23x+2=4x+5
32x-1=243
c) 23x + 2 = 4x + 5
d) 32x - 1 = 243
Tìm x, biết rằng: a) 2x = 16 b) 3x+1 = 9xc) 23x+2 = 4x+5d) 32x-1 = 243
a) 2x = 16 <=>x=8
b) 3x+1 = 9x <=>9x-3x=1
<=>6x=1 <=>x=1/6
c) 23x+2 = 4x+5 <=>23x-4x=5-2
<=>19x=3 <=>x=3/19
d) 32x-1 = 243 <=>32x=244
<=>x=61/8
a/ 2x=16
x=8
b/ 3x+1=9x
3x-9x=-1
-6x=-1
x=1/6
c/ 23x+2=4x
23x-4x=-2
19x=-2
x=-2/19
d/ 32x-1=243
32x=244
x=61/8
Bài 1 : Tìm x, biết :
a. 2x = 16 b. 3x+1 = 9x
c. 23x+2 = 4x+5 d. 32x-1 = 243
Bài 2 : So sánh :
a. 2225 và 3150 b. 291 và 535 c. 9920 và 999910
Bài 3 : Chứng minh các đẳng thức :
a. 128 . 912 = 1816 b. 7520 = 4510 . 530 .
\(1,\\ a,2^x=16=2^4\Rightarrow x=4\\ b,3^{x+1}=9^x=3^{2x}\\ \Rightarrow x+1=2x\Rightarrow x=1\\ c,2^{3x+2}=4^{x+5}=2^{2\left(x+5\right)}\\ \Rightarrow3x+2=2x+10\Rightarrow x=8\\ d,3^{2x-1}=243=3^5\\ \Rightarrow2x-1=5\Rightarrow x=3\\ 2,\\ a,2^{225}=8^{75}< 9^{75}=3^{150}\\ b,2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\\ c,99^{20}=\left(99^2\right)^{10}< \left(99\cdot101\right)^{10}=9999^{10}\\ 3,\\ a,12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}=\left(2\cdot3^2\right)^{16}=18^{16}\\ b,75^{20}=\left(3\cdot5^2\right)^{20}=3^{20}\cdot5^{40}=\left(3^{20}\cdot5^{10}\right)\cdot5^{30}=\left(3^2\cdot5\right)^{10}\cdot5^{30}=45^{10}\cdot5^{30}\)
Bài 1:
a) \(\Rightarrow2^x=2^4\Rightarrow x=4\)
b) \(\Rightarrow3^{x+1}=3^{2x}\Rightarrow x+1=2x\Rightarrow x=1\)
c) \(\Rightarrow2^{3x+2}=2^{2x+10}\Rightarrow3x+2=2x+10\Rightarrow x=8\)
d) \(\Rightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow x=3\)
Bài 2:
a) \(2^{225}=\left(2^3\right)^{75}=8^{75}< 9^{75}=\left(3^2\right)^{75}=3^{150}\)
b) \(2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\)
c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)
Bài 3:
a) \(12^8.9^{12}=\left(4.3\right)^8.9^{12}=4^8.3^8.9^{12}=2^{16}.9^4.9^{12}=2^{16}.9^{16}=\left(2.9\right)^{16}=18^{16}\)
b) \(75^{20}=\left(75^2\right)^{10}=5625^{10}=\left(45.125\right)^{10}=45^{10}.125^{10}=45^{10}.5^{30}\)
23x+2=4x+5
23x + 2 = 4x + 5
=> 23x - 4x = 5 - 2
=> 19x = 3
=> x = 3/19
23x+2=4x+5
23x-4x=5-2
19x=3
x=3:19
x=\(\frac{3}{9}\)
Vậy x=\(\frac{3}{9}\)là số cần tìm
Giải phương trình:
c) \(\dfrac{2x-1}{x^2+4x-5}+\dfrac{x-2}{x^2-10x+9}=\dfrac{3x-12}{x^2-4x-45}\)
d) \(\dfrac{3x-1}{18x^2+3x-28}-\dfrac{4x}{24x^2+23x-12}=\dfrac{3}{48x^2-74x+21}\)
c: =>\(\dfrac{2x-1}{\left(x+5\right)\left(x-1\right)}+\dfrac{x-2}{\left(x-1\right)\left(x-9\right)}=\dfrac{3x-12}{\left(x-9\right)\left(x+5\right)}\)
=>(2x-1)(x-9)+(x-2)(x+5)=(3x-12)(x-1)
=>2x^2-19x+9+x^2+3x-10=3x^2-15x+12
=>-16x-1=-15x+12
=>-x=13
=>x=-13
Giải phương trình:
d)\(\dfrac{3x-1}{18x^2+3x-28}-\dfrac{4x}{24x^2+23x-12}=\dfrac{3}{48x^2-74x+21}\)
\(\Leftrightarrow\dfrac{3x-1}{\left(6x-7\right)\left(3x+4\right)}-\dfrac{4x}{\left(8x-3\right)\left(3x+4\right)}=\dfrac{3}{\left(8x-3\right)\left(6x-7\right)}\)
=>(3x-1)(8x-3)-4x(6x-7)=3(3x+4)
=>24x^2-9x-8x+3-24x^2+28x=9x+12
=>11x+3=9x+12
=>2x=9
=>x=9/2
Tìm x,y biết:
a) 3 4 x + 1 5 = 1 6 b) 1 3 + 1 2 : x = − 4 c) c ) x − 2 3 x + 1 4 = 0
d) 2 x − 3 3 4 x + 1 = 0 d) 2 3 x + 1 2 − 5 3 = 0 e) 2 x − 0 , 3 = 1 , 5
f) x . − 3 5 3 = − 3 5 5 g) x : 1 3 2 = 1 3 3 h) x 26 = 21 39
k) 1 3 x = 2 3 1 6 l) 77 7 9 5 9 = x 0 , 03 m) 2 2 3 0 , 03 = 1 7 9 x
n) 2 x − 3 2 = 16 o) 3 x − 2 5 = − 243
Câu hỏi: Làm tính chia:
a/ (2x4 - 11x3 + 19x2 - 23x + 5) : (x2 - 4x + 1)
b/ (3x4 - 8x3 - 10x2 + 8x - 5) : (3x2 - 2x + 1)
c/ (2x4 - 23x + 19x2 -11x3 + 5) : (x2 - 4x + 1)
a, =2x2-3x+5
b,=x2-2x-5
c,2x2-19x+93(dư 368x-88)
Tìm x biết :23x+2=4x+5
23x+2=4x+5
\(\Rightarrow23x-4x=-2+5\)
\(\Rightarrow19x=3\)
\(\Rightarrow x=3:19=\dfrac{3}{19}\)
2x ^3 -5x^2+4x-1) : (2x+1)
(x63 -2x+4) ; (x+2)
(6x^3 - 19x^2+23x-12):(2x-3)
(x^4 - 2 x ^3 - 1+ 2 x ):(x^2 - 1)
(6x^3 - 5x^2 + 4x -1 ) : (2x^2-x+1)
(x^4 -5x^2+4):(x^2-3x+2)
d: \(\dfrac{x^4-2x^3+2x-1}{x^2-1}\)
\(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)
\(=x^2-2x+1\)
\(=\left(x-1\right)^2\)