K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

6 tháng 10 2019

(3x+2)^3:8=64

(3x+2)^3= 512

(3x+2)^3= 8^3

3x+2=8

3x=6

x=2

6 tháng 10 2019

( 3x+2):8 = 64

=> ( 3x+2) = 64x8=512=83

=> 3x+2=8

=>3x=6=>x=2

hc tốt nhá

12 tháng 10 2021

b: \(\left(17x-25\right):8+65=9^5:9^3\)

\(\Leftrightarrow\left(17x-25\right):8=9^2-65=81-65=16\)

\(\Leftrightarrow17x-25=128\)

hay x=9

 

12 tháng 10 2021

b: (17x−25):8+65=95:93(17x−25):8+65=95:93

⇔(17x−25):8=92−65=81−65=16⇔(17x−25):8=92−65=81−65=16

⇔17x−25=128⇔17x−25=128

hay x=9

 

HQ
Hà Quang Minh
Giáo viên
3 tháng 8 2023

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

12 tháng 11 2019

\(2.\left(3x-8\right)=64:2^3\)

\(2.\left(3x-8\right)=64:8\)

\(2.\left(3x-8\right)=8\)

\(3x-8=4\)

\(3x=12\)

\(x=4\)

12 tháng 11 2019

Ta có : \(2.\left(3x-8\right)=64:2^3\)

\(\Rightarrow3x-8=64:2^3:2\)

\(\Rightarrow3x-8=64:16\)

\(\Rightarrow3x-8=4\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

Vậy x = 4

14 tháng 5 2021

Đáp ấn:D

`(3x-1)^2=64`
`<=>` $\left[ \begin{array}{l}3x-1=8\\3x-1=-8\end{array} \right.$
`<=>` $\left[ \begin{array}{l}3x=9\\3x=-7\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=3\\x=-\dfrac73\end{array} \right.$

14 tháng 5 2021

D

\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\) =\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\) =\(a^2\) b)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\) =\(\left(3x+1\right)^2-2\left(3x+3-2\right)\left(3x+3+2\right)+\left(3x+5\right)^2\) =\(\left(3x+1\right)^2-2\left(\left(3x+3\right)^2-4\right)+\left(3x+5\right)^2\) =\(9x^2+6x+1-18x^2-36x-9+8+9x^2+30x+25\) =25 c)\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)\) =\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)....\left(2^{64}+1\right)\) =\(\left(2^2-1\right)\left(2^2+1\right)...\left(2^{64}+1\right)\) =... =\(\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\) \)

d)Tương tự

\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)

=\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\)

=\(a^2\)

b)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

=\(\left(3x+1\right)^2-2\left(3x+3-2\right)\left(3x+3+2\right)+\left(3x+5\right)^2\)

=\(\left(3x+1\right)^2-2\left(\left(3x+3\right)^2-4\right)+\left(3x+5\right)^2\)

=\(9x^2+6x+1-18x^2-36x-9+8+9x^2+30x+25\)

=25

c)\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)\)

=\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)....\left(2^{64}+1\right)\)

=\(\left(2^2-1\right)\left(2^2+1\right)...\left(2^{64}+1\right)\)

=...

=\(\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\)

d)Tương tự

5 tháng 9 2017

cảm ơn

a: \(=\left[a-\left(b-c\right)\right]^2-\left(b-c\right)^2+2ab-2ac\)

\(=a^2-2a\left(b-c\right)+\left(b-c\right)^2-\left(b-c\right)^2+2ab-2ac\)

\(=a^2-2ab+2ac+2ab-2ac=a^2\)

b: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left(3x+1-3x-5\right)^2\)

\(=\left(-4\right)^2=16\)

c: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)\left(2^{64}+1\right)\)

\(=2^{128}-1\)

d: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{3^{64}-1}{2}\)

12 tháng 2 2017

Tớ mới học lớp 5 mà

12 tháng 12 2017

a)(2x+4).8-40=24

(2x+4).8=24+40

(2x+4).8=64

2x+4=64:8

2x+4=8

2x=8+4

2x=12

x=12:2

x=6

b)(2x+42)-8=64:2

(2x+16)-8=32

2x+16=32+8

2x+16=40

2x=40-16

2x=24

x=24:2

x=12

c)(3x+2).4-20=24

(3x+2).4=24+20

(3x+2).4=44

3x+2=44:4

3x+2=11

3x=11-2

3x=9

x=9:3

x=3