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a, 12 - (2\(x^2\) - 3) = 7
2\(x^2\) - 3 = 12 - 7
2\(x^2\) - 3 = 5
2\(x^2\) = 8
\(x^2\) = 4
\(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
2x+69.2=69.4
2x+138=276
2x = 276-138
2x = 138
x = 138:2
x = 69
2x-12-x = 0
<=> 2x-x-12 =0
(2x-x)-12=0
=> x-12=0
x = 0+12
x =12
(x-7)(2x-8)=0
\(\Rightarrow\hept{\begin{cases}x-7=0\\2x-8=0\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\x=4\end{cases}}\)
Vậy ...
a, 2X+69.2=69.4
2X+138=276
2X=276-138=138
X=138:2=69
b,2X-12-x=0
2X-X=12-0
x=12
mik chi biet vay thoi con cau cuoi mik chiu
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
\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)
1. Ta có
2x - 12 - x = 0
\(\Leftrightarrow\)( 2x - x ) - 12 = 0
\(\Leftrightarrow\)x - 12 =0
\(\Leftrightarrow\)x = 0 + 12
\(\Leftrightarrow\)x = 12
Vậy x = 12
2. Vì ( x - 7 )( 2x - 8 ) = 0\(\Rightarrow\)\(\orbr{\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}x=0+7\\2x=0+8\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=7\\2x=8\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=7\\x=8:2\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}\)
\(\text{Mình không phải là chế nên không thèm trả lời}\) >_<
7.(x - 1) + 2x.(x - 1) = 0
(x - 1).(7 + 2x) = 0
=> x - 1 = 0 hoặc 7 + 2x = 0
=> x = 1 hoặc 2x = -7
=> x = 1 hoặc x = -7/2
Vậy x thuộc {1 ; -7/2}
7( x - 1 ) + 2( x - 1 ) = 0
7x - 7 + 2x - 2 = 0
=> x = 1
\(2x.\left(x-\frac{1}{7}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-\frac{1}{7}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}}\)
Vậy x = 0 hoặc x = 1/7
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 ...
\(2x-12-x=0\)
\(\Leftrightarrow2x-x=12\Leftrightarrow x=12\)
\(\left(x-7\right)\left(2x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\2x=8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)