\[\left| {2x - 3} \right| > x + 1\\ \Leftrightarrow \left| {2x - 3} \right| - x > 1\\ T{H_1}:2x - 3 \ge 0 \Rightarrow x \ge {3 \over 2}\\ 2x - 3 - x > 1\\ \Leftrightarrow x - 3 > 1\\ \Leftrightarrow x > 4\left( {TM} \right)\\ T{H_2}:2x - 3 < 0 \Rightarrow x < {3 \over 2}\\ - \left( {2x - 3} \right) - x > 1\\ \Leftrightarrow - 2x + 3 - x > 1\\ \Leftrightarrow - 3x > - 2\\ \Leftrightarrow x < {2 \over 3}\left( {TM} \right)\]

a, 15x - 6 = 12x + 3

\(\Leftrightarrow\) 15x - 12x = 3 + 6

\(\Leftrightarrow\) 3x = 9

\(\Leftrightarrow\) x = 3

Vậy S = {3}

b, \(\frac{x+2}{2}-\frac{2x-3}{5}=10x+\frac{13}{10}\)

\(\Leftrightarrow\) \(\frac{5\left(x+2\right)}{10}-\frac{2\left(2x-3\right)}{10}=\frac{100x}{10}+\frac{13}{10}\)

\(\Leftrightarrow\) 5(x + 2) - 2(2x - 3) - 100x - 13 = 0

\(\Leftrightarrow\) 5x + 10 - 4x + 6 - 100x - 13 = 0

\(\Leftrightarrow\) -99x + 3 = 0

\(\Leftrightarrow\) x = \(\frac{1}{33}\)

Vậy S = {\(\frac{1}{33}\)}

d, (3x + 2)(4x - 5) = 0

\(\Leftrightarrow\) 3x + 2 = 0 hoặc 4x - 5 = 0

\(\Leftrightarrow\) x = \(\frac{-2}{3}\) và x = \(\frac{5}{4}\)

Vậy S = {\(\frac{-2}{3}\); \(\frac{5}{4}\)}

Phần c với phần e bạn viết vậy mình ko hiểu, bn viết lại đi!

Chúc bn học tốt!!

a) 3x² + 2x - 1 = 0

<=> 3x² + 3x - x - 1 = 0

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

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

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

b) x² - 5x + 6 = 0

<=> x² - 2x - 3x + 6 = 0

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

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

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

c) x² - 3x + 2 = 0

<=> x² - x - 2x + 2 = 0

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

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

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

d) 2x² - 6x + 1 = 0

<=> 2( x² - 3x + 9/4 ) - 7/2 = 0

<=> 2( x - 3/2 )² = 7/2

<=> ( x - 3/2 )² = 7/4

<=> \(\left(x-\frac{3}{2}\right)=\left(\pm\sqrt{\frac{7}{4}}\right)^2=\left(\pm\frac{\sqrt{7}}{2}\right)^2\)

<=> \(\orbr{\begin{cases}x-\frac{3}{2}=\frac{\sqrt{7}}{2}\\x-\frac{3}{2}=\frac{-\sqrt{7}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{7}}{2}\\x=\frac{3-\sqrt{7}}{2}\end{cases}}\)

a.

\(2\left(x+5\right)-x^2-5x=0\)

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

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

\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow x=2\) hoặc \(x=-5\)

a,\(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Vậy...

b,\(2x^2+3x-5=0\)

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

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

\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\)

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

Vậy...

b.

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

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

\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=1\end{matrix}\right.\)

mk mới lớp 6 thôi ,lớp 9 mình .......mình.........chịu (I VERY SORRY YOU!!)

sorry, i cant do it

mình lớp 9 nhưng mình lười giải vì " QUÁ NHIỀU " lười viết

1) \(x^4-6x^3-x^2+54x-72=0\)

\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)

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

Tự làm nốt...

2) \(x^4-5x^2+4=0\)

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

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

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

Tự làm nốt...

\(x^4-2x^3-6x^2+8x+8=0\)

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

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

\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+2\right)-2x\left(x+2\right)-2\left(x+2\right)\right]=0\)

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

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

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

...

\(2x^4-13x^3+20x^2-3x-2=0\)

\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)

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

Bí

\(2x^3-9x^2+2x+1\)

\(=2x^3-x^2-8x^2+4x-2x+1\)

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

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

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

\(=\left(2x-1\right)\left[\left(x-2\right)^2-5\right]\)

.......

a)

\(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2-x+3x-1=0\)

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

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

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

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

b)

\(x^2-5x+6=0\)

\(\Leftrightarrow x^2-3x-2x+6=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

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

\(\Rightarrow3x^2-x+3x-1=0\)

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

\(\Rightarrow x.\left(3x-1\right)+\left(3x-1\right)=0\)

\(\Rightarrow\left(3x-1\right).\left(x+1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}3x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x=1\\x=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\)

Vậy......

b, \(x^2-5x+6=0\)

\(\Rightarrow x^2-3x-2x+6=0\)

\(\Rightarrow\left(x^2-3x\right)-\left(2x-6\right)=0\)

\(\Rightarrow x.\left(x-3\right)-2.\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right).\left(x-2\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

Vậy......

Chúc bạn học tốt!!!

\(a,3x^2+2x-1=0\Leftrightarrow x^2-1+2x^2-2x\)

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

\(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\\dfrac{-1}{3}\end{matrix}\right.\)Vậy phương trình có tập nghiệm S=\(\left\{1;\dfrac{-1}{3}\right\}\)

\(b,x^2-5x+6=0\Leftrightarrow x^2-2x-3x+6=0\)\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

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

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

Vậy phương trình có tập nghiệm S=\(\left\{2;3\right\}\)

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

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

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

Vậy phương trình có tập nghiệm S = \(\left\{1;2\right\}\)

a, (3x - 1)(5x + 3) = (2x + 3)(3x - 1)

⇔ 5x + 3 = 2x + 3

⇔ 3x = 0

⇔ x = 0

Vậy phương trình có nghiệm là x = 0

Mình làm lại rồi nhé!

a, (3x - 1)(5x + 3) = (2x + 3)(3x - 1)

⇔ 5x + 3 = 2x + 3

⇔ 3x = 0

⇔ x = 0

Vậy phương trình có nghiệm là x = 3.

b, 9x² - 1 = (3x + 1)(2x - 1)

⇔ (3x + 1)(3x - 1) = (3x + 1)(2x - 1)

⇔ 3x - 1 = 2x - 1

⇔ x = 0

Vậy phương trình có nghiệm là x = 0