27+8)=1

=>4x+35=1

=>4x=1-35=-34

=>x=-34:4=8.5

lx-3l+5=-27

=>lx-3l=-27-5=-32

mà lx-3l>0 với mọi x

=>k co gt x t/m

-3lx-1l=-27

=>lx-1l=-27:(-3)=9

=>x-1=+9

+)x-1=9=>x=10

+)x-1=-9=>x=-8

27*x là j z bn?Tran Thuy vy

\(=27x^3+27-27x^3+1=28\)

Ta có: 

\(C=1-x-\frac{x^2}{3}-\frac{x^3}{27}\)

\(C=1-\left(-27\right)-\frac{\left(-27\right)^2}{3}-\frac{\left(-27\right)^3}{27}\) tại x = -27

\(C=1+27-\frac{3^6}{3}+27^2\)

\(C=1+27-243+729\)

\(C=514\)

\(1-27x^3\)

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

\(=\left(1-3x\right)\left(1+3x+9x^2\right)\)

\(---\)

\(x-3^3+27\)

\(=x-27+27=x\) 

\(---\)

\(27x^3+27x^2+9x+1\)

\(=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3\)

\(=\left(3x+1\right)^3\)

\(---\)

\(\dfrac{x^6}{27}-\dfrac{x^4y}{3}+x^2y^2-y^3\) (sửa đề)

\(=\left(\dfrac{x^2}{3}\right)^3-3\cdot\left(\dfrac{x^2}{3}\right)^2\cdot y+3\cdot\dfrac{x^2}{3}\cdot y^2-y^3\)

 \(=\left(\dfrac{x^2}{3}-y\right)^3\)

#Ayumu

1-27x\(^3\)

=(1-3x)(1+3x+9x\(^2\)