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a) 2x = 64
=> 2x = 26
=> x = 6
b) 5x = 7x
=> 7x - 5x = 0
=> 5x(2x - 1) = 0
=> \(\orbr{\begin{cases}5^x=0\\2^x-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x\in\varnothing\\2^x=1\end{cases}}\)\(\Rightarrow2^x=2^0\Rightarrow x=0\)
c) 5x . 53 = 125
=> 5x + 3 = 53
=> x + 3 = 3
=> x = 0
d) 3x - 17 = 64
=> 3x = 64 + 17
=> 3x = 81
=> 3x = 34
=> x = 4
e) (3x - 5)3 = 52 . 24 + 600
=> (3x - 5)3 = 25.16 + 600
=> (3x - 5)3 = 400+ 600
=> (3x - 5)3 = 1000
=> (3x - 5)3 = 103
=> 3x - 5 = 10
=> 3x = 15
=> x = 15 : 3
=> x = 5
g) (5x - 15)3 = (5x - 15)7
=> (5x - 15)7 - (5x - 15)3 = 0
=> (5x - 15)3. [(5x - 15)4 - 1] = 0
=> \(\orbr{\begin{cases}\left(5x-15\right)^3=0\\\left(5x-15\right)^4-1=0\end{cases}\Rightarrow\orbr{\begin{cases}\left(5x-15\right)=0\\\left(5x-15\right)^4=1\end{cases}\Rightarrow}\orbr{\begin{cases}5x-15=0\\5x-15=\pm1\end{cases}}}\)
Nếu 5x - 15 =0
=> 5x = 15
=> x = 3
Nếu 5x - 15 = 1
=> 5x = 16
=> x = 16 : 5
=> x = 16/5
Nếu 5x - 16 = -1
=> 5x = 14
=> x = 14 : 5
=> x = 14/5
Vậy \(x\in\left\{3;\frac{16}{5};\frac{14}{5}\right\}\)
a) 2x = 64
Vì 26 = 64 nên x = 6
Vậy x = 6
b) 5x = 7x
Vì 50 = 1 và 70 = 1
=> x = 0
Vậy, x = 0
c) 5x . 53 = 625
Ta có 625 = 54
nên 5x . 53 = 54
5x+3 = 54
=> x = 1
Vậy x = 1
d) 3x - 17 = 64
3x = 64 + 17 = 81 = 34
=> x = 4
Vậy x = 4
e) ( 3x - 5 ) 3 = 52 . 24 + 600
( 3x - 5 ) 3 = 25 . 16 + 600 = 1000 = 103
=> 3x - 5 = 10
3x = 10 + 5 = 15
x = 15 : 3 = 5
Vậy x = 5
g) ( 5x - 15 ) 3 = ( 5x - 15 ) 7
=> (5x - 15 ) 3 : ( 5x - 15 ) 7 = 1
( 5x - 15 ) 3 - 7 = 1
( 5x - 15 ) -4 = 1 = 1-4 = -1-4
=> 5x - 15 = 1 hoặc 5x - 15 = -1
5x = 1 + 15 hoặc 5x = -1 + 15
5x = 16 hoặc 5x = 14
\(x=\frac{16}{5}\) hoặc \(x=\frac{14}{5}\)
Vậy, \(x\in\left\{\frac{16}{5};\frac{14}{5}\right\}\)
Cbht
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
\(x=7^{10}\div7^8\)
\(x=7^{10-8}\)
\(x=7^2\)
\(x=49\)
1)\(79-5\left(11-x\right)=34\)
\(\Rightarrow79-55+5x=34\)
\(\Rightarrow24+5x=34\)
\(\Rightarrow5x=-10\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
2)\(32+2\left(7-x\right)=40\)
\(\Rightarrow32+14-2x=40\)
\(\Rightarrow46-2x=40\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
3)\(\left(166-2x\right).8^9=2.8^{11}\)
\(\Rightarrow\left(83-x\right).2.8^9=2.8^{11}\)
\(\Rightarrow83-x=8^3\)
\(\Rightarrow83-x=512\)
\(\Rightarrow x=-429\)
Vậy \(x=-429\)
4)\(5^2.x-2^3.x=51\)
\(\Rightarrow x\left(5^2-2^3\right)=51\)
\(\Rightarrow x\left(25-8\right)=51\)
\(\Rightarrow17x=51\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
5)\(3^x+4.3^x=5.3^7\)
\(\Rightarrow3^x\left(1+4\right)=5.3^7\)
\(\Rightarrow5.3^x=5.3^7\)
\(\Rightarrow3^x=3^7\)
\(\Rightarrow x=7\)
Vậy \(x=7\)
6)\(7.2^x-2^x=6.32\)
\(\Rightarrow2^x\left(7-1\right)=6.2^5\)
\(\Rightarrow6.2^x=6.2^5\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
7)\(15^{3-x}=225\)
\(\Rightarrow15^{3-x}=15^2\)
\(\Rightarrow3-x=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
8)\(4.5^x-3=97\)
\(\Rightarrow4.5^x=100\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
9)\(171-3.2^x=123\)
\(\Rightarrow3.2^x=48\)
\(\Rightarrow2^x=16\)
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
10)\(180-4.x^5=32\)
\(\Rightarrow4.x^5=148\)
\(\Rightarrow x^5=37\)//Đề có lỗi không ???
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
s1=1+2+3+...+99
s1=99+98+...+1
2s1=100+100+....+100
2s1=100.99
s1=100.99:2=4950(mấy bài sau lam tương tự nha)
4+4^2+4^3+...+4^90 chia hết cho 21
=(4+4^2+4^3)+...+(4^88+4^89+4^90)
=84.1+(4^4+4^5+4^6+...+4^90)
vì 84 chia hết cho 21 suy ra tổng trên chia hét cho 21 (ĐPCM)
Làm mẫu 1 phần ko hiểu thì bảo mình làm típ
a) Ta có: \(\left(x+3\right)^2\ge0;\forall x\)
\(\left(x+3\right)^2-2\ge0-2;\forall x\)
Hay \(A\ge-2;\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+3\right)^2=0\)
\(\Leftrightarrow x=-3\)
VẬY MIN A=-2 \(\Leftrightarrow x=-3\)
\(a)x^{15}=x\)
\(\Rightarrow x^{15}-x=0\)
\(\Leftrightarrow x\left(x^{14}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy....
\(b)2^x-15=17\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy...
\(c)\left(2x+1\right)^3=125\)
\(\Leftrightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Leftrightarrow2x=4\Rightarrow x=2\)
Vậy...
_Y nguyệt_
\(a)x^{15}=x\)
\(\Rightarrow x=1\)
\(b)2^x-15=17\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\Rightarrow x=5\)
\(c)\left(2x+1\right)^3=125\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow x=2\)
giúp tôi nhá
a) 5(x+7) - 10 = 2^3 . 5
5(x+7 ) -10 = 8 . 5 = 40
5(x+7) = 40 + 10 = 50
x + 7 = 50 : 5 = 10
x = 10 - 7 = 3