a,x+x2+x3+....+x9=459-9

<=>x(1+2+3+...+9)=450

<=>x45=450

<=>x=10

b ,thì tương tự nhé b

28 giây trước (15:57)

Tìm x , biết

a)x+2x+3x+...+9x=459 - 9

45x = 450

x= 10

---------

b)x+(x+1)+(x+2)+...+(x+30)=1240

31x + (1 +2 + 3 ....+ 30 ) =1240

31x + 456 = 1240

31x = 784

x= 784/31

a)\(x+2x+3x+...+9x=459-9\)

\(x+2x+3x+...+9x=450\)

\(x\left(1+2+3+...+9\right)=450\)

\(45x=450\)

\(x=450\div45\)

\(x=10\)

b)\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)

\(x+x+1+x+2+...+x+30=1240\)

\(\left(x+x+x+...+x\right)+\left(1+2+3+...+30\right)=1240\)

\(31x+465=1240\)

\(31x=1240-465\)

\(31x=775\)

\(x=775\div31\)

\(x=25\)

x + 2x +3x + ... + 9x = 459 - 3^2

x + 2x + 3x + ... + 9x = 459 - 9

x + 2x + 3x + ... + 9x = 450 .

x . ( 1 +2 + 3 + ... + 9 ) = 450

x . 45 = 450

x = 450 : 45

x = 10

\(x+2x+3x+...+9x=459-3^2\) 

\(\Rightarrow9x+\left(1+2+3+...+9\right)=450\) 

\(\Rightarrow9x+\frac{\left[\left(9+1\right).9\right]}{2}=450\) 

\(\Rightarrow9x+45=450\) 

\(\Rightarrow9x=450-45\) 

\(\Rightarrow x=\frac{450-45}{9}=\frac{405}{9}=45\)

\(2^{x+3}+2^x=144\) 

\(\Rightarrow2^x\left(2^3+1\right)=144\) 

\(\Rightarrow2^x.9=144\) 

\(\Rightarrow2^x=\frac{144}{9}=16=2^4\) 

Vậy \(x=4\)

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giai giup minh vs

Địt bố đéo biết

=15 nha bn

giúp minh vs

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