12x - 33 = 32 . 33
Nhớ giải ra nhé!!!
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: 3A = 3.(1+3+32+33+...+399+3100)
3A = 3+32+33+...+3100+3101
Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)
2A = 3101−1
⇒ A = 3101−1
2
Vậy A = 3101−1
2
1.
$2(-2x+1)\leq -x+3$
$\Leftrightarrow -4x+2\leq -x+3$
$\Leftrightarrow -1\leq 3x$
$\Leftrightarrow x\geq \frac{-1}{3}$
2.
$2(x+1)\leq -x+3$
$\Leftrightarrow 2x+2\leq -x+3$
$\Leftrightarrow 3x\leq 1$
$\Leftrightarrow x\leq \frac{1}{3}$
3.
$5-3(x-1)>2$
$\Leftrightarrow 5-(3x-3)>2$
$\Leftrightarrow 8-3x>2$
$\Leftrightarrow 8-3x-2>0$
$\Leftrightarrow 6-3x>0$
$\Leftrightarrow 6>3x$
$\Leftrightarrow x< 2$
4.
$x^2-12x+3-(x-3)^2>0$
$\Leftrightarrow x^2-12x+3-(x^2-6x+9)>0$
$\Leftrightarrow -6x-6>0$
$\Leftrightarrow -6>6x$
$\Leftrightarrow x< -1$
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
Đặt \(A=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=30.\frac{1}{60}=\frac{1}{2}\)
\(B=\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{90}>\frac{1}{90}+\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}=30.\frac{1}{90}=\frac{1}{3}\)
\(=>Q=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}=A+B>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
Vậy \(Q>\frac{5}{6}\)
a: \(B=3+3^2+3^3+...+3^{120}\)
\(=3\left(1+3+3^2+...+3^{119}\right)⋮3\)
b: \(B=3+3^2+3^3+3^4+...+3^{2020}\)
\(=3\left(1+3\right)+...+3^{2019}\left(1+3\right)\)
\(=4\cdot\left(3+...+3^{2019}\right)⋮4\)
a) 96 - 3 (x - 1) = 42
3 (x - 1) = 96 - 42
3 (x -1) = 54
x - 1 = 54 : 3
x - 1 = 18
x = 18 + 1
x = 19
b) 12x - 33 = 32 . 33
12x - 33 = 35
12x - 33 = 243
12x = 243 + 33
12x = 276
x = 276 : 12
x = 23
Bạn nhớ kiểm tra lại nha dinh nguyen
96 - 3( x+ 1 ) = 42
3( x + 1 ) = 96 - 42
3( x + 1 ) = 54
x + 1 = 54 : 3
x + 1 = 18
x = 18 - 1
x = 17
12x - 33 = 3 ^ 2 . 3 ^ 3
12x - 33 = 9 . 27
12x - 33 = 243
12x = 243 + 33
12x = 276
x = 276 : 12
x = 23
=16.2.(-39) + 16.(-22)
=16.(-78)+16.(-22)
=16.(-78+(-22))
=16.(-100)
=-1600
32.(-39)+16.(-22)
=16.2.(-39)+16.(-22)
=16.(-78)+16.(-22)
=16.(-78+-22)
=16.(-100)
=-1600
\(2x:32=0\)
\(2x=0\times32\)
\(2x=0\)
\(x=0:2\)
\(\Rightarrow x=0\)
\(12x-33=3^2.3^3\)
\(\Leftrightarrow4x-11=3^{\text{4}}\)
\(\Leftrightarrow4x=81+11\)
\(\Leftrightarrow4x=92\)
\(\Leftrightarrow x=92\div4\)
\(\Leftrightarrow x=23\)
Học hỏi nha :))
~ Good luck ~
12x - 33 = 32.33
=> 12x - 33 =9. 27
=> 12x - 33 = 243
=> 12x = 243+ 33
=> 12x= 276
=> x= 276 : 12
=> x = 23
Vậy..