Tìm x là số TN biết:
a) 22 . 23 . x = 82
b) 5x-1 . 52 = 252
c) (2x - 1)3 = 125
d) 8x . 2x = 44
e) ( x2 . 53 )2 = 56
Làm nhanh nhất và chính xác nhất mình tick nhé
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Những câu hỏi liên quan
LT
Lương Thị Vân Anh
CTVHS
16 tháng 8 2023
a) Ta có A = 21 + 22 + 23 + ... + 22022
2A = 22 + 23 + 24 + ... + 22023
2A - A = ( 22 + 23 + 24 + ... + 22023 ) - ( 21 + 22 + 23 + ... + 22022 )
A = 22023 - 2
Lại có B = 5 + 52 + 53 + ... + 52022
5B = 52 + 53 + 54 + ... + 52023
5B - B = ( 52 + 53 + 54 + ... + 52023 ) - ( 5 + 52 + 53 + ... + 52022 )
4B = 52023 - 5
B = \(\dfrac{5^{2023}-5}{4}\)
b) Ta có : A + 2 = 2x
⇒ 22023 - 2 + 2 = 2x
⇒ 22023 = 2x
Vậy x = 2023
Lại có : 4B + 5 = 5x
⇒ 4 . \(\dfrac{5^{2023}-5}{4}\) + 5 = 5x
⇒ 52023 - 5 + 5 = 5x
⇒ 52023 = 5x
Vậy x = 2023
10 tháng 2 2019
Ta có -2-(x-17)=34-(-x+25)
-2-x+17=34+x+25
-2 +(-x)+17+(-x)=34+25
[(-x)+(-x)]+[(-2)+17]=29
(-x).2+15=29
(-x).2=29-15
(-x).2=14
(-x)=14:2
(-x)=7
x=-7
9 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
VT
2
2 tháng 10 2021
a) \(\Rightarrow\left(x-1\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b) \(\Rightarrow25x^2+10x+1-25x^2+9=30\)
\(\Rightarrow10x=20\Rightarrow x=2\)
HP
2 tháng 10 2021
a. x2 - 2x + 1 = 25
<=> x2 - 2x - 24 = 0
<=> x2 - 6x + 4x - 24 = 0
<=> x(x - 6) + 4(x - 6) = 0
<=> (x + 4)(x - 6) = 0
<=> \(\left[{}\begin{matrix}x+4=0\\x-6=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-4\\x=6\end{matrix}\right.\)
b. (5x + 1)2 - (5x - 3)(5x + 3) = 30
<=> 25x2 + 10x + 1 - 25x2 + 9 = 30
<=> 25x2 - 25x2 + 10x = 30 - 1 - 9
<=> 10x = 20
<=> x = 2
H9
HT.Phong (9A5)
CTVHS
9 tháng 7 2023
a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)
\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)
\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)
\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)
\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)
Vậy: \(1+2^2+2^3+...+2^{10}=2045\)
b)
a] \(60-3\left(x-1\right)=2^3\cdot3\)
\(\Rightarrow60-3\left(x-1\right)=24\)
\(\Rightarrow3\left(x-1\right)=36\)
\(\Rightarrow x-1=12\)
\(\Rightarrow x=13\)
b] \(\left(3x-2\right)^3=2\cdot2^5\)
\(\Rightarrow\left(3x-2\right)^3=2^6\)
\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)
\(\Rightarrow3x-2=2^2\)
\(\Rightarrow3x=6\)
\(x=2\)
c] \(5^{x+1}-5^x=500\)
\(\Rightarrow5^x\left(5-1\right)=500\)
\(\Rightarrow5^x\cdot4=500\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
d] \(x^2=x^4\)
\(\Rightarrow x=x^2\)
\(\Rightarrow x-x^2=0\)
\(\Rightarrow x\left(1-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
30 tháng 6 2021
a) Ta có: \(8x\left(2x-3\right)-4x\left(4x+3\right)=72\)
\(\Leftrightarrow16x^2-24x-16x^2-12x=72\)
\(\Leftrightarrow-36x=72\)
hay x=-2
b) Ta có: \(\left(x+2\right)\left(x+4\right)-x\left(x+2\right)=104\)
\(\Leftrightarrow x^2+6x+8-x^2-2x=104\)
\(\Leftrightarrow4x=96\)
hay x=24
c) Ta có: \(\left(x-1\right)\left(x+4\right)-x\left(x-1\right)=308\)
\(\Leftrightarrow x^2+3x-4-x^2+x=308\)
\(\Leftrightarrow4x=312\)
hay x=78
d) Ta có: \(15x\left(2x-3\right)-\left(5x+2\right)\left(6x-5\right)=-22\)
\(\Leftrightarrow30x^2-45x-30x^2+25x-12x+10=-22\)
\(\Leftrightarrow-32x=-32\)
hay x=1
a) 22.23.x=82
22.23.x= 26
25.x = 26
x = 26:25
x = 2
b) 5x-1.52= 252
5x-1.52 = 54
5x-1 = 54: 52
5x-1 = 52
=> x-1=2
x = 2+1
x = 3
c)(2x-1)3=125
(2x-1)3= 53
=> 2x-1=5
2x = 5+1
2x = 6
x = 6:2
x = 3
d) 8x.2x=44
(8.2)x= 44
16x = 44
16x = 256
16x = 162
=> x=2
e) (x2.53)2 = 56
x2.2. 53.2= 56
x4 . 56 = 56
x4 = 56:56
x4 = 50
x4 =1
=> x=1