a: (2x-3)(3x+6)>0

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

=>x<-2 hoặc x>3/2

b: (3x+4)(2x-6)<0

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

=>-4/3<x<3

c: (3x+5)(2x+4)>4

\(\Leftrightarrow6x^2+12x+10x+20-4>0\)

\(\Leftrightarrow6x^2+22x+16>0\)

=>\(6x^2+6x+16x+16>0\)

=>(x+1)(3x+8)>0

=>x>-1 hoặc x<-8/3

f: (4x-8)(2x+5)<0

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

=>-5/2<x<2

h: (3x-7)(x+1)<=0

=>x+1>=0 và 3x-7<=0

=>-1<=x<=7/3

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

\(\Rightarrow-3x+2-5+3x=-3\)

\(\Rightarrow-3x+3x=-3+5-2\)

\(\Rightarrow0x=0\Rightarrow x\in Z\)

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

\(\Rightarrow3+x-3x+1=6-2x\)

\(\Rightarrow x-3x+2x=6-1-3\)

\(\Rightarrow0x=2\left(loại\right)\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{4}{3}\end{cases}}}\)

\(7x\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}}\)

khó hiểu quá

bạn ghi bằng số luôn đừng ghi phần

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

\(\Rightarrow-3x+2-5+3x=-3\)

\(\Rightarrow-3=-3\)

\(\forall x\in Z\)

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

\(\Rightarrow2=6\left(vl\right)\)

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

\(\Rightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)

\(7x\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)

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

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

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}\)

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

w

a) 3x(4x - 3) - 2x(5 - 6x) = 0

=> 6x² - 9x - 10x + 12x² = 0

=> 18x² - 19x = 0

=> x(18x - 19) = 0

=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)

b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0

=> 10x - 15 + 4x² - 8x + 6x - 4x² = 0

=> 8x - 15 = 0

=> 8x = 15

=> x = 15 : 8 = 15/8

c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)

=> 6x - 3x² + 2x² - 2x = 5x² + 15x

=> 4x - x² - 5x² - 15x = 0

=> -6x² - 11x = 0

=> -x(6x - 11) = 0

=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)

a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)

b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)

\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)

\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)

d) \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)

\(\Leftrightarrow14x^2+18=0\)

Mà \(14x^2+18>0\)nên pt vô nghiệm

a) Ta có: 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{19}{24}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{19}{24}\right\}\)

b) Ta có: \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)

\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

hay \(x=\frac{15}{8}\)

Vậy: \(x=\frac{15}{8}\)

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

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\)

\(\Leftrightarrow-x^2+4x-5x^2-15x=0\)

\(\Leftrightarrow-6x^2-11x=0\)

\(\Leftrightarrow6x^2+11x=0\)

\(\Leftrightarrow x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-11}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{-11}{6}\right\}\)

d) Ta có: \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)

\(\Leftrightarrow14x^2+18=0\)

\(\Leftrightarrow14x^2=-18\)

mà \(14x^2\ge0\forall x\)

nên \(x\in\varnothing\)

Vậy: \(x\in\varnothing\)

ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0

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

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

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0

=>(x-1)(x^3+4x^2+2x+3)=0

=>x-1=0

=>x=1

Ta có (a³ + b³) + c³ - 3abc = 0

<=> (a + b)³ - 3ab(a + b) + c³ - 3abc = 0

<=> (a + b + c)[(a + b)² - (a + b)c + c²] - 3ab(a + b + c) = 0

<=> (a + b + c)(a² + b² + c² - ab - ac - bc) = 0

<=> (a + b + c).(2a² + 2b² + 2c² - 2ab - 2ac - 2bc) = 0

<=> (a + b + c)[(a - b)² + (b - c)² + (c - a)²] = 0 (1)

Áp dụng (1) cho bài toán ta được 

(x - 1)³ + (2x - 3)³ + (3x - 5)³ - 3(x - 1)(2x - 3)(3x - 5) = 0

<=> (6x - 9)[(x - 2)² + (x - 2)² + (2x - 4)²] = 0

<=> \(\left[{}\begin{matrix}6x-9=0\\\left(x-2\right)^2+\left(x-2\right)^2+\left(2x-4\right)^2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\6.\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)

<=> (a + b)³ - 3ab(a + b) + c³ - 3abc = 0

<=> (a + b + c)[(a + b)² - (a + b)c + c²] - 3ab(a + b + c) = 0

<=> (a + b + c)(a² + b² + c² - ab - ac - bc) = 0

<=> (a + b + c).(2a² + 2b² + 2c² - 2ab - 2ac - 2bc) = 0

<=> (a + b + c)[(a - b)² + (b - c)² + (c - a)²] = 0 (1)

Áp dụng (1) cho bài toán ta được 

(x - 1)³ + (2x - 3)³ + (3x - 5)³ - 3(x - 1)(2x - 3)(3x - 5) = 0

<=> (6x - 9)[(x - 2)² + (x - 2)² + (2x - 4)²] = 0

<=> 

<=>