a, 4x+1=13-2x <-->6x=12 <-->x=2

b, (2x-5)(x-4)=0 <-->x=5/2  hoặc x=4

c,Đề bài -->x(x-2)+6(x+2)=2x+12 -->x^2+2x=0 -->x=0  hoặc x=-2

d,|x-3|=9-2x -->TH1: x-3=9-2x -->x=x=4           TH2:3-x=9-2x -->x=6

a) Ta có: \(3\left(2-x\right)+1=4-2x\)

\(\Leftrightarrow6-3x+1-4+2x=0\)

\(\Leftrightarrow-x+3=0\)

\(\Leftrightarrow-x=-3\)

hay x=3

Vậy: S={3}

b) Ta có: \(2\left(x+4\right)=3-x\)

\(\Leftrightarrow2x+8-3+x=0\)

\(\Leftrightarrow3x+5=0\)

\(\Leftrightarrow3x=-5\)

hay \(x=-\dfrac{5}{3}\)

Vậy: \(S=\left\{-\dfrac{5}{3}\right\}\)

c) Ta có: \(7-3x=x-5\)

\(\Leftrightarrow7-3x-x+5=0\)

\(\Leftrightarrow-4x+12=0\)

\(\Leftrightarrow-4x=-12\)

hay x=3

Vậy: S={3}

d) Ta có: \(5x-\left(x-1\right)=7\)

\(\Leftrightarrow5x-x+1=7\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)

`2/(4-x^2)+1/(x^2-2x)=(x-4)/(x^2+2x)(x ne 0,+-2)`

`<=>(2x)/(4x-x^3)+(x+2)/(x^3-4x)=(x^2-6x+8)/(x^3-4x)`

`<=>-2x+x+2=x^2-6x+8`

`<=>x^2-7x+10=0`

`<=>x^2-2x-5x+10=0`

`<=>x(x-2)-5(x-2)=0`

`<=>(x-2)(x-5)=0`

Vì `x ne 2=>x-2 ne 0`

`=>x-5=0`

`=>x=5`

Vậy `S={5}`

b) ĐKXĐ: \(x\ne1\)

Ta có: \(\dfrac{2}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{x}{x^2+x+1}\)

\(\Leftrightarrow\dfrac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

Suy ra: \(2x^2+2x+1-3x^2-x^2+x=0\)

\(\Leftrightarrow-2x^2+x+1=0\)

\(\Leftrightarrow-2x^2+2x-x+1=0\)

\(\Leftrightarrow-2x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

\(a,\Rightarrow\left[{}\begin{matrix}x-1=2x\\1-x=2x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Rightarrow\left[{}\begin{matrix}x+x-2=2\left(x\ge2\right)\\x+2-x=2\left(0\le x< 2\right)\\-x+2-x=2\left(x< 0\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\left(x\ge2\right)\left(tm\right)\\x=0\left(0\le x< 2\right)\left(tm\right)\\x=0\left(x< 0\right)\left(ktm\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

a: Ta có: \(\left|x-1\right|=2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=2x\left(x\ge1\right)\\x-1=-2x\left(x< 1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=\dfrac{1}{3}\left(nhận\right)\end{matrix}\right.\)

Em hc bảng xét dáu chx ??

a) Ta có: (x-1)(x-4)>0

\(\Leftrightarrow\left[{}\begin{matrix}x-4>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\)

b) Ta có: (x-6)(x-7)<0

\(\Leftrightarrow\left\{{}\begin{matrix}x-6>0\\x-7< 0\end{matrix}\right.\Leftrightarrow6< x< 7\)

c) Ta có: \(\left(x-1\right)\left(x-2\right)\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x-2\le0\end{matrix}\right.\Leftrightarrow1\le x\le2\)

d) Ta có: \(\left(x-2\right)\left(x-\dfrac{2}{3}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2\ge0\\x-\dfrac{2}{3}\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le\dfrac{2}{3}\end{matrix}\right.\)

a) Ta có: (x-1)(x-4)>0

⇔[x−4>0x−1<0⇔[x>4x<1⇔[x−4>0x−1<0⇔[x>4x<1

b) Ta có: (x-6)(x-7)<0

⇔{x−6>0x−7<0⇔6<x<7⇔{x−6>0x−7<0⇔6<x<7

c) Ta có: (x−1)(x−2)≤0(x−1)(x−2)≤0

⇔{x−1≥0x−2≤0⇔1≤x≤2⇔{x−1≥0x−2≤0⇔1≤x≤2

d) Ta có: (x−2)(x−23)≥0(x−2)(x−23)≥0

⇔⎡⎣x−2≥0x−23≤0⇔⎡⎣x≥2x≤23

a)

Cho A(X) = 0

 -18+2x =0

2x = 18

x = 9

Vậy nghiệm của đa thức A(x) là9

b)

CHo B(x) = 0

(x+1)(x-2) =0

TH1)

x+1= 0

x = -1

TH2)

x-2 =0

x = 2

Vậy nghiệm của đa thức B(x) = -1 hoặc 2

a) choA(x) = 0

\(=>-18+2x=0\)

\(=>2x=18=>x=9\)

b) cho B(x) = 0

\(=>\left(x+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

a) A(x) = -18+2x

   A(x) = 0

-18 + 2x = 0

         2x = 0 + (-18)

         2x = -18

           x = -18 :2

           x = -9

Vậy nghiệm của A(x) là: x=-9.

b) B(x) = (x+1)(x-2)

    B(x) = 0

(x+1)(x-2) = 0

TH1: (x+1) = 0

          x = -1

TH2: (x-2) = 0

          x = 2 

Vậy nghiệm của B(x) là: x ∈ {-1;2}.

a: (x-2)(x+2)-(x+1)²=1

=>\(x^2-4-\left(x^2+2x+1\right)=1\)

=>\(x^2-4-x^2-2x-1=1\)

=>-2x-5=1

=>-2x=6

=>\(x=\dfrac{6}{-2}=-3\)

b: Sửa đề:\(x^3-8-\left(x-2\right)\left(x-4\right)=0\)

=>\(\left(x^3-8\right)-\left(x-2\right)\left(x-4\right)=0\)

=>\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-4\right)=0\)

=>\(\left(x-2\right)\left(x^2+2x+4-x+4\right)=0\)

=>\(\left(x-2\right)\left(x^2+x\right)=0\)

=>x(x+1)(x-2)=0

=>\(\left[{}\begin{matrix}x=0\\x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=2\end{matrix}\right.\)

c: 3x(x-1)+1-x=0

=>3x(x-1)-(x-1)=0

=>(x-1)(3x-1)=0

=>\(\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

a.   2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0

2x=- 4/5 hoặc 3x=1/2

x=-2/5 hoặc x=\(\dfrac{1}{6}\)

b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0

x=2/5 hoặc x=-\(\dfrac{4}{7}\)

d. x(1+5/8-12/16)=1

\(\dfrac{7}{8}\)x=1=> x=8/7

Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

a: \(A=\dfrac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\dfrac{-6}{\left(x+2\right)}\cdot\dfrac{-\left(x+1\right)}{6\left(x+2\right)}=\dfrac{\left(x+1\right)}{\left(x+2\right)^2}\)

b: A>0

=>x+1>0

=>x>-1

c: x^2+3x+2=0

=>(x+1)(x+2)=0

=>x=-2(loại) hoặc x=-1(loại)

Do đó: Khi x^2+3x+2=0 thì A ko có giá trị