tìm x : I x + 3 I+ 10 - 5x = 0
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`|x+3|+10-5x=0`
`<=>|x+3|=5x-10(x>=2)`
`+)x+3=5x-10`
`<=>4x=13`
`<=>x=13/4(tm)`
`+)x-3=10-5x`
`<=>6x=13`
`<=>x=13/6(tm)`
Vậy `S={13/4,13/6}`
TH1: `x+3>=0 <=> x>=-3`
`x+3+10-5x=0`
`-4x=-13`
`x=13/4` (TM)
TH2: `x+3<0 <=> x<-3`
`-x-3+10-5x=0`
`-6x=-7`
`x=7/6` (L)
Vậy `x=13/4`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`|5x-1|-|x-4|=0`
`<=>|5x-1|=|x-4|`
`+)5x-1=x-4`
`<=>4x=-3`
`<=>x=-3/4`
`+)5x-1=4-x`
`<=>6x=5`
`<=>x=5/6`
Vậy `S={-3/4,5/6}`
Giải:
\(\left|5x-1\right|-\left|x-4\right|=0\)
\(\Rightarrow\left|5x-1\right|=\left|x-4\right|\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=x-4\\5x-1=4-x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-3}{4}\\x=\dfrac{5}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-3}{4};\dfrac{5}{6}\right\}\)
Chúc bạn học tốt!
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Bài 1 :
a) 2x3-3+3x2+8=0
b) x3-1=0
Bài 2 :
a) (x2-5x)2 + 10.(x2-5x)+24=0
b) (x+2)(x+3)(x-5)(x-6)=180
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) Bạn xem lại đề
b)
\(x^3-1=0\)
\(\Leftrightarrow (x-1)(x^2+x+1)=0\)
Vì \(x^2+x+1=x^2+2.\frac{1}{2}x+(\frac{1}{2})^2+\frac{3}{4}=(x+\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}>0\)
\(\Rightarrow x^2+x+1\neq 0\)
Do đó: \(x-1=0\Rightarrow x=1\) là nghiệm duy nhất
Bài 2:
a) \((x^2-5x)^2+10(x^2-5x)+24=0\)
\(\Leftrightarrow (x^2-5x)^2+2.5(x^2-5x)+5^2-1=0\)
\(\Leftrightarrow (x^2-5x+5)^2-1=0\)
\(\Leftrightarrow (x^2-5x+5-1)(x^2-5x+5+1)=0\)
\(\Leftrightarrow (x^2-5x+4)(x^2-5x+6)=0\)
\(\Leftrightarrow (x-1)(x-4)(x-2)(x-3)=0\)
\(\Rightarrow \left[\begin{matrix} x-1=0\\ x-4=0\\ x-2=0\\ x-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=1\\ x=4\\ x=2\\ x=3\end{matrix}\right.\)
b)
\((x+2)(x+3)(x-5)(x-6)=180\)
\(\Leftrightarrow [(x+2)(x-5)][(x+3)(x-6)]=180\)
\(\Leftrightarrow (x^2-3x-10)(x^2-3x-18)=180\)
\(\Leftrightarrow a(a-8)=180\) (đặt \(x^2-3x-10=a\) )
\(\Leftrightarrow a^2-8a+16-196=0\)
\(\Leftrightarrow (a-4)^2-14^2=0\)
\(\Leftrightarrow (a-4-14)(a-4+14)=0\Leftrightarrow (a-18)(a+10)=0\)
\(\Rightarrow a=18\) hoặc $a=-10$
+) Nếu $a=18$ thì \(x^2-3x-10=18\)
\(\Leftrightarrow x^2-3x-28=0\)
\(\Leftrightarrow (x-7)(x+4)=0\Rightarrow \left[\begin{matrix} x=7\\ x=-4\end{matrix}\right.\)
+) Nếu $a=-10$ thì \(x^2-3x-10=-10\Leftrightarrow x^2-3x=0\Leftrightarrow x(x-3)=0\)
\(\Leftrightarrow \left[\begin{matrix} x=0\\ x=3\end{matrix}\right.\)
Vậy pt có 4 nghiệm \(x\in \left\{7;-4;0;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
1. \(\left(x-1\right)\left(x+2\right)+5x-5=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+2+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.......................
2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
Vậy........................
3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy............................
4, 5 tương tự nhé bn!
bài 3
1 (x-1)(x+2)+5x-5=0
=>(x-1)(x+2)+(5x-5)=o
=>(x-1)(x+2)+5(x-1)=0
=>(x-1)(x+2+5)=0
=>(x-1)(x+7)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
vậy x=1 hoặc x=-7
2. (3x+5)(x-3)-6x-10=0
=>(3x+5)(x-3)-(6x+10)=0
=>(3x+5)(x-3)-2(3x+5)=0
=>(3x+5)(x-3-2)=0
=>(3x+5)(x-5)=0
=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
`|x+3|+10-5x=0`
`<=>|x+3|=5x-10(x>=2)`
`+)x+3=5x-10`
`<=>4x=13`
`<=>x=13/4(tm)`
`+)x-3=10-5x`
`<=>6x=13`
`<=>x=13/6(tm)`
Vậy `S={13/4,13/6}`
\(\left|x+3\right|+10-5x=0\)
\(\Leftrightarrow\left|x+3\right|=5x-10\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x-10\ge0\\\left[{}\begin{matrix}x+3=5x-10\\x+3=10-5x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=\dfrac{13}{4}\left(N\right)\\x=\dfrac{7}{6}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)