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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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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
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\(\left(x-7\right).\left(2x-8\right)=0\)
\(x-7=0\)
\(x=0+7\)
\(x=7\)
\(2x-8=0\)
\(2x=0+8\)
\(2x=8\)
\(x=8:2\)
\(x=4\)
\(\Rightarrow x\in\left\{4;7\right\}\)
\(\left(x-7\right)\times\left(2x-8\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-7=0\\2x-8=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=7\\x=4\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x-12-x=0\)
\(\Leftrightarrow2x-x=12\Leftrightarrow x=12\)
\(\left(x-7\right)\left(2x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\2x=8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Để tích trên bằng 0 thì 2x - 8 hoặc x - 7 phải bằng 0.
Nếu x - 7 = 0 => x = 7. Thay vào ta có: 0. (2 x 7 - 8) = 0. 6 = 0(hợp lí)
Nếu 2x - 8 = 0 thì 2x = 8 => x = 4. Vô lí vì 4 - 7 không phải số tự nhiên.
Vậy x = 7.
(x-7)(2x-8)=0
=> x-7=0 hoặc 2x-8=0
* x-7=0 * 2x-8=0
x = 0+7 2x = 0+8
x = 7 2x = 8
x = 8:2
x = 4
Vậy x thuộc { 7 ; 4 }
(Sorry vì ko có dấu thuộc)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x=1;x=2;x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( 4 - x) (x + 3) = 0
=> \(\hept{\begin{cases}4-x=0\\x+3=0\end{cases}}\) => \(\hept{\begin{cases}x=4\\x=-3\end{cases}}\)
b) 7(2x - 8) = 0
=> 2x - 8 = 0
=> 2x = 8
=> x = 4
c) -x(8 + x) = 0
=> 8 + x = 0
=> x = -8
câu c) ko chéc nghen!!
565454785698532275475568679760351352636547765768843
a) ( 4 - x ) . ( x + 3 ) = 0
Ta có các trường hợp sau đây :
\(\orbr{\begin{cases}4-x=0\\x+3=0\end{cases}}\orbr{\begin{cases}x=4-0=4\\x=0-3=-3\end{cases}}\)
==> x = -3 ; 4
b) 7. ( 2x - 8 ) = 0
2x - 8 = 0 : 7
2x - 8 = 0
2x = 0 + 8 =8
==> x = 8 : 2 = 4
c) -x . ( 8 + x ) = 0
Ta có các trường hợp sau đây :
\(\orbr{\begin{cases}-x=0\\8+x=0\end{cases}}\orbr{\begin{cases}x=0\\x=-8\end{cases}}\)
MÌNH HK CHẮC CÂU c) ÂU NHA, MÌNH THẤY CÁI NÀO CŨNG LẠ MÀ 0 CÓ THỂ ĐỂ LÀ -0 ĐƯỢC HK CÁC BẠN
TH1: x - 7 = 0
x = 7
TH2: 2x - 8 = 0
2x = 8
x = 4
\(\left(x-7\right)\left(2x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}\)