Bài 1:

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

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

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

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

m) \(=\left(x^4+4x^2+4\right)-9=\left(x^2+2\right)^2-9=\left(x^2-1\right)\left(x^2+5\right)=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

Bài 2:

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

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

f) giống câu a

g) \(=x^2-2xy=x\left(x-2y\right)\)

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

k) \(=\left(x+1\right)^3-27z^3=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

l) \(=\left(2x+1\right)^2-9y^2=\left(2x+1-3y\right)\left(2x+1+3y\right)\)

a. (3x - 1)² - 16

= (3x - 1)² - 4²

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

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

= 3(3x - 5)(x + 1)

b. (5x - 4)² - 49x²

= (5x - 4)² - (7x)²

= (5x - 4 - 7x)(5x - 4 + 7x)

= \(\left(-2x-4\right)\left(2x-4\right)\)

= 2(-x - 2) . 2(x - 2)

= 4(-x - 2)(x - 2)

c. (2x + 5)² - (x - 9)²

= (2x + 5 - x + 9)(2x + 5 + x - 9)

= (x + 14)(3x - 4)

2. 

a)(3x-1)²-16 =  (3x-1)²-4²

                        =(3x-1-4).(3x-1+4)= (3x-5).(3x+3)

b)(5x-4)²-49x²  =(5x-4)²-(7x)²

                        =  (5x-4-7x).(5x-4+7x) = (-2x-4).(12x-4)

c)(2x+5)²-(x-9)²  =(2x+5-x+9).(2x+5+x-9)  =(x+14).(3x-4)

d)(3x+1)²-4.(x-2)² =(3x+1)²-[2.(x-2)]²  = [3x+1-2.(x-2)].[3x+1+2.(x-2)]

                                                            = (3x+1-2x+4).(3x+1+2x-4)

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

e)9.(2x+3)²-4.(x+1)²  =[3.(2x+3)]²-[2.(x+1)]² 

 =[3.(2x+3)-2.(x+1)].[3.(2x+3)+2.(x+1)]

=(6x+9-2x-2).(6x+9+2x+2)

=(4x+7).(8x+11)

a: Ta có: \(\left(4x-6\right)^2-16\left(x-3\right)^2\)

\(=16x^2-48x+36-16x^2+96x-144\)

\(=48x-108\)

\(=48\cdot\left(-7\right)-108\)

\(=-444\)

b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

\(=x^3-1+x^3+27\)

\(=2x^3+26\)

\(=-2+26=24\)

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)

c: \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)

d: \(3x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(3x+5\right)\)

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

f: \(-3x-6xy+9xz=-3x\left(1+2y-3z\right)\)

\(a,=\left(6x+1-6x+1\right)^2=4\\ b,=3x^2-6x-5x+5x^2-8x^2-24=-11x-24\\ c,=14x^2+x-3-5x^2-18x+8-9x^2+17x=5\\ d,=6x^2+43x-40-6x^2-7x+3-36x+27=-10\)

a) \(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2=\left(6x+1-6x+1\right)^2=2^2=4\)

b) \(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)

c) \(\left(7x-3\right)\left(2x+1\right)-\left(5x-2\right)\left(x+4\right)-9x^2+17x=\left(7x-3\right).2x+\left(7x-3\right)-\left[\left(5x-2\right).x+4\left(5x-2\right)\right]-9x^2+17x=14x^2-6x+7x-3-\left(5x^2-2x+20x-8\right)-9x^2+17x=5x^2+18x-3-\left(5x^2+18x-8\right)=5x^2+18x-3-5x^2-18x+8=5\)

d) \(\left(6x-5\right)\left(x+8\right)-\left(3x-1\right)\left(2x+3\right)-9\left(4x-3\right)=\left(6x-5\right).x+8\left(6x-5\right)-\left[\left(3x-1\right).2x+3\left(3x-1\right)\right]-36x+27=6x^2-5x+48x-40-\left(6x^2-2x+9x-3\right)-36x+27=6x^2+7x-13-\left(6x^2+7x-3\right)=6x^2+7x-13-6x^2-7x+3=-10\)

a: Ta có: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

c: Ta có: \(4-x=2\left(x-4\right)^2\)

\(\Leftrightarrow2\left(x-4\right)^2+x-4=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)

a) ta có AD vg góc DC

            BC vg góc DC

             EG vg góc DC

=> AD//BC//EG

Xét tg ABCD có AD//BC=> ABCD là hình thang

Xét hình thang ABCD có EG //BC//AD

                                        AE= EB (gt)

=> G là td DC (EG là dg tb hthang ABCD)

=> DG=GC=5cm=> X =5 cm

b) Thêm dk là HI//OJ//NK//M...

Xét tg HINK có HI//NK(gt) => HINK là hthang 

Xét Hthang HINK có OH =ON (gt)

                                  IJ=IK(gt)

=>OJ là dg tb hình thang HINK

=> OJ= (HI+NK):2=(8+16):2=12 cm=> x=12 cm

Xét tg OJM... có OJ//M.. => OJM.. là hình thang

(cmt tương tự như ở trên)

=> M..= NK.2-OJ=16.2-12=20 cm

Bạn xem lại chỗ .... là gì nhé!

a, văng vẳng, quang quác quác, tè te te

Tác dụng: Cho thấy sự huyên náo của các loài vật trong rừng núi, nó làm bức tranh thiên nhiên trở nên sinh động và đẹp hơn

b, đùng đùng, vù vù, lộp độp

Tác dụng: Cho thấy những dấu hiệu trước cơn mưa, cơn mưa diễn ra nhanh chóng và mạnh