<=>4x-8=0 

<=>4x=8 

=.x=2(nhan)

Giải pt :

a) \(2x\left(x+5\right)-\left(x-3\right)^2=x^2+6\)

\(\Leftrightarrow2x^2+10x-x^2+6x-9-x^2-6=0\)

\(\Leftrightarrow16x-15=0\)

\(\Leftrightarrow x=\frac{15}{16}\)

b) \(6\left(x-3\right)+\left(x-1\right)^2-\left(x+1\right)^2=2x\)

\(\Leftrightarrow2x-18=2x\)

\(\Leftrightarrow-18=0\)( vô lí )

=> x thuộc rỗng

c)d) tương tự

e) \(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)

\(\Leftrightarrow\frac{5x-2}{6}+\frac{9-12x}{6}=\frac{12}{6}-\frac{2x+14}{6}\)

\(\Leftrightarrow5x-2+9-12x=12-2x-14\)

\(\Leftrightarrow-5x+9=0\)

\(\Leftrightarrow x=\frac{9}{5}\)

f) \(\frac{2x-1}{2}=\frac{2x+1}{4}-\frac{1-2x}{8}\)

\(\Leftrightarrow\frac{4\left(2x-1\right)}{8}=\frac{2\left(2x+1\right)}{8}-\frac{1-2x}{8}\)

\(\Leftrightarrow8x-4=4x+2-1+2x\)

\(\Leftrightarrow2x-5=0\)

\(\Leftrightarrow x=\frac{5}{2}\)

Tìm x :

a) \(3x^3-27x=0\)

\(\Leftrightarrow3x\left(x^2-9\right)=0\)

\(\Leftrightarrow3x\left(x-3\right)\left(x+3\right)=0\)

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

b) \(2x^3-12x^2+18x=0\)

\(\Leftrightarrow2x\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)^2=0\)

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

a) 3x(x - 1) + 2(x - 1) = 0

<=> (3x + 2)(x - 1) = 0

<=> \(\orbr{\begin{cases}3x+2=0\\x-1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=1\end{cases}}\)

Vậy S = {-2/3; 1}

b) x² - 1 - (x + 5)(2 - x) = 0

<=> x² - 1 - 2x + x² - 10 + 5x = 0

<=> 2x² + 3x - 11 = 0

<=> 2(x² + 3/2x + 9/16 - 97/16) = 0

<=> (x + 3/4)² - 97/16 = 0

<=> \(\orbr{\begin{cases}x+\frac{3}{4}=\frac{\sqrt{97}}{4}\\x+\frac{3}{4}=-\frac{\sqrt{97}}{4}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{\sqrt{97}-3}{4}\\x=-\frac{\sqrt{97}-3}{4}\end{cases}}\)

Vậy S = {\(\frac{\sqrt{97}-3}{4}\); \(-\frac{\sqrt{97}-3}{4}\)

d) x(2x - 3) - 4x + 6 = 0

<=> x(2x - 3) - 2(2x - 3) = 0

<=> (x - 2)(2x - 3) = 0

<=> \(\orbr{\begin{cases}x-2=0\\2x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=\frac{3}{2}\end{cases}}\)

Vậy  S = {2; 3/2}

e)  x³ - 1 = x(x - 1)

<=> (x - 1)(x² + x + 1) - x(x - 1) = 0

<=> (x - 1)(x² + x +  1 - x) = 0

<=> (x - 1)(x² + 1) = 0

<=> x - 1 = 0

<=> x = 1

Vậy S = {1}

f) (2x - 5)² - x² - 4x - 4 = 0

<=> (2x - 5)² - (x + 2)² = 0

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

<=> (x - 7)(3x - 3) = 0

<=> \(\orbr{\begin{cases}x-7=0\\3x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=7\\x=1\end{cases}}\)

Vậy S = {7; 1}

h) (x - 2)(x² + 3x - 2) - x³ + 8 = 0

<=> (x - 2)(x² + 3x - 2) - (x- 2)(x² + 2x + 4) = 0

<=> (x - 2)(x² + 3x - 2 - x² - 2x - 4) = 0

<=> (x - 2)(x - 6) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

Vậy S = {2; 6}

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

\(3x.x-3x+2x-2=0\)

\(2x-2=0\)

\(2x=2\)

\(x=1\)

\(\frac{x-2}{18}-\frac{2x+5}{12}>\frac{x+6}{9}-\frac{x-3}{6}\)

\(\Leftrightarrow\frac{2\left(x-2\right)}{36}-\frac{3\left(2x+5\right)}{36}>\frac{4\left(x+6\right)}{36}-\frac{6\left(x-3\right)}{36}\)

\(\Leftrightarrow2x-4-6x-15>4x+24-6x+18\)

\(\Leftrightarrow2x-6x-4x+6x>24+18+4+15\)

\(\Leftrightarrow-2x>61\)

\(\Leftrightarrow x< -\frac{61}{2}\)

Vậy nghiệm của bất phương trình là \(x< -\frac{61}{2}\)

Bài b và c làm cách mình thì dễ hiểu hơn nhiều :3

\(\left(2x-2\right)\left(2x+3\right)\le0\)

TH1 : \(\hept{\begin{cases}2x-3\le0\\2x+3\ge0\end{cases}< =>\hept{\begin{cases}2x\le3\\2x\ge-3\end{cases}}}\)

\(< =>\hept{\begin{cases}x\le\frac{3}{2}\\x\ge-\frac{3}{2}\end{cases}}\)

TH2 : \(\hept{\begin{cases}2x-3\ge0\\2x+3\le0\end{cases}< =>\hept{\begin{cases}2x\ge3\\2x\le-3\end{cases}}}\)

\(< =>\hept{\begin{cases}x\ge\frac{3}{2}\\x\le-\frac{3}{2}\end{cases}}\)

Vậy ...

\(\left(3-2x\right)\left(4x+8\right)\ge0\)

TH1 : \(\hept{\begin{cases}3-2x\ge0\\4x+8\ge0\end{cases}}\)

\(< =>\hept{\begin{cases}3\ge2x\\4x\ge-8\end{cases}< =>\hept{\begin{cases}\frac{3}{2}\ge x\\x\ge-\frac{8}{4}=-2\end{cases}}}\)

TH2 : \(\hept{\begin{cases}3-2x\le0\\4x+8\le0\end{cases}}\)

\(< =>\hept{\begin{cases}3\le2x\\4x\le-8\end{cases}}< =>\hept{\begin{cases}x\ge\frac{3}{2}\\x\ge-2\end{cases}}\)

Vậy ...

\(a,\)( sửa lại xíu đề cho đúng nhé )

\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=-\frac{2x}{x^2+x+1}\)

\(\Rightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Rightarrow x^2+x+1-3x^2=-2x^2+2x\)

\(\Rightarrow x=1\)

\(g,\)\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)=-16\)

\(\Rightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)=-16\)

Đặt \(x^2+10x+16=a\)

\(\Rightarrow a\left(a+8\right)=-16\)

\(\Rightarrow a^2+8a+16=0\)

\(\Rightarrow\left(a+4\right)^2=0\)

\(\Leftrightarrow\left(x^2+10x+20\right)^2=0\)

\(\Rightarrow x^2+10x+25-25=0\)

\(\Rightarrow\left(x+5\right)^2-\left(\sqrt{5}\right)^2=0\)

\(\Rightarrow\left(x+5-\sqrt{5}\right)\left(x+5+\sqrt{5}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-5+\sqrt{5}\\x=-5-\sqrt{5}\end{cases}}\)

\(h,\)\(x^3+3x^2+3x+2=0\)

\(\Rightarrow x^3+2x^2+x^2+2x+x+2=0\)

\(\Rightarrow x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x^2+x+1\right)=0\)

Vì \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow x+2=0\Leftrightarrow x=-2\)

Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) đẻ được hỗ trợ tốt hơn. Viết như thế kia rất khó đọc => khả năng bị bỏ qua bài cao.

a: =>3x=3

=>x=1

b: =>12x-2(5x-1)=3(8-3x)

=>12x-10x+2=24-9x

=>2x+2=24-9x

=>11x=22

=>x=2

c: =>2x-3(2x+1)=x-6x

=>-5x=2x-6x-3=-4x-3

=>-x=-3

=>x=3

d: =>2x-5=0 hoặc x+3=0

=>x=5/2 hoặc x=-3

e: =>x+2=0

=>x=-2

hết cứu đi mà làm

a)Ta có \(\left(2x+1\right)\left(x^2+2\right)=0\)<=>

2x+1=0<=>x=\(-\frac{1}{2}\)

hoặc \(x^2+2=0\)<=>\(x^2=-2\)(Vô lí)

Vậy tập nghiệm của pt S=(\(-\frac{1}{2}\))

b)\(\left(x^2+4\right)\left(7x-3\right)=0\)

<=>\(\left[{}\begin{matrix}x^2+4=0\\7x-3=0\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x^2=-4\\x=\frac{3}{7}\end{matrix}\right.\)

\(x^2=-4\) vô lí

Vậy ..........

c)\(\left(x^2+x+1\right)\left(6-2x\right)=0\)

<=>\(\left[{}\begin{matrix}x^2+x+1=0\\6-2x=0\end{matrix}\right.\)

Vì \(x^2+x+1>0\)(dễ dàng c/m)

=>6-2x=0=>x=3

Vậy...

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

<=>8x-4=0,x=\(\frac{1}{2}\)

hoặc \(x^2+2x+2=0\)(vô lí)

Vậy .....