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B1:
Ta có: \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}-2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
\(=\frac{2.\left(-4\right)}{3.5}=-\frac{8}{15}\)
B2:
Ta có: \(1+3+5+...+x=1600\)
\(\Leftrightarrow\frac{\left(x+1\right)\cdot\left(\frac{x-1}{2}+1\right)}{2}=1600\)
\(\Leftrightarrow\left(x+1\right)\cdot\frac{x+1}{2}=3200\)
\(\Leftrightarrow\left(x+1\right)^2=6400\)
Xét theo dãy tăng tiến ta thấy được giá trị của x càng tăng
=> x dương => x + 1 dương
\(\Rightarrow x+1=80\)
\(\Rightarrow x=79\)
a. ( x - 1 )2 = 25
<=> \(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)
<=>\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
b. 2x + 2 - 2x = 96
<=> 2x.4-2x=96
<=> 2x.3=96
<=> 2x=32=25
<=> x=5
c. 2x . 7 = 224
<=> 2x=32=25
<=> x=5
d. ( 7x - 11 )3 = 25 . 52 = 200 (xem lại đề)
e. 9 < 3x < 81
<=> 32<3x<34
<=> 2<x<4
\(35-3^{x+1}=8\)
\(\Rightarrow3^{x+1}=35-8=27\)
Mà \(27=3^3\)
\(\Rightarrow x+1=3\Rightarrow x=3-1=2\)
Vậy \(x=2\)
\(a.35-3^{x+1}=8\)
\(3^x.3^1=35-8\)
\(3^x.3^1=27\)
\(3^x.3^1=3^3\)
\(3^x=3^3:3^1\)
\(3^x=3^{3-1}\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
\(b.3.2^{x-1}+5=29\)
\(3.2^x:2^1=29-5\)
\(3.2^x:2^1=24\)
\(2^x:2^1=24:3\)
\(2^x:2^1=8\)
\(2^x:2^1=2^3\)
\(2^x=2^3.2^1\)
\(2^x=2^{3+1}\)
\(2^x=2^4\)
\(\Rightarrow x=4\)
\(c.9:3^x+15=18\)
\(3^2:3^x+15=18-15\)
\(3^2:3^x=3\)
\(3^x=3^2:3\)
\(3^x=3^{2-1}\)
\(3^x=3^1\)
\(\Rightarrow x=1\)
Chúc bạn học tốt!
a) 52.x = 62 + 82
=> 25 .x = 36 + 64
=> 25.x = 100
=> x = 100 : 25
=> x = 4
b) (22 + 42).x + 24 . 5x = 100
=> (4 + 16).x + 16.5x = 100
=> 20x + 80x = 100
=> 100x = 100
=> x = 100 : 100 = 1
c) 24 : x = 26
=> x = 24 : 26
=> x = 2-2 = 1/4
d) 33x + 23x = 102
=> 27x + 8x = 100
=> 35x = 100
=> x = 100 : 35
=> x = 20/7
a. ( 3x + 9 ).( 1 - 3x ) = 0
\(\Leftrightarrow\orbr{\begin{cases}3x+9=0\\1-3x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=-9\\3x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
Vậy \(x\in\left\{-3;\frac{1}{3}\right\}\)
b, \(\left(x^2+1\right)\left(81-x^2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\81-x^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\\x^2=81\end{cases}}\) ( vô lí ở trg hợp 1 nha )
<=> \(x^2=81\)
\(\Leftrightarrow\) \(x\in\left\{-9;9\right\}\)
Vậy \(x\in\left\{-9;9\right\}\)
a.(3x+9).(1-3x)=0
\(\Rightarrow\orbr{\begin{cases}3x+9=0\\1-3x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=-9\Rightarrow x=-9:3=-3\\3x=1\Rightarrow x=\frac{1}{3}\end{cases}}\)
Vậy...........................................................
a) x-12=(-28)
x=(-28)+12
x=(-16)
Vậy x=(-16)
b)20+8|x-3|=52.4
20+8|x-3|=100
8|x-3|=100-20
8|x-3|=80
|x-3|=80:8
|x-3|=10
=>x-3=10 hoặc x-3=(-10)
x=10+3 x=(-10)+3
x=13 x=(-7)
Vậy x thuộc {13;-7}
c) 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
Vậy x=17
|x-3|=7-(-2)
|x-3|=9
=>x-3=9 hoặc x-3=(-9)
x=9+3 x=(-9)+3
x=12 x=(-6)
Vậy...
e) (2x-1)3=125
(2x-1)3=53
=>2x-1=5
...
Còn lại tự lm nha
Câu g tương tự câu e
a) 5^x=5^78:5^14(lấy 78-14)
5^x=5^64
=> x=64
b) 7^x.7^2=7^21
7^x=7^21:7^2
7^x=7^19
=> x=19
Bài 1:
a) \(x^{10}=1^x\Rightarrow\orbr{\begin{cases}x=1\\x=10\end{cases}}\)
b) \(x^{10}=x\Rightarrow x=1\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\left(2x-15\right)^5.\left(2x-15\right)^3=\left(2x-15\right)^3\)
\(\left(2x-15\right)^2=1\Rightarrow x=8\)
Bài 2:
\(a;2^{16}=2^{13}\cdot2^3=2^{13}\cdot8>7\cdot2^{13}\)
\(b;49^8\cdot27^5=7^{16}\cdot3^{15}=21^{15}\cdot7>21^5\)
C;Ta có:\(199^{20}< 200^{20}=2^{20}\cdot10^{40}=2^{15}\cdot10^{40}\cdot2^5\)
\(2003^{15}>2000^{15}=2^{15}\cdot10^{45}=2^{15}\cdot10^{40}\cdot10^5\)
Vì 25<105 nên 19920<200315
\(d;3^{39}< 3^{40}=9^{20}< 11^{20}< 11^{21}\)
a) 52 . x = 62 + 82
\(5^2\cdot x=36+64\)
\(5^2\cdot x=100\)
\(x=100\div5^2\)
\(x=100\div25\)
\(x=4\)
b) ( 22 + 42 ) . x + 24 . 5 . x = 102
\(\left(4+16\right)\cdot x+16\cdot5\cdot x=100\)
\(x\cdot\left(20+80\right)=100\)
\(x\cdot100=100\)
\(x=100\div100\)
\(x=1\)
c ) 24 . x = 26
\(x=2^6\div2^4\)
\(x=2^{6-4}\)
\(x=2^2\)
\(x=4\)
d) 33 . x + 23 . x = 102
\(x\cdot\left(23+27\right)=100\)
\(x\cdot50=100\)
\(x=100\div50\)
\(x=2\)
e) 78 . x = 710
\(x=7^{10}\div7^8\)
\(x=7^{10-8}\)
\(x=7^2\)
\(x=49\)
a) x : 123 = 12
x = 12 . 123
x = 124
b) 81 . x = 94
81 . x = 6561
x = 6561 : 81
x = 81
c) 2x - 26 = 6
2x = 6 + 26
2x = 32
2x = 25
x = 5
d) 3x = 81
3x = 34
x = 4
b, \(81.x=9^4\)
\(\Rightarrow81.x=81\)
\(x=81:81\)
\(\Rightarrow x=1\)
Vậy : \(x=1\)