Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. 25 . 3x-3 = 2025
3x-3 = 2025 : 25
3x-3 = 81
3x-3 = 34
=> x - 3 = 4
x = 4 + 3
x = 7
Vậy x = 7
2. Chứng minh:
M = 2 + 22 + 23 +...+298
M = ( 2 + 22 ) + ( 23 + 24 ) +...+ ( 297 + 298 )
M = 2.( 1 + 2 ) + 23.( 1 + 2 ) +...+ 297.( 1 + 2 )
M = 2.3 + 23.3 +...+ 297.3 \(⋮\)3
=> M\(⋮\)3
Answer :
\(\Rightarrow A+1=1+1+2+2^2+...+2^{2021}\)
\(\Rightarrow A+1=2+2+2^2+...+2^{2021}\)
\(\Rightarrow A+1=2^2+2^2+2^3+...+2^{2021}\)
\(\Rightarrow A+1=2^3+2^3+2^4+...+2^{2021}\)
....
\(\Rightarrow A+1=2^{2021}+2^{2021}=2^{2022}\)
Mà \(2^x=A+1\Rightarrow2^x=2^{2022}\Rightarrow x=2022\)
Nhiều câu quá >.<
a/ \(2x\left(x+5\right)=\left(x+3\right)^2+\left(x-1\right)^2+20.\)
\(2x^2+10x=x^2+6x+9+x^2-2x+1+20.\)
\(10x=4x+30\)
\(6x=30\Rightarrow x=5\)
các câu còn lại tương tự
\(a,2x\left(x+5\right)=\left(x+3\right)^2+\left(x-1\right)^2+20\)
\(\Leftrightarrow2x^2+10x=x^2+6x+9+x^2-2x+1+20\)
\(\Leftrightarrow2x^2+10x=2x^2+4x+30\)
\(\Leftrightarrow2x^2+10x-2x^2-4x=30\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\)
Vậy ...........
\(b,\left(2x-2\right)^2=\left(x+1\right)^2+3\left(x-2\right)\left(x+5\right)\)
\(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3x^2+15x-6x-30\)
\(\Leftrightarrow4x^2-8x+4=4x^2+11x-29\)
\(\Leftrightarrow4x^2-8x-4x^2-11x=-29-4\)
\(\Leftrightarrow-19x=-33\)
\(\Leftrightarrow x=\frac{33}{19}\)
Vậy...........
\(c,\left(x-1\right)^2+\left(x+3\right)^2=2\left(x-2\right)\left(x+1\right)+38\)
\(\Leftrightarrow x^2-2x+1+x^2+6x+9=2x^2+2x-4x-4+38\)
\(\Leftrightarrow2x^2+4x+10=2x^2-2x+34\)
\(\Leftrightarrow2x^2+4x-2x^2+2x=34-10\)
\(\Leftrightarrow6x=24\)
\(\Leftrightarrow x=4\)
Vậy.............
\(d,\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-18\)
\(\Leftrightarrow x^3+6x+12x+8-\left(x^3-6x+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x+12x+8-x^3+6x-12x+8=12x^2-12x-8\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy............
2.
\(\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left(65\cdot111-13\cdot15\cdot37\right)\\ =\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left(65\cdot111-13\cdot5\cdot3\cdot37\right)\\=\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left[65\cdot111-\left(13\cdot5\right)\cdot\left(3\cdot37\right)\right]\\ =\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left[65\cdot111-65\cdot111\right]\\ =\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot0\\ =0\)
b) \(2025^x=9^4\cdot5^4\)
\(\left(45^2\right)^x=\left(9\cdot5\right)^4\)
\(45^{2x}=45^4\)
\(\Rightarrow2x=4\)
\(x=4:2\)
\(x=2\)
Vậy x = 2
=))
1) (x+2)2=36
(x+2)2=62
=>x+2=6 hoặc x+2=-6
x=6-2 hoặc x=-6-2
x=4 hoặc x=-8
2)\(\left(\frac{3}{4}\right)^x=\frac{2^8}{3^4}\)
\(\left(\frac{3}{4}\right)^x:\frac{2^8}{3^4}=0\)
\(\left(\frac{3}{4}\right)^x.\frac{3^4}{2^8}=0\)
không có giá trị x nào thỏa mãn vì \(\left(\frac{3}{4}\right)^x>0;\frac{3^4}{2^8}>0\)
5(x-2)(x+3)=1
5(x-5)(x+3)=50
=>(x-2)(x+3)=0
=>x-2=0 hoặc x+3=0
x=2 hoặc x=-3
(x-2)8=(x-2)6
vì 8 là mũ chẵn nên (x-2)8=(x-2)6 khi:
x-2=0 hoặc x-2=1 hoặc x-2=-1
x=2 hoặc x=1/2 hoặc x=1