\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)

\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)

a. x=0

b.x=1,7

c.x=5,3

G.X=7

h.x=6

Mk làm vậy thôi

hok tốt

Professor minhmama

a;\(x\left(x+0\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+0=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=0\end{cases}}}\)

\(b,\left(x-1\right)\left(7-x\right)=0\)

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

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

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

\(d,\left(x+5\right)+\left(x-9\right)=13\)

\(\Rightarrow x+5+x-9=13\)

\(\Rightarrow2x=17\)

\(\Rightarrow x=\frac{17}{2}\)

\(e;\left(4+x\right)+\left(x-7\right)=x+2\)

\(\Rightarrow4+x+x-7=x+2\)

\(\Rightarrow x=5\)

\(f,\left(3x+5\right)-\left(2x-7\right)=4-x\)

\(\Rightarrow3x+5-2x+7=4-x\)

\(\Rightarrow2x=-8\Rightarrow x=-4\)

\(g,\left(x-1\right)^2=36\)

\(\Rightarrow\left(x-1\right)^2=\left(\pm6\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=6\\x-1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=5\end{cases}}}\)

\(h,\left(3-x\right)^3=-27\)

\(\Rightarrow\left(3-x\right)^3=\left(-3\right)^3\)

\(\Rightarrow3-x=-3\)

\(\Rightarrow x=6\)

g)(x-1)²=36

=>(x-1)²=6²

=>x-1=6

=>x=7

h) (3-x)³= -27

=>(3-x)³=3³

=>3-x=30

=>x=0

Mọi người giúp mình nhanh nhé ! Mình đang cần gấp

      c.   x^2-5x +6 = 0

<=> x^2 - 5x = -6

<=> - 4x = -6

<=> x= -6/-4

 Mình chỉ phân tích đa thức thành nhân tử thôi , phần còn lại bạn tự tính nha keo dài lắm

A)  2x²(x+3) - x(x+3) = 0  <=> x(x - 3)(2x-1)=0

B)  (2x+5)² - (x+2)²=0  <=>  (x+3)(3x+7)=0

C)  (x²-2x) - (3x-6)=0  <=> (x-2)(x-3)=0

D)  (2x-7)(2x-7-6x+18)=0   <=> (2x-7)(-4x+11)=0

E)  (x-2)(x+1) - (x-2)(x+2)=0   <=>  (x-2)*(-1)=0   <=> x-2=0

G)  (2x-3)(2x+2-5x)=0  <=> (2x-3)(-3x+2)=0

H)  (1-x)(5x+3+3x-7)=0     <=>  (1-x)(8x-4)=0

F)   (x+6)*3x=0

I)  (x-3)(4x-1-5x-2)=0  <=>  (x-3)(-x-3)=0

K)   (x+4)(5x+8)=0

H)  (x+3)(4x-9)=0

B> <2X+5>²-<X+2>²=0

<2X+5-X-2><2X+X+2>=0

<X+3><3X+7>=0

X+3=0 HOẶC 3X+7=0

X=-3 HOẶC X=-7/3

C>X²-5X+6=0

X²-4X+4-X+2=0

<X-2>²-<X-2>=0

<X-2.><X-3>=0

X-2=0 HOẶC X-3=0

X=2 HOẶC X=3

D> <2X-7><2X-7-6<X-3>>=0

<2X-7><-4X+11>=0

2X-7=0 HOẶC -4X+11=0

X=7/2 HOẶC X=11/4

E><X-2><X+1>=X²-4

<X-2><X+1>-<X²-4>=0

<X-2><X+1>-<X-2><X+2>=0

-X+2=0

X=2

CÒN NHIÊU TỰ LÀM ĐI MỆT WA

c. x^2-5x+6=0

<=> x^2-5x=-6

<=> -4x=-6

<=> x=-6/-4

vậy tập nghiệm của pt là s={-6/-4}

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

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

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

e, 3x(2-x) =15(x-2)

\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)

\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Vậy..

f, (x+5)(x+4)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)

Vậy..

g, x(x+4)=0

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

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

\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)

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

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

i, (x+1/5)(2x-3)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)

k, x²-4x=0

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

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

m, 4x²-1=0

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

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

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

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

n, x²-6x+9=0

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

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

<=> x=3

l, (3x-5)²-(x+4)²=0

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

\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy ..

o, 7x(x+2)-5(x+2)=0

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)

Vậy....

p, 3x(2x-5)-4x+10=0

\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)

\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)

Vậy...

q, (2-2x)-x²+1=0

\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)

\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)

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

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

Vậy ....

r, x(1-3x)=5(1-3x)

\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)

\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)

s, 2x-3/4+x+1/6=3

\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)

r, x(1-3x)=5(1-3x)

➜x(1-3x)-5(1-3x)=0

➜(x-5)(1-3x)=0

➜\(\left[{}\begin{matrix}x-5=0\\1-3x=0\end{matrix}\right.\text{➜}\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)

Mk lười lắm mai nha!!!~~~~~~~~~~~~

Làm dần:

a, 3x+6=0

➜3x=-6

➜x=2

b, 2x-10=0

➜2x=10

➜x=5

c, 3x-7=11

➜3x=11+7

➜3x=18

➜x=6

d, 3x-9=0

➜3x=9

➜x=3