c) \(5x^2+3y+15x+xy=5x\left(x+3\right)+y\left(x+3\right)=\left(x+3\right)\left(5x+y\right)\)

d) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)

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

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

c: \(5x^2+15x+3y+xy\)

\(=5x\left(x+3\right)+y\left(x+3\right)\)

\(=\left(x+3\right)\left(5x+y\right)\)

d: \(x^2+6x+9-y^2\)

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

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

e: \(x^2+2x+1-y^2\)

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

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

f: \(x^2-2xy+y^2-9\)

\(=\left(x-y\right)^2-9\)

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

\(2\left(x+5\right)-x^2-5x\)

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

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

\(y^2-6y+9-z^2\)

\(=\left(y-3\right)^2-z^2\)

\(=\left(y-3-z\right)\left(y-3+z\right)\)

Bài 1:

c. \(a^3-a^2x-ay+xy\)

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

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

Bài 2:

a. \(x^2-6x=0\)

\(x\left(x-6\right)=0\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

b) \(x^3-2x2+x=0\)

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

\(\Rightarrow\)\(\hept{\begin{cases}x=0\\x^2=3\end{cases}}\)

\(\Rightarrow\)\(\hept{\begin{cases}x=0\\x=\sqrt{3}\\x=-\sqrt{3}\end{cases}}\)

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

\(=\left(x+3\right)^3\)

d.

\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)

e.

\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

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

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2

\(a,3x-15xy=3x\left(1-5y\right)\\ ---\\ 8x^2+6x-4=2\left(4x^2+3x-2\right)\\ ---\\ 5x^2+25xy+10y^2=5\left(x^2+5xy+2y^2\right)\\ ---\\ 9x^2y^2+6x^2y-\dfrac{1}{2}xy^2=\dfrac{1}{2}xy\left(18xy+12x-y\right)\)

a) (x - 2)(x - 3).                        b) 3(x - 2)(x + 5).

c) (x - 2)(3x + 1).                     d) (x-2y)(x - 5y).

e) (x + l)(x + 2)(x - 3).             g) (x-1)(x + 3)( x 2  + 3).

h) (x + y - 3)(x - y + 1).

a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)

b: \(=25-\left(x-2y\right)^2\)

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

\(b,\left(x+2\right)^2-25\)

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

\(=\left(x-3\right)\left(x+7\right)\)

\(c,36\left(x-y\right)^2\)

\(=36\left(x^2-2xy+y^2\right)\)

\(=36x^2-72xy+36y^2\)

\(d,x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)

\(=x^2+2.x.\dfrac{1}{4}+\dfrac{1}{4}^2\)

\(=\left(x+\dfrac{1}{4}\right)^2\)

\(e,2x^4y^3-3x^2y^4+5x^3y^4\)

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

Các câu còn lại làm tương tự, chú ý sd HĐT

Nếu ol thì tham khảo nah nguoiemtinhthong.

1.1

2x2+5x−1=7x3−1−−−−−√2x2+5x−1=7x3−1

⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)−−−−−−−−−−−−−−−√(1)⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)(1)

Đặt a=x−1−−−−−√;b=x2+x+1−−−−−−−−√;a≥0;b>0a=x−1;b=x2+x+1;a≥0;b>0

pt (1) trở thành 3a2+2b2−7ab=03a2+2b2−7ab=0

a=2ba=2b v a=13ba=13b

Các bạn tự giải quyết tiếp nhé.

1.2

TXĐ D=[1;+∞)D=[1;+∞)

đặt a=x−1−−−−−√4;b=x+1−−−−−√4;a,b≥0a=x−14;b=x+14;a,b≥0

pt (2) trở thành 3a2+2b2−5ab=03a2+2b2−5ab=0

⇔a=b⇔a=b v a=23ba=23b

...

1.3

D=[3;+∞)D=[3;+∞)

Đặt a=x2+4x−5−−−−−−−−−√;b=x−3−−−−−√;a,b≥0a=x2+4x−5;b=x−3;a,b≥0

pt (3) trở thành 3a+b=11a2−19b2−−−−−−−−−√3a+b=11a2−19b2

⇔2a2−6ab−20b2=0⇔2a2−6ab−20b2=0

⇒a=5b⇒a=5b
...

1.4

ĐK

⇔2x2−2x+2=3(x−2)x(x+1)−−−−−−−−−−−−√2x2−2x+2=3(x−2)x(x+1)

⇔(x2−2x)+2(x+1)=3(x2−2x)(x+1)−−−−−−−−−−−−−√2(x2−2x)+2(x+1)=3(x2−2x)(x+1)

Đặt x2−2x−−−−−−√=ax2−2x=a; x+1−−−−−√=bx+1=b (a;b\geq0)

⇔2a2+2b2=3ab

1.5

Đặt 4x2−4x−10=t4x2−4x−10=t (t \geq 0)

⇔t=t+4x2−2x−−−−−−−−−−√t=t+4x2−2x

⇔t2−t−4x2+2x=0t2−t−4x2+2x=0

Δ=1−4(2x−4x2)=(4x−1)2Δ=1−4(2x−4x2)=(4x−1)2

⇒t=1−2xt=1−2x hoặc t=2xt=2x

1.1

2.2+5.-1=7.3-1-----v2.2+5.-1=7.3-1

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

3a+b=11a2-19b2

tóm tắt

\(âP\left(x\right)=13x^3+4x^2-11x-2\)

\(b.Q\left(x\right)=x^3+9x-5\)
\(c.A\left(x\right)=14x^3-x^2+10x+14\)
\(d.B\left(x\right)=2x^2+x+3\)