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12 + ( 5 + x ) = 20 5.22 + ( x + 3 ) = 52 23 + ( x + 3 ) = 52 43 - ( x - 2 ) = 52
17 + x = 20 5.4 + x + 3 = 25 8 + x + 3 = 25 64 - x + 2 = 25
x = 20 - 17 20 + 3 + x = 25 11 + x = 25 66 - x = 25
x = 3 23 + x = 25 x = 25 - 11 x = 66 - 25
x = 25 - 23 x = 14 x = 41
x = 2
Đăng nhìu v bn :) Đáng quan ngại đây :)
1) \(2^x-15=17\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
2) \(\left(7x-11\right)^3=25\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=825\)
\(\Leftrightarrow7x-11=\sqrt[3]{825}\)
\(\Leftrightarrow7x=11+\sqrt[3]{825}\)
\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)
3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)
\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
4) \(4x+5\left(x+3\right)=105\)
\(\Leftrightarrow9x+15=105\)
\(\Leftrightarrow9x=90\)
\(\Rightarrow x=10\)
5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
\(x^4\cdot x^7\cdot...\cdot x^{100}\)
\(=x^{4+7+...+100}\)
\(=x^{52\cdot33}=x^{1716}\)
\(x^1\cdot x^2\cdot x^3\cdot...\cdot x^{2006}\)
Ta có : \(x^1\cdot x^2=x^{1+2}=x^3\)
Tương tự : \(x^1\cdot x^2\cdot x^3=x^{1+2+3}=x^6\)
Áp dụng vào bài toán :
\(x^1\cdot x^2\cdot x^3\cdot...\cdot x^{2006}=x^{1+2+3+...+2006}\)
\(\Rightarrow x^{1+2+3+...+2006}=x^{2013021}\)
a, 100 - 7 ( x - 5 ) = 31 + 33
100 - 7 ( x - 5 ) = 31 + 27
100 - 7 ( x - 5 ) = 58
7 ( x - 5 ) = 100 - 58
7 ( x - 5 ) = 42
x - 5 = 42 : 7
x - 5 = 6
=> x = 6 +5
=> x = 11
Vậy x = 11
b, 12 ( x - 1 ) : 3 = 43 + 23
12 ( x - 1 ) : 3 = 64 + 8
12 ( x - 1 ) : 3 = 72
12 ( x - 1 ) = 72 . 3
12 ( x - 1 ) = 216
x - 1 = 216 : 12
x - 1 = 18
=> x = 18 + 1
=> x = 19
Vậy x = 19
c, 24 + 5x = 75 : 73
24 + 5x = 72
24 + 5x = 49
5x = 49 - 24
5x = 25
=> x = 25 : 5
=> x = 5
Vậy x = 5
d, 5x - 206 = 24 . 4
5x - 206 = 16 . 4
5x - 206 = 64
5x = 64 + 206
5x = 270
=> x = 270 : 5
=> x = 54
Vậy x = 54
e, 125 = x3
53 = x3
=> x = 5
Vậy x = 5
g, 64 = x2
82 = x2
=> x = 8
Vậy x = 8
\(a,\left|x\right|-2=7-\left(-8\right)\)
\(\Rightarrow\left|x\right|-2=15\)
\(\Rightarrow\left|x\right|=17\)
\(\Rightarrow\orbr{\begin{cases}x=-17\\x=17\end{cases}}\)
\(b,5^{2x-3}-2.5^2=5^2.3\)
\(\Rightarrow5^{2x-3}-50=75\)
\(\Rightarrow5^{2x-3}=125\)
\(\Rightarrow5^{2x-3}=5^3\)
\(\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
bài 1) thực hiện phép tính
a) \(4^5-81:3^2=1024-9=1015\)
b) \(3^2.22-3^2.12=3^2.2\left(11-6\right)=18.5=90\)
c) \(2^3.15-\left[115-\left(12-5\right)\right]=120-\left(115-12+5\right)=120-115+12-5=12\)
d) \(3.3^2-19^{21}:19^{20}+2010^0=27-19+1=9\)
e)\(7^{25}:\left(7^{21}.46+7^{21}.3\right)=7^{25}:\left(7^{21}.49\right)=\frac{7^{21}.7^4}{7^{21}.49}=\frac{2401}{49}=49\)
bài 2)Tim x
a) 716 - (x - 143) = 695
<=> x - 143 = 716 - 695
<=> x - 143 = 21
<=> x = 21 + 143
<=> x = 164
vậy x = 164
Ta có:
\(x^7=\frac{1}{100}\cdot x^5\)
=>\(x^5\cdot x^2=x^5\cdot\left(\frac{1}{10}\right)^2\)hoặc \(=x^5\cdot\left(-\frac{1}{10}\right)^2\)
Từ đó có thể kết luận được x=1/10 hoCJW -1/10.
vẬY........
\(x^7=\frac{x^5}{100}\)
\(\Rightarrow\)\(\frac{x^7}{x^5}=\frac{1}{100}\)
\(\Rightarrow\)\(x^2=\frac{1}{100}\)
\(\Rightarrow\)\(x^2=\left(\frac{1}{10}\right)^2\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{1}{10}\\x=-\frac{1}{10}\end{cases}}\)