\(a,\left(x-1\right)\left(x-2\right)>\left(x-1\right)^2\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)-\left(x-1\right)^2>0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x-2\right)-\left(x-1\right)\right]>0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2-x+1\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left(-1\right)>0\)

\(\Leftrightarrow x-1< 0\)

\(\Leftrightarrow x< 1\)

Vậy \(S=\left\{x|x< 1\right\}\)

\(b,\left(4x-1\right)\left(x^2+1\right)\left(-x+4\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1>0\\x^2+1>0\forall x\left(x^2\ge0\forall x\right)\\-x+4>0\end{matrix}\right.\)

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

Vậy \(S=\left\{x|\dfrac{1}{4}< x< 4\right\}\)

a)

`(2x-3)(x+4)>2(x^2 +1)`

`<=>2x^2 +8x-3x-12>2x^2 +2`

`<=>5x>14`

`<=>x>14/5`

//////////////////////l/////////////l---------->

                    `0`       `14/5`

b)

`(3x-1)/2-(5x+1)/3>4`

`<=>(9x-3)/6-(10x+2)/6>24/6`

`<=>9x-3-10x-2>24`

`<=>-x>29`

`<=>x<-29`

-------------l////////////////l////////////////////

               -29            0

A=   +  +   +     

   =   +  +   +     

   =   + 2. + 2  +     

   = 3. + 2  +

`@` `\text {Ans}`

`\downarrow`

`x(1-x) + (x-1)^2`

`= x-x^2 + x^2 - 2x + 1`

`= (x-2x) + (-x^2 + x^2) + 1`

`= -x+1`

x ( 1 - x ) + ( x - 1 )² = x - x² + x² - 2x + 1 = -x + 1 = 1 - x

Xem lại biểu thức P.

Mình phải đi ăn nên chiều mình làm nốt câu d nhé

Đề yc gì em?

\(\dfrac{x-1}{x+2}+\dfrac{6x}{x^2-4}=\dfrac{x+1}{2-x}\left(dkxd:x\ne\pm2\right)\)

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

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)+6x+\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow x^2-2x-x+2+6x+x^2+2x+x+2=0\)

\(\Leftrightarrow2x^2+6x+4=0\)

\(\Leftrightarrow2x^2+2x+4x+4=0\)

\(\Leftrightarrow2x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x+4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{-1\right\}\)

A B C D

\(\widehat{A}+\widehat{D}=70^o+110^o=180^o\) 

=> ABCD là tứ giác nội tiếp (tứ giác có tổng 2 góc đối =180 là tứ giác nt)

\(\widehat{ABD}=\widehat{ACD}\) (góc nt cùng chắn cung AD) (1)

\(\widehat{CBD}=\widehat{CAD}\) (góc nt cùng chắn cung CD) (2)

Tg ADC cân tại D \(\Rightarrow\widehat{ACD}=\widehat{CAD}\) (3)

Từ (1) (2) (3) \(\Rightarrow\widehat{ABD}=\widehat{CBD}\)

\(a,\left(x-2\right)\left(3x-1\right)=x\left(2-x\right)\)

\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)+x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-1+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{1}{4};2\right\}\)

\(b,\left|2x+3\right|=4x+1\)

\(TH_1:x\ge-\dfrac{3}{2}\)

\(2x+3=4x+1\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\left(tm\right)\)

\(TH_2:x< -\dfrac{3}{2}\)

\(-2x-3=4x+1\\ \Leftrightarrow-6x=4\Leftrightarrow x=-\dfrac{2}{3}\left(ktm\right)\)

Vậy \(S=\left\{1\right\}\)

\(c,\dfrac{x+1}{3}+1=3-\dfrac{5x}{10}\\ \Leftrightarrow\dfrac{10\left(x+1\right)+30-90+15x}{30}=0\\ \Leftrightarrow10x+10-60+15x=0\\ \Leftrightarrow25x=50\\ \Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

\(d,\dfrac{1}{x+2}+\dfrac{3}{2-x}=\dfrac{2x-3}{x^2-4}\left(dk:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{x-2-x-2-2x+3}{x^2-4}=0\)

\(\Leftrightarrow-2x=1\)

\(\Leftrightarrow x=-\dfrac{1}{2}\left(tmdk\right)\)

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

Câu d, Sửa từ dòng 2 :

\(\Leftrightarrow\dfrac{x-2-3x-6-2x+3}{x^2-4}=0\)

\(\Leftrightarrow-4x=5\)

\(\Leftrightarrow x=-\dfrac{5}{4}\left(tm\right)\)

Vậy ...