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-16 + 23 + x = -16
7 + x = -16
x = -16 - 7
x = -23
2x + 35 = -15
2x = -15 - 35
2x = -50
x = -25
-13 x |x | = -26
x = -26 : (-13)
x = 2
Vậy x = 2 hoặc -2
|2x - 5| = 13
2x = 13 + 5
2x = 18
x = 9
(x - 3) x (x + 2) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
2)2x+35=-15
2x =-15-35
2x = -50
x =-25
1)-16+23+x=-16
7+x =-16
x =-16-7
x =-23
3)-13 * lxl =-26
lxl = -26:-13
lxl =2
<=>x=-2 hoac x=2
5) (x-3).(x+2) =0
<=>\(\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\) <=>\(\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
Toán lớp 6Tìm x
Trả lời Câu hỏi tương tự
Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !
a, ( x + 1 ) + ( x + 2 ) + ... + ( x + 199 ) = 0
x + 1 + x + 2 + ... + x + 199 = 0
( x + x + ... + x ) + ( 1 + 2 + ... + 199 ) = 0
199x + 19900 = 0
199x = 0 - 19900
199x = -19900
x = -19900 : 199
x = -100
Vậy ...
b, ( x - 30 ) + ( x - 29 ) + ( x - 28 ) = 11
x - 30 + x - 29 + x - 28 = 11
( x + x + x ) - ( 30 + 29 + 28 ) = 11
3x - 87 = 11
3x = 11 + 87
3x = 98
x = \(\frac{98}{3}\)
Vậy ...
128 - 3 ( x + 4 ) = 23
3 ( x + 4 ) = 128 - 23
3 ( x + 4 ) = 105
x + 4 = 105 : 5
x + 4 = 35
x = 35 - 4
x = 31
128 - 3 ( x + 4 ) = 23
3 ( x + 4 ) = 128 - 23
3 ( x + 4 ) = 105
x + 4 = 105 : 5
x + 4 = 35
x = 35 - 4
x = 31
k mik nha
a) \(\frac{-x}{2}+\frac{2x}{3}+x+\frac{1}{4}+2x+\frac{1}{6}=\frac{3}{8}.\)
\(\frac{-x}{2}+\frac{2x}{3}+3x+\frac{5}{12}=\frac{3}{8}\)
\(x.\left(-\frac{1}{2}+\frac{2}{3}+3\right)+\frac{5}{12}=\frac{3}{8}\)
\(x\cdot\frac{19}{6}=-\frac{1}{24}\)
x = -1/76
b) \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
\(\frac{3}{2x+1}+\frac{2.5}{2.\left(2x+1\right)}-\frac{2.3}{3.\left(2x+1\right)}=\frac{6}{13}\)
\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\frac{3+5-2}{2x+1}=\frac{6}{13}\)
\(\frac{6}{2x+1}=\frac{6}{13}\)
=> 2x + 1 = 13
2x = 12
x = 6
a) 2x(x-5)-x(3+2x)=26
<=> \(2x^2-10x-3x-2x^2=26\)
<=> -13x=26
=<=> x=-2
b) (x-1)(2x-1)-(x+8)(x-1)=0
<=> (x-1)(2x-1-x-8)=0
<=>(x-1)(x-9)=0
> x=1 hoặc x=9
a ) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=26:-13\)
\(\Leftrightarrow x=-2\)
b ) \(\left(x-1\right)\left(2x-1\right)-\left(x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1-x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+7=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-7\end{array}\right.\)
Vậy \(x\in\left\{1;-7\right\}\)
a/ 2x(x-5)-x(3+2x)=26
=>2x2-10x-2x2-3x=26
=>-13x=26
=>x=-2
b/ 49x2-81=0
=>49x2=81
=>x2=\(\frac{49}{81}\)
\(\Rightarrow x^2=\left(-\frac{9}{7}\right)^2\)hoặc \(\left(\frac{9}{7}\right)^2\)
\(\Rightarrow x=\pm\frac{9}{7}\)