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Nhác rồi :(
Dễ thấy GTTĐ ko âm nên:
5x ko âm => x ko âm
=> /x/+/x+1/+/x+2/+/x+3/=x+1+x+2+x+3+x=4x+6=5x=>x=6
Vậy: x=6
3x.9+3x+1=22.39
3x.9+3x.3=4.19683
3x.(9+3)=78732
3x.12=78732
3x=78732:12
3x=6561
3x=38
=>x=8
Vậy x=8
Bài làm
( 3 + 6 + x ) . 9 = ( 2 - x ) . 2
27 + 54 + 9x = 4 - 2x
9x + 2x = 4 - 27 - 54
11x = -77
x = -7
Vậy x = -7
b) ( 6 - x ) . 3 = ( 9 + x ) . 3
18 - 3x = 27 + 3x
-3x - 3x = 27 - 18
-6x = 9
x = 9/-6
x = -3/2
Vậy x = -3/2
a,\(\text{(3+6+x) . 9 = (2-x).2}\)
\(27+54+9x=4-2x\)
\(9x+2x=4-54-27\)
\(11x=-77\)
\(\Rightarrow x=-7\)
b, \(\text{(6-x) . 3 = (9+x) . 3}\)
\(18-3x=27+3x\)
\(-3x-3x=27-18\)
\(-6x=9\)
\(\Rightarrow x=\frac{9}{-6}\)
học tốt
\(\frac{1.5.18+2.10.36+3.15.54}{1.3.9+2.6.18+3.9.27}=\frac{1.5.18.\left(1+2.2.2+3.3.3\right)}{1.3.9.\left(1+2.2.2+3.3.3\right)}\)
\(=\frac{1.5.18}{1.3.9}=\frac{10}{3}\)
Ta có :\(x+2x+3x+...+9x=459-9\)
\(\Rightarrow\)\(x\left(1+2+3+...+9\right)=450\)
\(\frac{\Rightarrow x.\left[\left(9-1\right):1+1\right].\left(9+1\right)}{2}\)\(=450\)
\(\frac{\Rightarrow x.90}{2}=450\)
\(\Rightarrow x.90=450.2=900\)
\(\Rightarrow x=\frac{900}{90}\)
\(\Rightarrow x=10\)
\(\text{Tìm }x\text{ : }\)
\(\text{Ta có : }\)
\(x+2x+3x+...+9x=459-9\)
\(\Rightarrow\text{ }x+2x+3x+...+9x=450\)
\(\Rightarrow\text{ }x\cdot\left(\left(9-1\right)\text{ : }1+1\right)\cdot\left(\left(9+1\right)\text{ : }2\right)=450\)
\(\Rightarrow\text{ }x\cdot\left(8\text{ : }1+1\right)\cdot\left(\left(9+1\right)\text{ : }2\right)=450\)
\(\Rightarrow\text{ }x\cdot\left(8+1\right)\cdot\left(\left(9+1\right)\text{ : }2\right)=450\)
\(\Rightarrow\text{ }x\cdot9\cdot\left(\left(9+1\right)\text{ : }2\right)=450\)
\(x\cdot9\cdot\left(10\text{ : }2\right)=450\)
\(x\cdot9\cdot5=450\)
\(x\cdot45=450\)
\(x=450\text{ : }45\)
\(x=10\)
TL:
3\(^x\)= 9
\(\Rightarrow\)3\(^x\)=3\(^2\)(3^2 =9)
\(\Rightarrow\)x = 2
Ta có : 3 * 3 = 9 = 32
Vậy x = 2