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25 tháng 3 2023

\(\dfrac{x}{2}\left(4x-3\right)+2\left(3-x\right)\left(x+4\right)\le0\)

\(\Leftrightarrow\dfrac{4x^2}{2}-\dfrac{3x}{2}+2\left(3x+12-x^2-4x\right)\le0\)

\(\Leftrightarrow\dfrac{4x^2-3x}{2}+6x+24-2x^2-8x\le0\)

\(\Leftrightarrow\dfrac{4x^2-3x+2\left(6x+24-2x^2-8x\right)}{2}\le0\)

\(\Leftrightarrow4x^2-3x+12x+48-4x^2-16x\le0\)

\(\Leftrightarrow-7x\le-48\)

\(\Leftrightarrow x\ge\dfrac{48}{7}\)

25 tháng 3 2023

=>-7x+48≤0

<=>-7x≤-48

<=>(-7x)(-1)≥(-48)(-1)

<=>\(\dfrac{7x}{7}\)\(\dfrac{48}{7}\)

<=>x≥\(\dfrac{48}{7}\)

4 tháng 10 2015

<=>x(x^2-1)=0

<=>x=0 hoặc x^2-1=0

<=>x=0 hoặc x^2=1

<=>x=0 hoặc x=1 hoặc x=-1

 

a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)

Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)

b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)

\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(4x-1-x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)

c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left[\left(x+3\right)^2-\dfrac{1}{4}\right]=0\)

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

\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)

a: =>x^2+4x-4x+1=0

=>x^2+1=0

=>Loại

b: =>2x-6+4=2x+2

=>-2=2(loại)

c: =>2(x+3)-2x-1=1

=>6-1=1

=>5=1(loại)

d =>x+3=0

=>x=-3(loại)

e: =>x^2-3x^2+3x-3x-2=0

=>-2x^2-2=0

=>x^2+1=0

=>Loại

7 tháng 8 2017

a)Dat \(x^2-4x+3=a;x^2-7x+6=b \Rightarrow a+b=2x^2-11x+9\)

....

a: \(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)

=>x=0

b: \(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{x+1}{x-3}=\dfrac{1}{2}\)

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

\(\Leftrightarrow x^2-4x+3=2x^2-6x-2x^2+2=-6x+2\)

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

=>x=-1(nhận)

22 tháng 2 2022

\(a,2x^3+4x^2+10x=0\\ \Leftrightarrow2x\left(x^2+2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x^2+2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x^2+2x+1\right)+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2+4=0\left(vô..lí\right)\end{matrix}\right.\)

\(b,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne3\\x\ne4\end{matrix}\right.\\ \dfrac{x^2-4x}{x^2-5x+4}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x}{x-1}-\dfrac{1}{2}-\dfrac{x+1}{x-3}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{2\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)}-\dfrac{x^2-4x+3}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2x^2-2}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2-6x-x^2+4x-3-2x^2+2}{2\left(x-1\right)\left(x-3\right)}=0\)

\(\Rightarrow-x^2-2x-1=0\)

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

 

7 tháng 3 2022

a, đk : x khác -2 ; 2 

\(\left(x+2\right)^2-8x=0\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)(ktm) 

pt vô nghiệm 

b, đk : x khác -1 ; 1 

\(x\left(x+1\right)-5x+3=0\Leftrightarrow x^2-4x+3=0\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow x=1\left(ktm\right);x=3\left(tm\right)\)

24 tháng 9 2018

\(x^4-2x^3+4x^2-3x+2=0\)

\(\Leftrightarrow x^4-2x^3+x^2+3x^2-3x+\dfrac{9}{4}-1=0\)

\(\Leftrightarrow\left(x^2-x\right)^2+3\left(x^2-x\right)+\dfrac{9}{4}-1=0\)

\(\Leftrightarrow\left(x^2-x+\dfrac{3}{2}\right)^2-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+\dfrac{3}{2}=1\\x^2-x+\dfrac{3}{2}=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+\dfrac{1}{4}+\dfrac{5}{4}=1\\x^2-x+\dfrac{1}{4}+\dfrac{5}{4}=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}=1\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=-\dfrac{1}{4}\\\left(x-\dfrac{1}{2}\right)^2=-\dfrac{9}{4}\end{matrix}\right.\)

\(\Rightarrow\) Vô lý ( vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\) )

\(\Rightarrow PT\) vô nghiệm .

21 tháng 12 2018

GIÚP MÌNH VỚI MAI LÀ NỘP BÀI RỒI

23 tháng 12 2018

câu a) và b) thì sử dụng tính chất nếu tích =0 thì có ít nhất 1 thừa số =0

c)4x^2+4x+1=0

(2x+1)^2=0

2x+1=0

x=-1/2

23 tháng 8 2021

3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)

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

23 tháng 8 2021

4x.(x+1)-8(x+1)=0

(4x-8)(x+1)=0

suy ra x=2 hoặc x=-1