a: Ta có: \(\left(x-5\right)\left(x+3\right)=x\left(x-3\right)\)

\(\Leftrightarrow x^2-2x-15-x^2+3x=0\)

\(\Leftrightarrow x=15\)

b: Ta có: \(\left(x+2\right)^2=\left(x-1\right)\left(x+2\right)\)

\(\Leftrightarrow x+2=0\)

hay x=-2

c: Ta có: \(\left(x-6\right)\left(x+6\right)=x^2\)

\(\Leftrightarrow x^2-36=x^2\)(vô lý)

a. (x - 5)(x + 3) = x(x - 3)

<=> x² + 3x - 5x - 15 = x² - 3x

<=> x² - x² + 3x - 5x + 3x - 15 = 0

<=> x = 15

b. (x + 2)² = (x - 1)(x + 2)

<=> x² + 4x + 4 = x² + 2x - x - 2

<=> x² - x² + 4x - 2x + x = -2 - 4

<=> 3x = -5

<=> \(x=\dfrac{-5}{3}\)

c. (x - 6)(x + 6) = x²

<=> x² - 36 - x² = 0

<=> x² - x² = 36

<=> 0 = 36 (vô lí)

Vậy nghiệm của PT là \(S=\varnothing\)

d. (2x - 3)² = 4x² - 8 

<=> 4x² - 12x + 9 - 4x² + 8 = 0

<=> 4x² - 4x² - 12x = -8 - 9

<=> -12x = -17

<=> \(x=\dfrac{17}{12}\)

a. (2x + 1)² - 4x² + 2x² - 2 = 0

<=> (2x + 1 - 2x)(2x + 1 + 2x) + 2(x² - 1) = 0

<=> (4x + 1) + 2x² - 2 = 0

<=> 4x + 1 + 2x² - 2 = 0

<=> 2x² + 4x - 2 + 1 = 0

<=> 2x² + 4x - 1 = 0

<=> 2x² + 4x = 1

<=> 2x(x + 2) = 1

Vì 1 chỉ có tích là 1 . 1 nên:

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

\(a,\Leftrightarrow4x^2+4x+1-4x^2+2x^2-2=0\\ \Leftrightarrow2x^2+4x-1=0\\ \Leftrightarrow2\left(x^2+2x+1\right)-3=0\\ \Leftrightarrow2\left(x+1\right)^2-3=0\\ \Leftrightarrow\left(x+1\right)^2=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{3}{2}}\\x+1=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{6}}{2}\\x=\dfrac{-2+\sqrt{6}}{2}\end{matrix}\right.\)

\(b,\left(x-2\right)\left(x+2\right)-\left(x+3\right)^2-2x-5=0\\ \Leftrightarrow x^2-4-x^2-6x-9-2x-5=0\\ \Leftrightarrow-8x=18\\ \Leftrightarrow x=-\dfrac{9}{4}\)

Bài 2:

a: \(\Leftrightarrow x-1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

=>\(x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)

b: \(\Leftrightarrow x+3\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

=>\(x\in\left\{-2;-4;0;-6;2;-8;12;-18\right\}\)

c: \(\Leftrightarrow x+3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(x\in\left\{-2;-4;-1;-5;0;-6;1;-7;3;-9;9;-15\right\}\)

d: =>x+1+15 chia hết cho x+1

=>\(x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

=>\(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

1.

=x²+2x+1+x²+x-6-4x

=2x²-x-5

2.

=x²+5x-24-x²+12x

=17x-24

Bài 3: 

Gọi số nhóm là x

Theo đề, ta có: \(x\in\left\{1;2;3;4;6;9;12;18;36\right\}\)

mà 2<x<6

nên \(x\in\left\{3;4\right\}\)

Vậy: Có 2 cách chia nhóm

a) 25 - x = 12 + 6  =18

x=25-18=7 Vậy x=7

b) 7 + 2 x ( x -3 ) = 11   

2.(x-3)=11-7=4

x-3=4:2=2

x=3+2=5                          

c) 102 : ( 2.x + 13) : 4) = 6   

(2.x+13):4=102:6=17

2.x+13=17.4=68

2.x=68-13=55

x=27,5 Vậy x=27,5

Bài 3: 

Gọi số nhóm là x

Theo đề, ta có: 

mà 2<x<6

nên 

Vậy: Có 2 cách chia nhóm

còn bài 1 chắc bn làm đc nha tick mk nha

a: x*3/4=1/5

=>x=1/5:3/4=1/5*4/3=4/15

b: x*3/7=2/5

=>x=2/5:3/7=2/5*7/3=14/15

c: 1/3+2/9=2/12x

=>1/6x=3/9+2/9=5/9

=>x=5/9*6=30/9=10/3

d: 4/15*x-2/3=1/5

=>4/15*x=2/3+1/5=10/15+3/15=13/15

=>4x=13

=>x=13/4

e: x:1/7=2/3

=>x=2/3*1/7=2/21

f: 1/9:x=7/3

=>x=1/9:7/3=1/9*3/7=3/63=1/21

j: 1/4+5/12=8/3:x

=>8/3:x=3/12+5/12=8/12=2/3

=>x=4

h: =>7/4:x=1/5+1/2=7/10

=>x=7/4:7/10=10/4=5/2

b) x(x-4) - 2x+8 = 0
    x(x-4) - 2(x-4) = 0
    (x-2) (x-4) = 0
TH1: x-2=0              TH2: x-4=0
            x=2                          x=4
Vậy x\(\in\){2;4}

\(b,\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ c,\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\\ d,\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ e,\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ f,\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-12\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow x=2\)

b) x(x-4)-2x+8=0

x(x-4)-2(x-4)=0

(x-4)(x-2)=0

th1: x-4=0

x=4

th2: x-2=0

x=2

Vậy x thuộc tập hợp 4;-2

ở oooo

e: \(\Leftrightarrow3x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

a) \(\left(2x-1\right)^3-4x^2\left(2x-3\right)=5\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)

\(\Leftrightarrow6x-1=5\Leftrightarrow6x=6\Leftrightarrow x=1\)

b) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x+1\right)^2=-10\)

\(\Leftrightarrow\left(x+1-x+1\right)\left[\left(x^2+2x+1+x^2-2x+1+\left(x^2-1\right)\right)\right]-6\left(x^2+2x+1\right)=-10\)

\(\Leftrightarrow2\left(3x^2+1\right)-6x^2-12x-6=-10\)

\(\Leftrightarrow6x^2+2-6x^2-12x-6=-10\)

\(\Leftrightarrow-12x-4=-10\Leftrightarrow12x=-6\Leftrightarrow x=\dfrac{1}{2}\)

c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x=0\)

\(\Leftrightarrow x\left(3x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)

\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)