a: x=1

=>\(f\left(x\right)=1^2+2\cdot1^2+3\cdot1^3+\cdots+2018\cdot1^{2018}+2019\cdot1^{2019}\)

=1+2+3+...+2019

\(=\frac{2019\left(2019+1\right)}{2}=2019\cdot\frac{2020}{2}=2019\cdot1010\)

=2039190

b: Khi x=1 thì ta có:

\(g\left(1\right)=2\cdot1+4\cdot1^2+6\cdot1^3+\cdots+200\cdot1^{100}+202\cdot1^{101}\)

=2+4+...+202

=2(1+2+...+101)

\(=2\cdot101\cdot\frac{102}{2}=101\cdot102=10302\)

Bạn có thể viết rõ đề ra được không?

Ví dụ 2 (60s): Số dư của phép chia (6x3 − 4x2 + 3x + 7): (2x2 + 1) là:  A. 4x2 + 7                B. 9                  C. 5                 D. 4x + 7

Ta có:

- 4 x 2 ( 6 x 3   +   5 x 2   –   3 x   +   1 )     =   ( - 4 x 2 ) . 6 x 3   +   ( - 4 x 2 ) . 5 x 2   +   ( - 4 x 2 ) . ( - 3 x )   +   ( - 4 x 2 ) . 1     =   - 24 x 5   –   20 x 4   +   12 x 3   –   4 x 2

Đáp án cần chọn là: C

a: \(=\dfrac{2x\left(3x^2+2\right)+3x^2+2}{3x^2+2}=2x+1\)

b: 

Sửa đề: 6x^4-4x^3+3x-2/3x-2

\(=\dfrac{6x^4-4x^3+3x-2}{3x-2}\)

\(=\dfrac{2x^3\left(3x-2\right)+3x-2}{3x-2}=2x^3+1\)

where is đề

a) 20x-5y=5(4x-y)

b) 5x(x-1)-3x(x-1)=(5x-3x)(x-1)=2x(x-1)

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

d) 6x³-9x²=3x²(2x-3)

e) 4x²y - 8xy² + 10x²y²=2xy(2x-4y+5xy)

g) 20x²y-12x³=4x²(5y-3x)

h) 8x⁴+ 12x²y⁴- 16x³y⁴ = 4x²(2x²+3y⁴-4xy⁴)

k) 4xy² + 8xyz=4xy(y+2z)

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

m) 4x²-1=(2x-1)(2x+1)

o) 9-(x-y)²=(3-x+y)(3+x-y)

p) x³+27=(x+3)(x²+3x+9)

n) (x-y)²-4=(x-y-2)(x-y+2)

r) x⁴ + 2x² + 1=(x⁴+x²)+(x²+1)=x²(x²+1)+(x²+1)=(x²+1)²

s) 4x²  - 12xy + 9y²  = (2x-3y)²

t) x² - x - y² - y =(x²-y²)-(x+y)=(x-y)(x+y)-(x+y)=(x+y)(x-y-1)

v) x³ - x + y³ - y = (x³+y³)-(x+y)=(x+y)(x²-xy+y²)-(x+y)=(x+y)(x²-xy+y²-1)

Câu 4:

D=(x+1)(x+3)(x+5)(x+7)+15

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+105+15\)

\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+120\)

\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(=\left(x+2\right)\left(x+6\right)x^2+8x+10\)

Câu 2:

b: \(4x^2-12x+9\)

\(=\left(2x\right)^2-2\cdot2x\cdot3+3^2\)

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

Câu 1:

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

b: \(\left(3x+y\right)^3=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot y+3\cdot3x\cdot y^2+y^3\)

\(=27x^3+27x^2y+9xy^2+y^3\)

\(\dfrac{3}{4}\times\dfrac{5}{7}=\dfrac{15}{28};\dfrac{5}{8}\times\dfrac{4}{15}=\dfrac{20}{120}=\dfrac{1}{6};\dfrac{7}{12}\times\dfrac{4}{9}=\dfrac{28}{108}=\dfrac{7}{27};\)

\(\dfrac{1}{6}\times\dfrac{3}{5}=\dfrac{3}{30}=\dfrac{1}{10};\dfrac{12}{21}\times\dfrac{23}{8}=\dfrac{276}{168}=\dfrac{23}{14};\dfrac{13}{4}\times\dfrac{5}{39}=\dfrac{65}{156}=\dfrac{5}{12}\)

\(\dfrac{7}{42}\times\dfrac{13}{21}=\dfrac{91}{882}=\dfrac{13}{126};\dfrac{3}{16}\times\dfrac{4}{15}=\dfrac{12}{240}=\dfrac{1}{20}\)