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\(2x+\left(1+2+3+...+100\right)=15150\)
\(2x+\left[\left(1+100\right)+\left(2+99\right)+...+\left(50+51\right)\right]=15150\)
\(2x+\left[101+101+...+101\right]=15150\)CÓ 50 SỐ 101
\(2x+\left[101\times50\right]=15150\)
\(2x=15150:5050\)
\(2x=3\)
\(x=3:2\)
\(x=1.5\)
a, 2x + (1+2+3+4+...+100) = 15150
=> 2x + \(\frac{\left(1+100\right).\left[\left(100-1\right)+1\right]}{2}\)= 15150
=> 2x + \(\frac{101.100}{2}\)= 15150
=> 2x + 5050 = 15150
=> 2x = 15150 - 5050
=> 2x = 10100
=> x = 10100 : 2
=> x = 5050
Vậy x = 5050
b, .(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8)=36
=> (x + x + x + x +x + x +x +x ) + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 36
=> 8x + 36 = 36
=> 8x = 0
=> x = 0
Vậy x = 0
c, 0+0+4+6+8+...+2x=110
Sửa đề :0 + 2 + 4 + 6 + 8 + ... + 2x = 110 = 2 + 4 + 6 + 8 + ... + 2x = 110
SSH : \(\frac{\left(2\text{x}-2\right)}{2}+1=x-1+1=x\)
Tổng : \(\frac{\left(2\text{x}+2\right).x}{2}=110\Leftrightarrow\frac{2.\left(x+1\right).x}{2}=110\)
\(\Leftrightarrow\left(x+1\right)x=110\)
\(\Leftrightarrow\left(10+1\right).10=110\)
=> x = 10
Vậy x = 10
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Bài 1 Tìm x biết:
a)65-(29-x)=32
65 -29+x=31
x=31-65+29
x=-5
b)(x+5)-(x+23)=x-34
x+5 -x +23 = x-34
(x-x)+ (23+5)=x-34
0+28=x-34
28=x-34
28+34=x
62=x
=>x=62
c)(16-x)+(x-38)=x+44
16-x+x-38=x+44
-x+x-x=44-16+38
-x=36
=>x=-36
d)-12+3(-x+7)=-18
3(-x+7)=-18+12
3(-x+7)=-6
-x+7=-6:3
-x+7=-2
-x=-2-7
-x=-9
=>x=9
Baif 2
d)|7-x|=10
=> \(\left[{}\begin{matrix}7-x=10\\7-x=-10\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=7-10\\x=-10-7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=-3\\x=-17\end{matrix}\right.\)
e)(x-6).(7-2x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}x-6=0\\7-2x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+6\\2x=7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=7:2\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=3,5\end{matrix}\right.\)
f)(9-x).(2x+8)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}9-x=0\\2x+8=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+9\\2x=-8\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)
g)x(-x+8).(-3x-18)=0
\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\-x+8=0\\-3x-18=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=0+8\\-3x=0+18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=8\\-3x=18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=18:\left(-3\right)\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=-6\end{matrix}\right.\)
h)(-x+8).(x-54).(-24-x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)
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Ta có : \(\left|2x+4\right|+\left|4x+8\right|=0\left|2x+4\right|+\left|4x+8\right|=0\)
\(\Rightarrow\left|2x+4\right|+2.\left|2x+4\right|=\left|4x+8\right|=0\)
\(\Rightarrow\left|2x+4\right|\left(1+2\right)=0\)
=> |2x + 4| = 0
=> 2x + 4 = 0
=> 2x = -4
=> x = -2
1. Đề đúng phải là thế này: \(\left|2x+4\right|+\left|4x+8\right|=0\)
\(\Rightarrow\left|2x+4\right|=\left|4x+8\right|=0\)
\(\Rightarrow2x+4=4x+8=0\)
\(\Rightarrow x=-\frac{4}{2}=-\frac{8}{4}\)
\(\Rightarrow x=-2\)
2. Sửa lại đề : \(\left|x-5\right|-\left|x-7\right|=0\)
\(\Rightarrow\left|x-5\right|=\left|x-7\right|\)
\(\Rightarrow\orbr{\begin{cases}x-5=x-7\\x-5=-\left(x-7\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-5=-7\\x-5=-x+7\end{cases}}\)
( Loại trường hợp 1)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
3. \(\left|x+8\right|-\left|2x+2\right|=0\)
\(\Rightarrow\left|x+8\right|=\left|2x+2\right|\)
\(\Rightarrow\orbr{\begin{cases}x+8=2x+2\\x+8=-\left(2x+2\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+2=8\\x+8=-2x-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\3x=-10\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-\frac{10}{3}\end{cases}}\)
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Bài 1L
a) \(\left(x-7\right)\left(x+3\right)< 0\)
TH1:
\(\hept{\begin{cases}x-7>0\\x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>7\\x< -3\end{cases}}}\)( loại )
TH2:
\(\hept{\begin{cases}x-7< 0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 7\\x>-3\end{cases}\Leftrightarrow}-3< x< 7}\)( chọn )
Vậy \(-3< x< 7\)
Bài 2:
a) \(\left(5x+8\right)-\left(2x-15\right)+21=2x-5\)
\(\Leftrightarrow5x+8-2x+15+21=2x-5\)
\(\Leftrightarrow5x-2x-2x=-5-21-8-15\)
\(\Leftrightarrow x=-49\)
Vậy ...
a) \(\left(x-7\right)\left(2x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}\)
b) \(\left(x-1\right)^6=\left(x-1\right)^8\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-1=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
\(\left(x-7\right)\left(2x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)
Vậy \(x=7\) hoặc \(x=4\)
\(\left(x-1\right)^6=\left(x-1\right)^8\)
\(\Rightarrow\left(x-1\right)^8-\left(x-1\right)^6=0\)
\(\left(x-1\right)^6.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left(x-1\right)^6=0\) hoặc \(\left(x-1\right)^2-1=0\)
\(\Leftrightarrow x-1=0\) \(\left(x-1\right)^2=\left(\pm1\right)^2\)
\(\Leftrightarrow x=1\) \(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
Vậy \(x=1;x=2;x=0\)