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

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

d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

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

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

f) \(x^3-6x^2-x+30=0\)

\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)

\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)

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

\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)

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

\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)

a)2x.(3x+5)-x.(6x-1)=33

=>\(6x^2+10x-6x^2+x=33\)

=>11x=33

=>x=3

b)x(3x-1)+12x-4=0

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

=>(x-4)(3x-1)=0

=>x-4=0 hoặc 3x-1=0

+)x-4=0 +)3x-1=0

=>x=4 =>x=\(\frac{1}{3}\)

c)5x(2x+1)-12x-6=0

=>10x\(^2\)+5x-12x-6=0

=>10x\(^2\)-7x-6=0

=>(10x\(^2\)+5x)-(12x+6)=0

=>5x(2x+1)-6(2x+1)=0

=>(5x-6)(2x+1)=0

=>\(\left[{}\begin{matrix}5x-6=0\\2x+1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\frac{6}{5}\\x=\frac{-1}{2}\end{matrix}\right.\)

\(d,x\left(x-3\right)-7x+21=0\)

\(\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)

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

\(a,2x\left(x-7\right)+5x-35=0\)

 \(\Leftrightarrow2x\left(x-7\right)+5\left(x-7\right)=0\)

 \(\Leftrightarrow\left(x-7\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{5}{2}\end{cases}}}\)

\(c,4x^2+12x+9=0\)

\(\Leftrightarrow4x^2+6x+6x+9=0\)

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

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

\(\Leftrightarrow2x+3=0\)

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

a) 2x(x-7)+5x-35=0
<=> 2x(x-7)+5(x-7)=0
<=>(2x+5)(x-7)=0
<=> (2x+5)=0 <=> x=-5/2
hoặc <=> x-7=0 <=> x=7

Bài 1:

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

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

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

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

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

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

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

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

tự tìm x nha

a) Ta có: \(3x^2+2x-1=0\)

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

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

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

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

Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)

b) Ta có: \(x^2-5x+6=0\)

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

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

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

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

Vậy: S={2;3}

c) Ta có: \(x^2-3x+2=0\)

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

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

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

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

Vậy: S={1;2}

d) Ta có: \(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)

mà \(2\ne0\)

nên \(x^2-3x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)

e) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

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

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

Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)

cho vào máy tính là ra hết

a) x+5x²=0

=>x(1+5x)=0

=>x=0 hoặc 1+5x=0 =>x=\(\dfrac{1}{5}\)

b)x+1=(x+1)²

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

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

=>-x(x+1)=0

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

các câu còn lại làm tương tự

a, \(x^2-12x-2x+24=0\Leftrightarrow x^2-14x+24=0\Leftrightarrow\left(x-12\right)\left(x-2\right)=0\)

TH1 : x = 12 ; TH2 : x = 2 

b, \(x^2-5x-24=0\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

TH1 : x = 8 ; TH2 : x = -3 

c, \(4x^2-12x-7=0\Leftrightarrow\left(2x+1\right)\left(2x-7\right)=0\)

TH1 : x = -1/2 ; TH2 : x = 7/2

d, \(x^3+6x^2+12x+8=0\Leftrightarrow\left(x+2\right)^3=0\Leftrightarrow x=-2\)

Tương tự HĐT thôi :)

a) x² - 12x - 2x + 24 = 0

<=> x( x - 12 ) - 2( x - 12 ) = 0

<=> ( x - 12 )( x - 2 ) = 0

<=> \(\orbr{\begin{cases}x-12=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=12\\x=2\end{cases}}\)

b) x² - 5x - 24 = 0

<=> x² + 3x - 8x - 24 = 0

<=> x( x + 3 ) - 8( x + 3 ) = 0

<=> ( x + 3 )( x - 8 ) = 0

<=> \(\orbr{\begin{cases}x+3=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)

c) 4x² - 12x - 7 = 0

<=> 4x² + 2x - 14x - 7 = 0

<=> 2x( 2x + 1 ) - 7( 2x + 1 ) = 0

<=> ( 2x + 1 )( 2x - 7 ) = 0

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

d) x³ + 6x² + 12x + 8 = 0

<=> ( x + 2 )³ = 0

<=> x + 2 = 0

<=> x = -2

e) ( x + 2 )² - x² + 4 = 0

<=> x² + 4x + 4 - x² + 4 = 0

<=> 4x + 8 = 0

<=> 4x = -8

<=> x = -2

f) 2( x + 5 ) = x² + 5x

<=> x² + 5x - 2x - 10 = 0

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

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

<=> \(\orbr{\begin{cases}x+5=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

m) 16( 2x - 3 )² - 25( x - 5 )² = 0

<=> 4²( 2x - 3 )² - 5²( x - 5 )² = 0

<=> [ 4( 2x - 3 ) ]² - [ 5( x - 5 ) ]² = 0

<=> ( 8x - 12 )² - ( 5x - 25 )² = 0

<=> [ 8x - 12 - ( 5x - 25 ) ][ 8x - 12 + ( 5x - 25 ) ] = 0

<=> ( 8x - 12 - 5x + 25 )( 8x - 12 + 5x - 25 ) = 0

<=> ( 3x + 13 )( 13x - 37 ) = 0

<=> \(\orbr{\begin{cases}3x+13=0\\13x-37=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{3}\\x=\frac{37}{13}\end{cases}}\)

n) x² - 6x + 4 = 0

<=> ( x² - 6x + 9 ) - 5 = 0

<=> ( x - 3 )² - ( √5 )² = 0

<=> ( x - 3 - √5 )( x - 3 + √5 ) = 0

<=> \(\orbr{\begin{cases}x-3-\sqrt{5}=0\\x-3+\sqrt{5}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{5}\\x=3-\sqrt{5}\end{cases}}\)