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\(\text{#040911}\)
\(x\cdot\left(x+1\right)\div2=153\)
\(\Rightarrow x\cdot\left(x+1\right)=153\cdot2\)
\(\Rightarrow x\cdot\left(x+1\right)=306\)
\(\Rightarrow x^2+x=306\)
\(\Rightarrow x^2+x-306=0\)
\(\Rightarrow x^2+18x-17x-306=0\)
\(\Rightarrow\left(x^2+18x\right)-\left(17x+306\right)=0\\ \Rightarrow x\left(x+18\right)-17\left(x+18\right)=0\\ \Rightarrow\left(x-17\right)\left(x+18\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-17=0\\x+18=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=17\\x=-18\end{matrix}\right.\\ \text{Vậy, x }\in\left\{-18;17\right\}.\)
(\(x\) \(\times\) (\(x\) + 1)): 2 = 153
(\(x\) \(\times\) (\(x\) + 1)) = 153 \(\times\) 2
\(x\) \(\times\) (\(x\) + 1) = 306
\(x\)2 + \(x\) = 306
\(x^2\) + \(x\) - 306 = 0
\(x^2\) - 17\(x\) + 18\(x\) - 306 =0
\(x\) \(\times\) (\(x\) - 17) + 18 \(\times\) (\(x\) - 17) = 0
(\(x\) - 17)\(\times\) ( \(x\) + 18) = 0
\(\left[{}\begin{matrix}x-17=0\\x+18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=17\\x=-18\end{matrix}\right.\)
\(x\cdot\left(x+1\right):2=153\)
\(\Rightarrow x\left(x+1\right)=153\cdot2\)
\(\Rightarrow x^2+x=306\)
\(\Rightarrow x^2+x-306=0\)
\(\Rightarrow x^2+18x-17x-306=0\)
\(\Rightarrow x\left(x+18\right)-17\left(x+18\right)=0\)
\(\Rightarrow\left(x-17\right)\left(x+18\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=-18\end{matrix}\right.\)
Trường hợp 1 :
2.3x + 5.3x+1 = 153
=> 2.3x + 5.3x + 3 = 153
=> (2 + 5).3x = 150
=> 7.3x = 150
=> 3x = 150/7 => x không thỏa mãn
Trường hợp 2 :
2.3x + 5.3x + 1 = 153
=> (2 + 5).3x = 152
=> 7 . 3x = 152
=> 3x= 152/7 => x không thỏa mãn
Nếu bạn không gõ latex thì 2 trường hợp cũng sẽ xảy ra :((
1 + 2 + 3 + .... + x = 153
x(x+ 1) = 153 x 2 = 306
x(x + 1) = 17 x (17 + 1)
Vậy x= 17
\(\text{(x+153)-(48-193)=1-2-3-4}\)
\(\left(x+153\right)+145=-8\)
\(x+153=-8-145\)
\(x+153=-153\)
\(x=-153-153\)
\(x=-306\)
học tốt
(x+153)-(48-193)=1-2-3-4
(x+153)+145=-8
x+153=-8-145
x+135=-153
x=-153-135
x=-288
Vậy x=-288
A,(X+153)-(48+193)=1-2-3-4:
x+153-48-193=1-2-3-4
x+(-88)=-8
x-88=-8
x=-8+88=80
Ta có : \(\frac{1-x}{x}+1=x\)
Vậy \(\frac{\left(x+1\right).x}{2}=153\)
(x+1)x=153.2
(x+1)x=306
=> x=17
Ta có: \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2019\right)+2019=2019\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(0+1+2+...+2019\right)=0\)( có 2020 chữ x )
\(\Leftrightarrow2020x+2039190=0\)
\(\Leftrightarrow x=-1009,5\)
Ta có : \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2019\right)+2019=2019\)
\(\Rightarrow x+x+1+x+2+...+x+2019=0\)
\(\Rightarrow2020x+\left(1+2+3+...+2019\right)=0\)
\(\Rightarrow2020x+\frac{2019.2020}{2}=0\)
\(\Rightarrow2020x+2039190=0\)
\(\Rightarrow2020x=-2039190\)
\(\Rightarrow x=1009,5\)
Vậy \(x=1009,5\)
a, \(x\) + (\(x\) + 1) + (\(x\) + 3) = 153
\(x\) + \(x\) + 1 + \(x\) + 3 = 153
3\(x\) + 4 = 153
3\(x\) = 153 - 4
3\(x\) = 149
\(x\) = 149 : 3
\(x\) = \(\dfrac{149}{3}\)
b, (\(2x-7\)) - (\(x\) + 135) =0
2\(x\) - 7 - \(x\) - 135 = 0
\(x\) - 142 = 0
\(x\) = 142
\(\text{#040911}\)
\(\left[x\cdot\left(x+1\right)\right]\div2=153\\ \Rightarrow x\cdot\left(x+1\right)=153\cdot2\\ \Rightarrow x\cdot\left(x+1\right)=306\\ \Rightarrow x^2+x=306\\ \Rightarrow x^2+x-306=0\\ \Rightarrow x^2+18x-17x-306=0\\ \Rightarrow\left(x^2+18x\right)-\left(17x+306\right)=0\\ \Rightarrow x\left(x+18\right)-17\left(x+18\right)=0\\ \Rightarrow\left(x-17\right)\left(x+18\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-17=0\\x+18=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=17\\x=-18\end{matrix}\right.\\ \text{Vậy, x }\in\left\{-18;17\right\}.\)