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\(\left(3x-15\right)\left(2x-6\right)=2x-6\)
\(\Leftrightarrow\left(3x-15\right)\left(2x-6\right)-\left(2x-6\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(3x-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\3x-16=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{16}{3}\end{cases}}\)
(3x-15).(2x-6)=2x-6
3x-15 =(2x-6):(2x-6)
3x-15 =1
3x =1+15
3x =16
x =16:3
x = 8
Vậy x =8
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a, 3x-(2x+1)=6
3x-2x-1=6
x-1=6
x=7
b,2x-[(-15)+x]-6=16
2x-[(-15)+x]=22
2x-(-15)-x=22
x+15=22
x=7
c,(15-18)2+3x=2.(x-6)
(-3)2+3x=2x-12
9+3x=2x-12
3x-2x=-12-9
x=-21
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\(2x\left(2x-6\right)\left(3x-15\right)\left(x^2+1\right)=0\)
=> TH1: \(2x=0\Rightarrow x=0\)
TH2: \(2x-6=0\Rightarrow2x=6\Rightarrow x=3\)
TH3: \(3x-15=0\Rightarrow3x=15\Rightarrow x=5\)
TH4: \(x^2+1=0\Rightarrow x^2=-1\) ( vô lý => loại )
Vậy: x = 0 hoặc x = 3 hoặc x = 5
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\(a,x=3x^2\Rightarrow x-3x^2=0\Rightarrow x\left(1-3x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)
\(b,\left(2x-6\right)\left(x+4\right)+2\left(2x-6\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(x+4+2\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-6=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)
\(c,\left(2x-5\right)\left(x+9\right)+6x-15=0\)
\(\Rightarrow\left(2x-5\right)\left(x+9\right)+3\left(2x-5\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(x+9+3\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(x+12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-12\end{cases}}\)
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Câu 1:
(3\(x\) - 15).(10 - \(x\)) < 0
3\(x-15\) = 0 ⇒ 3\(x\) = 15 ⇒ \(x\) = 15 : 3 ⇒ \(x=5\)
10 - \(x\) = 0 ⇒ \(x=10\)
Lập bảng ta có:
\(x\) | 5 10 |
3\(x\) - 15 | - 0 + + |
10 - \(x\) | + + 0 - |
(3\(x\) - 15).(10 - \(x\)) | - 0 + 0 - |
Theo bảng trên ta có: \(x\) < 5 hoặc \(x\) > 10
Vậy \(x\) < 5 hoặc \(x\) > 10
(2\(x\) - 8).(6 - \(x\)) ≥ 0
2\(x\) - 8 = 0 ⇒ 2\(x\) = 8 ⇒ \(x=8:2\) ⇒ \(x=4\)
6 - \(x\) = 0 ⇒ \(x=6\)
Lập bảng ta có:
\(x\) | 4 6 |
2\(x-8\) | - 0 + | + |
6 - \(x\) | + | + 0 - |
(2\(x-8\)).(6 - \(x\)) | - 0 + | - |
Theo bảng trên ta có: 4 ≤ \(x\) ≤ 6
Vậy \(4\le x\le6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16
7+x=-16
x=-16-7
x=-23
2) 2x – 35 = 15
2x=15+35
2x=50
x=50:2
x=25
3) 3x + 17 = 12
3x=12-17
3x=-5
x=-5/3
4) (2x – 5) + 17 = 6
2x-5=6-17
2x-5=-11
2x=-11+5
2x=-6
x=-6:2
x=-3
5) 10 – 2(4 – 3x) = -4
2(4-3x)=10-(-4)
2(4-3x)=14
4-3x=14:2
4-3x=7
3x=4-7
3x=-3
x=-3:3
x=-1
6) - 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