=>3x^2+1/6x-3x^2-x-2=3

=>-5/6x=5

=>x=-6

  4 x 24 x 5 ^ 2 - ( 3 ^ 3 x 18 + 3 ^ 3 x 12)

=96 x 25 - (27 x 18 + 27 x 12)

=96 x 25 - [ (27 x (18 + 12) ]

=96 x 25 - 27 x 30

=2400 - 810

=1590

4 x 24 x 25 - ( 9 x 18 x 9 x 12 )

= 100 x 24 - ( 81 x 18 x 12 )

= 2400 - 17496

= -15096

4.24.5² - (3³.18 + 3³.12) = 96.25 - 3³.(18 + 12) = 2400 - 27.30 = 2400 - 810 = 1590

Hướng làm:

Thấy cả tử mẫu cộng lại đều bằng 2021 → Cộng thêm 1 rồi quy đồng với mỗi phân thức

\(\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1=\dfrac{x+4}{2017}+1+\dfrac{x}{2021}+1\\ \Leftrightarrow\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}-\dfrac{x+2021}{2021}=0\\ \Leftrightarrow\left(x+2021\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2021}\right)=0\\ \Leftrightarrow x+2021=0\Leftrightarrow x=-2021\)

\(< =>\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1=\dfrac{x+4}{2017}+1+\dfrac{x}{2021}+1\)

\(< =>\dfrac{x+2+2019}{2019}+\dfrac{x+3+2018}{2018}=\dfrac{x+4+2017}{2017}+\dfrac{x+2021}{2021}\)

\(< =>\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}-\dfrac{x+2021}{2021}=0\)

\(< =>\left(x+2021\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2021}=\right)=0\)

\(< =>x+2021=0< =>x=-2021\)

Vậy....

P=5x⁶-x⁴-10x⁵-2x³+x³+5x⁴-x²-5x³+x

P=5x⁶-10x⁵+4x⁴-6x³-x²+x

Vậy hệ số của x⁴ trong đa thức là 4
 

\(a,\dfrac{1}{2}x=3+2\)

\(\dfrac{1}{2}x=5\)

\(x=5\div\dfrac{1}{2}\)

\(x=10\)

\(b,\dfrac{1}{4}x^2-\sqrt{36}=10\)

\(\dfrac{1}{4}x^2-6=10\)

\(\dfrac{1}{4}x^2=10+6\)

\(\dfrac{1}{4}x^2=16\)

\(x^2=16\div\dfrac{1}{4}\)

\(x^2=64\)

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

\(\Rightarrow x=8\)

S = 1x2 + 2x3 + 3x4 + ... + 38x39 + 39x40

3S = 1x2x3 + 2x3x3 + 3x4x3 + ... + 38x39x3 + 39x40x3

3S = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + ... + 38x39x(40-37) + 39x40x(41-38)

3S = 1x2x3 + 2x3x4-1x2x3 + 3x4x5-2x3x4 + ... + 38x39x40-37x38x39 + 39x40x41-38x39x40

S = 39x40x41 : 3

S = 21320

\(3S=1.2.3+2.3.3+...+39.40.3\)

\(3S=1.2.\left(3-0\right)+2.3.\left(4-1\right)+...+39.40.\left(41-38\right)\)

\(3S=0.1.2-1.2.3+1.2.3-2.3.4+...+38.39.40-39.40.41\)

\(3S=30.40.41\)

\(S=10.40.41\)

<span class="label label-info"><i class="fa fa-exclamation-triangle fa-2"></i> Quản lý</span>

Ta có: S = 1 x 2 + 2 x 3 + 3 x 4 +...+ 38 x 39 + 39 x 40

=> 3S = 1 x 2 x (3 - 0) + 2 x 3 x (4 - 1) + ..... + 39 x 40 x (41 - 38)

=> 3S = 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 + ..... + 39 x 40 x 41

=> 3S = 39 x 40 x 41

=> S = 39 x 40 x 41 : 3

=> S = 21320

\(\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x}{2021}\)

\(\Leftrightarrow\frac{x+2}{2019}+1+\frac{x+3}{2018}+1=\frac{x+4}{2017}+1+\frac{x}{2021}+1\)

\(\Leftrightarrow\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2021}\)

\(\Leftrightarrow x+2021=0\)

\(\Leftrightarrow x=-2021\)

h) x/y = 9/10 ⇒  y/10 = x/9

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

y/10 = x/9 = (y - x)/(10 - 9) = 120/1 = 120

*) x/9 = 120 ⇒ x = 120.9 = 1080

*) y/10 = 120 ⇒ y = 120.10 = 1200

Vậy x = 1080; y = 1200

k) x/y = 3/4

⇒ x/3 = y/4

⇒ 5y/20 = 3x/9

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

5y/20 = 3x/9 = (5y - 3x)/(20 - 9) = 33/11 = 3

*) 3x/9 = 3 ⇒ x = 3.9:3 = 9

*) 5y/20 = 3 ⇒ y = 3.20:5 = 12

Vậy x = 9; y = 12

\(x^4-x^3-x^2+1\)

\(\text{ Phân tích thành nhân tử}\)

\(\left(x-1\right)\left(x^3-x-1\right)\)

\(-x-y^2+x^2-y\)

\(\text{ Phân tích thành nhân tử}\)

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

\(x^2-y^2-x-y\)

\(\text{ Phân tích thành nhân tử}\)

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

\(x^2-y^2+4-4x\)

\(\text{ Phân tích thành nhân tử}\)

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