`|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}`

\(\left|x+3\right|+10-5x=0\)

\(\Leftrightarrow\left|x+3\right|=5x-10\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-10\ge0\\\left[{}\begin{matrix}x+3=5x-10\\x+3=10-5x\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=\dfrac{13}{4}\left(N\right)\\x=\dfrac{7}{6}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)

Giải:

\(\left|x+3\right|+10-5x=0\) 

\(\Rightarrow\left|x+3\right|=5x-10\) 

\(\Rightarrow\left[{}\begin{matrix}5x-10=x+3\\5x-10=x-3\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{4}\\x=\dfrac{7}{4}\end{matrix}\right.\) 

Vậy \(x\in\left\{\dfrac{13}{4};\dfrac{7}{4}\right\}\) 

Chúc bạn học tốt!

`|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!

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 \(x\in \left\{7;-4;0;3\right\}\)

.

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

=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)