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a) \(x+2x+3x+...+100x=-213\)
\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)
\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)
\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)
d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)
\(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)
c) \(3\left(x-2\right)+2\left(x-1\right)=10\)
\(\Rightarrow3x-6+2x-2=10\)
\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)
a) \(x+2x+3x+4x+...+100x=-213\)
\(x.\left(1+2+3+4+...+100\right)=-213\)
\(x.5050=-213\)
\(x=-\frac{213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)
\(\frac{1}{2}x=-\frac{43}{12}\)
\(x=\frac{-43}{6}\)

Ta có:
\(\left(\right. a - \frac{1}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. c - 3 \left.\right) = 0\) (1)
Và: \(a + 1 = b + 2 = c + 3\)
\(\Rightarrow a = b + 2 - 1 = b + 1\)
Thay vào (1) ta có:
\(\left(\right. b + 1 - \frac{1}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. c - 3 \left.\right) = 0\)
\(\Rightarrow \left(\right. b + \frac{2}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. c - 3 \left.\right) = 0\) (2)
Mà: \(b + 2 = c + 3\)
\(\Rightarrow c = b + 2 - 3 = b - 1\)
Thay vào (2) ta có:
\(\left(\right. b + \frac{2}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. b - 1 - 3 \left.\right) = 0\)
\(\Rightarrow \left(\right. b + \frac{2}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. b - 4 \left.\right) = 0\)
\(\Rightarrow \left[\right. b = - \frac{2}{3} \\ b = - \frac{1}{2} \\ b = 4\)
TH1 khi b=\(- \frac{2}{3}\)
\(\Rightarrow a = b + 1 = - \frac{2}{3} + 1 = \frac{1}{3}\)
\(\Rightarrow c = b - 1 = - \frac{2}{3} - 1 = - \frac{5}{3}\)
TH2 khi \(b = - \frac{1}{2}\)
\(\Rightarrow a = b + 1 = - \frac{1}{2} + 1 = \frac{1}{2}\)
\(\Rightarrow c = b - 1 = - \frac{1}{2} - 1 = - \frac{3}{2}\)
TH3 khi \(b = 4\)
\(\Rightarrow a = b + 1 = 4 + 1 = 5\)
\(\Rightarrow c = b - 1 = 4 - 1 = 3\)
sai mình xin lỗi

câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )

\(a,\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}\)và x + y + z = 49
Ta có : \(\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}=\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{2}}=\frac{z}{\frac{5}{4}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{2}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{2}+\frac{5}{4}}=\frac{49}{\frac{19}{4}}=49\cdot\frac{4}{19}=\frac{196}{19}\)
Vậy : \(\hept{\begin{cases}\frac{x}{\frac{3}{2}}=\frac{196}{19}\\\frac{y}{\frac{4}{2}}=\frac{196}{19}\\\frac{z}{\frac{5}{4}}=\frac{169}{14}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{294}{19}\\y=\frac{392}{19}\\z=\frac{245}{19}\end{cases}}\)
\(b,\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\)và 2x + 3y - z = 186
Ta có : \(\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\Leftrightarrow\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20};\frac{y}{20}=\frac{z}{28}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
\(\Leftrightarrow\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{186}{62}=3\)
Vậy : \(\hept{\begin{cases}\frac{x}{15}=3\\\frac{y}{20}=3\\\frac{z}{28}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=45\\y=60\\z=84\end{cases}}\)

a, \(\left|2x-3\right|=\left|3x-7\right|\)
\(\Rightarrow\orbr{\begin{cases}2x-3=3x-7\\2x-3=7-3x\end{cases}\Rightarrow}\orbr{\begin{cases}-x=-4\\5x=10\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=2\end{cases}\Rightarrow}x=2\)
b, \(\left|7x-1\right|-\left|2x-5\right|=0\)
\(\left|7x-1\right|=\left|2x-5\right|\)
\(\Rightarrow\orbr{\begin{cases}7x-1=2x-5\\7x-1=5-2x\end{cases}\Rightarrow}\orbr{\begin{cases}5x=-4\\9x=6\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{2}{3}\end{cases}}\)
c, \(\left|3x-1\right|+\left|4+3x\right|=0\)
Vì \(\left|3x-1\right|\ge0\); \(\left|4+3x\right|\ge0\)
\(\Rightarrow\left|3x-1\right|+\left|4+3x\right|\ge0\)
Dấu " = " xảy ra <=> \(\hept{\begin{cases}3x-1=0\\4+3x=0\end{cases}\Rightarrow}\hept{\begin{cases}3x=1\\3x=-4\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{3}\\x=\frac{-4}{3}\end{cases}}\)(loại)
d, 2x + 1 = 25 => 2x = 24 => x = 12
đề là thế này?
(2x + 1)2 = 25
\(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}\Rightarrow}\orbr{\begin{cases}2x=4\\2x=-6\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

\(\left(3-4x\right)^2=25=5^2\)
\(\Rightarrow3-4x=5\)
\(\Rightarrow4x=3-5=-2\Rightarrow x=-\frac{1}{2}\)
\(\left(2x-\frac{1}{4}\right)^2=16=4^2\)
\(\Rightarrow2x-\frac{1}{4}=4\Rightarrow2x=4+\frac{1}{4}=\frac{17}{4}\)
\(\Rightarrow x=\frac{17}{4}:2=\frac{17}{4}.\frac{1}{2}=\frac{17}{8}\)
Đề số 3 bị sai.
\(\left(2x+5\right)^2=0\Rightarrow2x+5=0\Rightarrow2x=-5\Rightarrow x=-\frac{5}{2}\)
(3-4x)2=25
3-4x=5
4x=3-5
4x=-2
x=-2:4
x=-0,5
b)(2x-1/42)=16
2x-1/4=4
2x=4+1/4
2x=4,25
x=2,125
c) cái này x ở đâu vậy bn
d) (2x+5)2=0
2x+5=0
2x=0+5
2x=5
x=5:2
x=5/2
Nhớ k cho mk nha
\(\frac14-2x=5\)
\(2x=\frac14-5\)
\(2x=\frac{-19}{4}\)
\(x=-\frac{19}{4}:2\)
\(x=\frac{-19}{8}\)
\(\frac12x-\frac13=25\%\)
\(\frac12x-\frac13=\frac14\)
\(\frac12x=\frac14+\frac13\)
\(\frac12x=\frac{7}{12}\)
\(x=\frac{7}{12}:\frac12\)
\(x=\frac76\)