a: Ta có: \(8x+11-3=5x+x-3\)

\(\Leftrightarrow8x+8=6x-3\)

\(\Leftrightarrow2x=-11\)

hay \(x=-\dfrac{11}{2}\)

b: Ta có: \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

\(\Leftrightarrow2x\left(x^3+6x^2+12x+8\right)-8x^2=2\left(x^3-8\right)\)

\(\Leftrightarrow2x^4+12x^3+24x^2+16x-8x^2-2x^3+16=0\)

\(\Leftrightarrow2x^4+10x^3+16x^2+16x+16=0\)

\(\Leftrightarrow2x^4+4x^3+6x^3+12x^2+4x^2+8x+8x+16=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x^3+6x^2+4x+8\right)=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

c: Ta có: \(\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\)

\(\Leftrightarrow2x^2-3x+2x-3-2x^2-10x+x+5=0\)

\(\Leftrightarrow-10x+2=0\)

\(\Leftrightarrow-10x=-2\)

hay \(x=\dfrac{1}{5}\)

d: Ta có: \(\dfrac{1}{10}-2\cdot\left(\dfrac{1}{2}t-\dfrac{1}{10}\right)=2\left(t-\dfrac{5}{2}\right)-\dfrac{7}{10}\)

\(\Leftrightarrow\dfrac{1}{10}-t+\dfrac{1}{5}=2t-5-\dfrac{7}{10}\)

\(\Leftrightarrow-t-2t=-\dfrac{57}{10}-\dfrac{3}{10}=-6\)

hay t=2

Mình khuyên bạn thế này : 

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Như vậy các bạn sẽ dễ giúp

Và cũng có nhiều bạn giúp hơn !

Bài 1.

a) ( x - 3 )( x + 7 ) = 0

<=> x - 3 = 0 hoặc x + 7 = 0

<=> x = 3 hoặc x = -7

Vậy S = { 3 ; -7 }

b) ( x - 2 )² + ( x - 2 )( x - 3 ) = 0

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

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

<=> x - 2 = 0 hoặc 2x - 5 = 0

<=> x = 2 hoặc x = 5/2

Vậy S = { 2 ; 5/2 }

c) x² - 5x + 6 = 0

<=> x² - 2x - 3x + 6 = 0

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

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

<=> x - 2 = 0 hoặc x - 3 = 0

<=> x = 2 hoặc x = 3

Bài 2.

a) \(\frac{x}{x+1}-1=\frac{3}{2}x\)

ĐKXĐ : x khác -1

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

<=> \(\frac{-1}{x+1}=\frac{3x}{2}\)

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

<=> 3x² + 3x + 2 = 0

Vi 3x² + 3x + 2 = 3( x² + x + 1/4 ) + 5/4 = 3( x + 1/2 )² + 5/4 ≥ 5/4 > 0 ∀ x

=> phương trình vô nghiệm

b) \(\frac{4x}{x-2}-\frac{7}{x}=4\)

ĐKXĐ : x khác 0 ; x khác 2

<=> \(\frac{4x^2}{x\left(x-2\right)}-\frac{7x-14}{x\left(x-2\right)}=\frac{4x^2-8x}{x\left(x-2\right)}\)

=> 4x² - 7x + 14 = 4x² - 8x

<=> 4x² - 7x - 4x² + 8x = -14

<=> x = -14 ( tm )

Vậy phương trình có nghiệm x = -14

`a,(x+3)(x^2+2021)=0`

`x^2+2021>=2021>0`

`=>x+3=0`

`=>x=-3`

`2,x(x-3)+3(x-3)=0`

`=>(x-3)(x+3)=0`

`=>x=+-3`

`b,x^2-9+(x+3)(3-2x)=0`

`=>(x-3)(x+3)+(x+3)(3-2x)=0`

`=>(x+3)(-x)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$

`d,3x^2+3x=0`

`=>3x(x+1)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$

`e,x^2-4x+4=4`

`=>x^2-4x=0`

`=>x(x-4)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$

1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)

=> S={-3}

Bài 1: 

a) Ta có: \(\left(x+3\right)\left(x^2+2021\right)=0\)

mà \(x^2+2021>0\forall x\)

nên x+3=0

hay x=-3

Vậy: S={-3}

Bài 2: 

b) Ta có: \(x\left(x-3\right)+3\left(x-3\right)=0\)

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

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

Vậy: S={3;-3}

Vì hai bài giống nhau nên anh sẽ làm mẫu bài 1 nhé.

Bài 4 :

24 phút = \(\dfrac{24}{60} = \dfrac{2}{5}\) giờ

Gọi thời gian dự định đi từ A đến B là x(giờ) ; x > 0 

Suy ra quãng đường AB là 36x(km)

Khi vận tốc sau khi giảm là 36 -6 = 30(km/h)

Vì giảm vận tốc nên thời gian đi hết AB là x + \(\dfrac{2}{5}\)(giờ)

Ta có phương trình: 

\(36x = 30(x + \dfrac{2}{5})\\ \Leftrightarrow x = 2\)

Vậy quãng đường AB dài 36.2 = 72(km)

Bài 3 : 

\(A = -x^2 + 2x + 9 = -(x^2 -2x - 9) \\= -(x^2 - 2x + 1 + 10) = -(x^2 -2x + 1)+ 10\\=-(x-1)^2 + 10\)

Vì : \((x-1)^2 \geq 0\) ∀x \(\Leftrightarrow -(x-1)^2 \)≤ 0 ∀x \(\Leftrightarrow -(x-1)^2 + 10\) ≤ 10

Dấu "=" xảy ra khi và chỉ khi x - 1 = 0 ⇔ x = 1

Vậy giá trị nhỏ nhất của A là 10 khi x = 1

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

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

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

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

Bài 3: 

b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)

hay \(x\in\left\{0;-1\right\}\)

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

=>x-1=0

hay x=1

d: \(\Leftrightarrow6x^2-3x-4x+2=0\)

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

hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)

\(a,\left(2x-3\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=-2\end{matrix}\right.\\ b,2x-\left(3-5x\right)=4\left(x+3\right)\\ \Leftrightarrow2x-3+5x=4x+12\\ \Leftrightarrow7x-3-4x-12=0\\ \Leftrightarrow3x-15=0\\ \Leftrightarrow x=5\)

\(c,ĐKXĐ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x-2}-\dfrac{2}{x+1}=\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{x+1}{\left(x-2\right)\left(x+1\right)}-\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}-\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x+1-x+2-11+3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Rightarrow3x-8=0\\ \Leftrightarrow x=\dfrac{8}{3}\left(tm\right)\)

\(a,4+3x=25-4x\\ \Leftrightarrow7x=21\\ \Leftrightarrow x=3\\ b,\left(x-1\right)^2+\left(x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-1+x+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(2x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c, ĐKXĐ:\(x\ne-1,x\ne2\)

\(\dfrac{1}{x+1}+\dfrac{3}{x-2}=\dfrac{9}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}+\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{9}{\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x-2+3x+3-9}{\left(x+1\right)\left(x-2\right)}=0\\ \Rightarrow4x-8=0\\ \Leftrightarrow x=2\left(ktm\right)\)