a, b: ở trên.

c. (x+1)+(x+3)+...+(x+97)+(x+99)=0 (có 50 cặp) 

=> x+1+x+3+...+x+97+x+99 = 0

=> (x+x+...+x)+(1+3+...+97+99)=0

=> 50x+\(\frac{\left(99+1\right).50}{2}\)=0

=> 50x+2500=0

=> 50x=-2500

=> x=-2500:50

=> x=-50

sao ko ai giúp zị :(

\(1)|5-2x|=|x+4|\)

\(\Leftrightarrow\orbr{\begin{cases}5-2x=x+4\\5-2x=-x-4\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x-x=4-5\\-2x+x=-4-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}-3x=-1\\-x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=9\end{cases}}}\)

Vậy \(x=\frac{1}{3};x=9\)

\(2)|x-1|=|2x+5|\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=2x+5\\x-1=-2x-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x-2x=5+1\\x+2x=-5+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}-x=4\\3x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=-\frac{4}{3}\end{cases}}}\)

Vậy \(x=-4;x=-\frac{4}{3}\)

\(3)|x+1|+|x+2|+|x+3|=0\left(1\right)\)

Ta có: \(|x+1|\ge0\forall x;|x+2|\ge0\forall x;|x+3|\ge0\forall x\)

\(\Leftrightarrow|x+1|+|x+2|+|x+3|\ge0\forall x\)

\(\left(1\right)\Leftrightarrow|x+1|+|x+2|+|x+3|=0\)

\(\Leftrightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=0\)

\(\Leftrightarrow x+1+x+2+x+3=0\)

\(\Leftrightarrow\left(x+x+x\right)+\left(1+2+3\right)=0\)

\(\Leftrightarrow3x+6=0\)

\(\Leftrightarrow3x=-6\)

\(\Leftrightarrow x=-6:3\)

\(\Leftrightarrow x=-2\)

Vậy x=-2

1: x = 1/3 , x=9

2: x = 4 , x = -4/3

3: x=2

A. 5(x-11) + 4(x-3) =10
<=>5x -55 + 4x -12 =10
<=>9x -67 =10
<=>9x =10+67 =77
<=>x =77 : 9 =77/9
B.21 - 5 ( x-7) = 2-3(x-5)
<=>21-5x+35 = 2 - 3x +15
<=>-5x + 56 = -3x + 17
<=>-5x +3x=17-56=-39
<=> -2x =-39
<=> x = -39 : -2 =39/2
C.3(2x-1)-7(x+4)=8-9x
<=>6x -4 -7x-28=-9x+8
<=>-x + 9x = 8+4+28=40
<=> 8x = 40
<=> x = 40:8=5
D. l 4-x l =11
<=>\(\orbr{\begin{cases}4-x=11\\4-x=-11\end{cases}}\)<=> \(\orbr{\begin{cases}x=-7\\x=15\end{cases}}\)
E. 5-3 l x+2 l = -28
<=> 3 l x+2 L = -28 -5 =-33
<=> l x+2 l = -33 :3 =11
<=>\(\orbr{\begin{cases}x+2=11\\x+2=-11\end{cases}}\)<=> \(\orbr{\begin{cases}x=9\\-13\end{cases}}\)
 T I C K mk nhé!!!^_^

\(a,\Rightarrow\left(35x+3\right)\cdot19=152\\ \Rightarrow35x+3=8\\ \Rightarrow x=\dfrac{1}{7}\\ b,\Rightarrow3\left(x+7\right)=42\\ \Rightarrow x+7=14\Rightarrow x=7\\ c,\Rightarrow3\left(x+1\right)=48\\ \Rightarrow x+1=16\Rightarrow x=15\\ d,\Rightarrow120-5x+100\cdot2:5=4\cdot15\\ \Rightarrow120-5x+40=60\\ \Rightarrow5x=100\Rightarrow x=20\\ e,\Rightarrow4x-10=30\\ \Rightarrow4x=40\\ \Rightarrow x=10\\ g,\Rightarrow10x+10=70\\ \Rightarrow10x=60\\ \Rightarrow x=6\)

a) Ta có: \(\left(2x-5\right)^3=216\)

\(\Leftrightarrow2x-5=6\)

\(\Leftrightarrow2x=11\)

hay \(x=\dfrac{11}{2}\)

b) Ta có: \(2x-3⋮x+4\)

\(\Leftrightarrow-11⋮x+4\)

\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)

hay \(x\in\left\{-3;-5;7;-15\right\}\)

Alo, sugeni two wai phem. Si ga no, you woo be the me that nas te, ai gi da

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)