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: \(4x^2-12x+9=\left(2x-3\right)^2\)

b: \(4x^2+4x+1=\left(2x+1\right)^2\)

c: \(36x^2+12x+1=\left(6x+1\right)^2\)

d: \(9x^2-24xy+16y^2=\left(3x-4y\right)^2\)

a)\(=\left(2x\right)^2-2.2x.3+3^2\)

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

b)\(=\left(2x\right)^2+2.2x.1+1^2\)

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

c)\(=\left(6x\right)^2+2.6x.1+1^2\)

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

d)\(=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2\)

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

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

Bài 1.

a, PTPƯ: kẽm + axit clohidric → kẽm clorua + hidro

b, Theo ĐLBTKL ta có:

 \(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)

c, Ta có: \(m_{HCl}=m_{ZnCl_2}+m_{H_2}-m_{Zn}=27,2+0,4-13=14,6\left(g\right)\)

Bài 2:

a, PTPƯ: metan + oxi → cacbon dioxit + hơi nước

b, Theo ĐLBTKL ta có: 

 \(m_{CH_4}+m_{O_2}=m_{CO_2}+m_{H_2O}\)

c, Ta có: \(m_{O_2}=m_{CO_2}+m_{H_2O}-m_{CH_4}=132+108-48=192\left(g\right)\)  

 PT chữ: Metan + Oxi \(\xrightarrow[]{t^o}\) Cacbon đioxit + Nước

Bảo toàn khối lượng: \(m_{O_2}=m_{CO_2}+m_{H_2O}-m_{CH_4}=192\left(g\right)\)

Bài 1:

PT chữ: Kẽm + Axit clohidric \(--->\) Kẽm clorua + Hidro

Bảo toàn khối lượng: \(m_{HCl}=m_{ZnCl_2}+m_{H_2}-m_{Zn}=14,6\left(g\right)\)

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

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: \(4-x=2\left(x-4\right)^2\)

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

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

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

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) 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!