1,x^2-(x+1)(x-1)=0
x^2-x^2+1+0
1=0(vô lý)
2,5x^3+3x^2+3x+1=4x^2
x^3+3x^2+3x+1=0
(x+1)=0
x=-1
3,x^3+x^2=0
x^2(x+1)=0
x=0 or x=-1
4,2x^3-12x^2+18x=0
x^3-6x^2+9x=0
x(x^2-6x+9)=0
x(x-3)^2=0
x=0 or x=3
5,5x^2-4(x^2-2x+1)+20=0
5x^2-4x^2+8x-4+20=0
x^2+8x+16=0
(x+4)^2=0
x=-4
6,5x(x-3)+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
5x-7=0 or x-3=0
x=7/5 or x=3
7,2x^3-50x=0
2x(x^2-25)=0
2x(x-5)(x+5)=0
x=0 or x=5 or x=-5
8,(4x-1)^2-9(x+3)^2=0
(4x-1)^2-3^2*(x+3)^2=0
(4x-1)^2-(3x+9)^2=0
(4x-1-3x-9)(4x-1+3x+9)=0
(x-10)(7x+8)=0
x=10 or x=-8/7
9,3(x-2)^2-x+2=0
3*(x-2)*(x-2)-(x-2)=0
(3x-6)(x-2)-(x-2)=0
(x-2)(3x-6-1)=0
(x-2)(3x-7)=0
x=2 or x=7/3
10,9x^2+6x-8=0
9x^2+12x-6x-8=0
3x(3x-2)+4(3x-2)=0
(3x+4)(3x-2)=0
3x+4=0 or 3x-2=0
x=-4/3 or x=2/3

Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@

B1: Phân tích thành nhân tử:

a) \(6x^2+9x=3x\left(2x+3\right)\)

b) \(4x^2+8x=4x\left(x+2\right)\)

c) \(5x^2+10x=5x\left(x+2\right)\)

d) \(2x^2-8x=2x\left(x-4\right)\)

e) \(5x-15y=5\left(x-3y\right)\)

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

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

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

\(=\left(x-1-2y\right)\left(x-1+2y\right)\)

h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)

i) \(9x^2-18x+9=\left(3x-3\right)^2\)

k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)

l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)

m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)

\(=-\left(2x-y\right)^2\)

n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)

\(=\left(x-31\right)\left(x+1\right)\)

o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)

\(=\left(2+x\right)\left(8+x\right)\)

p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)

\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)

\(=\left(5x-5\right)\left(9x-3\right)\)

Bài 1 :

a ) \(6x^2+9x=3x\left(x+3\right)\)

b ) \(4x^2+8x=4x\left(x+2\right)\)

c ) \(5x^2+10x=5x\left(x+2\right)\)

d ) \(2x^2-8x=2x\left(x-4\right)\)

e ) \(5x-15y=5\left(x-3y\right)\)

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

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

h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)

i ) \(9x^2-18x+9=\left(3x-3\right)^2\)

k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)

l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)

m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)

n ) \(\left(x-15\right)^2=x^2-30x+15^2\)

o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)

p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)

Bài 2 :

a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)

b ) \(x^2-2x+xy-2y=x\left(x-2\right)+y\left(x-2\right)=\left(x-2\right)\left(x+y\right)\)

c ) \(2x+x^2-2y-2xy=......................\)

d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)

e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)

f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b,\)\(x^2+5x+4=0\)

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

\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)

\(c,\)\(9x-6x^2-3=0\)

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

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow2x^2-2x-x+1=0\)

\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)

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

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

\(d,\)\(2x^2+5x+2=0\)

\(\Leftrightarrow2x^2+4x+x+2=0\)

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

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

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

Lần sau đăng thì chia thành nhiều câu hỏi nhé

\(16^2-9.\left(x+1\right)^2=0\)

\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)

\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)

\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)

\(\left[13-3x\right].\left[19+3x\right]=0\)

\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)

KL:..............................

Nhiều câu hỏi mà bn ??

câu a, b, c dễ mà. Bạn áp dụng 7 hằng đẳng thúc là làm đc thoii!!

vd: a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)

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

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

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

\(\Rightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\) (bạn phá ngoặc ra rồi tính là ra bước này)

\(\Leftrightarrow3x+2=0\) hoặc \(x+1=0\) hoặc \(2x-1=0\) ( đến đây bạn chia làm 3 trường hợp r tự tính nhé)

Chúc bạn học tốt!!

d/

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

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

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

e/

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

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

\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)

Có làm theo hàng đẳng thức k bạn?

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

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

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

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

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

c )\(x^2-x-6=0\)

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

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

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

d ) \(x^2+x-12=0\)

\(\Leftrightarrow x^2-4x+3x-12\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

e ) \(15\left(x+9\right)\left(x-3\right)\left(x+21\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=3\\x=-21\end{matrix}\right.\)

g ) \(\left(x^2+1\right)\left(x^2+4x+4\right)=0\)

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

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

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=1\\x=-3\end{matrix}\right.\)

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

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

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

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

a) \(4x^2-8x=0\)

\(\Rightarrow4x\left(x-2\right)=0\)

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

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

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

Vậy \(x_1=0;x_2=2\)

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

\(\Rightarrow-3x^2-14x+5=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)

Vậy \(x_1=-5;x_2=\dfrac{1}{3}\)

\(a,4x^2-8x=0\Rightarrow4x\left(x-8\right)=0\Rightarrow\left[{}\begin{matrix}4x=0\\x-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)\(b,\left(x+5\right)-3x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(1-3x\right)=0\Rightarrow\left[{}\begin{matrix}x+5=0\\1-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\3x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)