Ta có :

  1 + 2 + 3 +...+ x = 45

=>  x . ( x + 1 ) : 2 = 45

=> x . ( x + 1 ) = 45 . 2 = 90 

Ta có : 

       x . ( x+ 1 ) = 90 = 9 x 10

=> x = 9

                      Vậy x = 9

Bạn dùng công thức tính tổng dãy số có quy luật là ra ak

Ta có \(1+2+3+....+x=45\)

\(\Rightarrow\frac{\left(x+1\right).x}{2}=45\)

\(\Rightarrow x.\left(x+1\right)=90\)

\(\Rightarrow x.\left(x+1\right)=9.10\)

\(\Rightarrow x=9\)

Vậy x = 9

theo công thức tính tổng dãy số cách đều ý bạn

(x+1).x:2=45

(x+1).x=90=10.9

=> x=9

1 + 2 + 3 + ... + x = 45

\(\Rightarrow\frac{x.\left(x+1\right)}{2}=45\)

\(\Rightarrow x.\left(x+1\right)=90\)

\(\Rightarrow x.\left(x+1\right)=9.10\)

\(\Rightarrow x=9\)

ta có 

\(1+2+3+4+...+x=45.\)

\(\frac{x\left(x+1\right)}{2}=45\)

\(x\left(x+1\right)=90\)

\(x\left(x+1\right)=9.10\)

=>\(x=9\)

2 x 3^x-1 = 45

      3^x-1 = 45 : 2 =24,5

=>Không tồn tại x.

2 . 3^{x - 1} = 45

3^{x - 1} = 45 : 2

x = không tồn tại

cảm ơn các bạn nha

1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )

\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)

\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)

Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)

2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)

​\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)

\(6^2-\left(x+3\right)=45\)

\(36-\left(x+3\right)=45\)

\(x+3=35-45\)

\(x+3=-10\)

\(x=-13\)

lớp 6 mà bạn

\(6^2-\left(x+3\right)=45\)

\(\Rightarrow36-\left(x+3\right)=45\)

\(\Rightarrow x+3=36-45\)

\(\Rightarrow x+3=-9\)

\(\Rightarrow x=-9-3=-12\)

\(125-5\left(3x-1\right)=5^5.5^3\)

\(\Rightarrow5^3-5\left(3x-1\right)=5^5.5^3\)

\(\Rightarrow5^3-15x+5=5^5.5^3\)

\(\Rightarrow5\left(5^2-3x+1\right)=5^5.5^3\)

\(\Rightarrow5^2-3x+1=5^5.5^3:5\)

\(\Rightarrow25-3x+1=5^7\)

Từ đây làm nốt nhé

\(4^x+1+4^0=65\)

\(\Rightarrow4^{x+1}+1=65\)

\(\Rightarrow4^{x+1}=64\)

\(\Rightarrow4^{x+1}=4^3\)

\(\Rightarrow x+1=3\)

\(\Rightarrow x=2\)

\(70-5\left(x-3\right)=5.3^2\)

\(\Rightarrow70-5\left(x-3\right)=5.9=45\)

\(\Rightarrow5\left(x-3\right)=70-45=25\)

\(\Rightarrow x-3=25:5=5\)

\(\Rightarrow x=8\)

Chúc em học tốt hơn nhé!!

a. \(\Leftrightarrow\) (3.x+4):12 = 19-17

   \(\Leftrightarrow\)(3.x+4):12 = 2

\(\Leftrightarrow\)3.x+4 = 24   \(\Leftrightarrow\)3.x = 20   \(\Leftrightarrow\)x = \(\frac{20}{3}\)

b. \(\Leftrightarrow\)45: (2.x+1) = 3 \(\Leftrightarrow\)2.x+1 = 15 \(\Leftrightarrow\)2.x = 14 \(\Leftrightarrow\)x = 7

c. \(\Leftrightarrow\)72.(x+1) = 360 \(\Leftrightarrow\)x+1 = 5\(\Leftrightarrow\)x = 4

d.\(\Leftrightarrow\)(2.x+3).11 = 253\(\Leftrightarrow\)2.x+3 = 23\(\Leftrightarrow\)2.x = 20\(\Leftrightarrow\)x = 10

a) Ta có: 302^(2x-6) +45 = 46

             => 302^(2x-6) = 1 = 302^0

             => 2x-6 = 0  

              => 2x = 6

              => x = 3

b) Ta có:  (x+1)+(x+2)+(x+3)+....+(x+100)=5950

           => 100x+ (1+2+3+4+...+100) = 5950

           => 100x + 5050 = 5950

           => 100x = 900

           => x = 9

Nhấn đúng cho mk nha!!!!!!!!

Ta có : 

\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)

\(\Leftrightarrow\)\(\left(\frac{x+1}{49}+1\right)+\left(\frac{x+2}{48}+1\right)+\left(\frac{x+3}{47}+1\right)+\left(\frac{x+4}{46}+1\right)+\left(\frac{x+5}{45}+1\right)=-5+5\)

\(\Leftrightarrow\)\(\frac{x+50}{49}+\frac{x+50}{48}+\frac{x+50}{47}+\frac{x+50}{46}+\frac{x+50}{45}=0\)

\(\Leftrightarrow\)\(\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)

Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\)

Nên \(x+50=0\)

\(\Rightarrow\)\(x=-50\)

Vậy \(x=-50\)

Chúc bạn học tốt ~ 

\(\frac{5}{x}=\frac{x}{45}\)

=> \(x^2=225\)

<=> \(x=15\)hoặc \(x=-15\)

\(x+\frac{1}{12}=\frac{3}{x+1}\)

<=> \(\frac{12x+1}{12}=\frac{3}{x+1}\)

=> \(12x^2+12x+x+1=36\)

<=> \(12x^2+13x+1=36\)

<=> \(12x^2+13x-35=0\)

<=> \(\left(x-\frac{5}{4}\right)\left(x+\frac{7}{3}\right)=0\)

<=> \(\hept{\begin{cases}x-\frac{5}{4}=0\\x+\frac{7}{3}=0\end{cases}}\)

<=>\(\hept{\begin{cases}x=\frac{5}{4}\\x=\frac{-7}{3}\end{cases}}\)

a)\(\frac{5}{x}=\frac{x}{45}\)

\(\Rightarrow x^2=5.45\)

\(\Rightarrow x^2=225=\orbr{\begin{cases}15^2\\\left(-15\right)^2\end{cases}}\)

\(\Rightarrow x=\orbr{\begin{cases}15\\-15\end{cases}}\)

b) \(\frac{x+1}{12}=\frac{3}{x+1}\)

\(\Rightarrow\left(x+1\right)^2=36\)

\(\Rightarrow x+1=\orbr{\begin{cases}6\\-6\end{cases}}\)

+) \(\frac{5}{x}=\frac{x}{45}\)

\(\Rightarrow x\times x=5\times45\)

\(\Rightarrow x^2=225\)

\(\Rightarrow\orbr{\begin{cases}x^2=15^2\\x^2=\left(-15\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

Vậy x = 15 hoặc x = -15

+) \(\frac{x+1}{12}=\frac{3}{x+1}\)

\(\Rightarrow\left(x+1\right)\times\left(x+1\right)=3\times12\)

\(\Rightarrow\left(x+1\right)^2=36\)

\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2=6^2\\\left(x+1\right)^2=\left(-6\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}\)

Vậy x = 5 hoặc x = -7

_Chúc bạn học tốt_