Thực hiện các phép tính sau: x 3 - 8 5 x + 20 . x 2 + 4 x x 2 + 2 x + 4
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Những câu hỏi liên quan
7 tháng 1 2022
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
20 tháng 9 2023
Tham khảo:
a) \((8{x^6} - 4{x^5} + 12{x^4} - 20{x^3}):4{x^3}\)
\( = (8{x^6}:4{x^3}) - (4{x^5}:4{x^3}) + (12{x^4}:4{x^3}) - (20{x^3}:4{x^3})\)
\( = 2{x^2} - {x^2} + 3x - 5\)
b)
Vậy \((2{x^2} - 5x + 3):(2x - 3)= x - 1\)
J
7
HQ
Hà Quang Minh
Giáo viên
27 tháng 11 2023
a)
160 : (4 × 8) = 160 : 32 = 5
160 : 4 : 8 = 40 : 8 = 5
Vậy 160 : (4 × 8) = 160 : 4 : 8.
96 : (3 × 8) = 96 : 24 = 4
96 : 3 : 8 = 32 : 8 = 4
Vậy 96 : (3 × 8) = 96 : 3 : 8.
105 : (5 × 7) = 105 : 35 = 3
105 : 5 : 7 = 21 : 7 = 3
Vậy 105 : (5 × 7) = 105 : 5 : 7.
Nhận xét: a : (b × c) = a : b : c
b)
270 : (9 × 6) = 270 : 9 : 6 = 30 : 6 = 5
420 : (7 × 3) = 420 : 7 : 3 = 60 : 3 = 20
144 : (2 × 8) = 144 : 2 : 8 = 72 : 8 = 9
2 tháng 8 2023
\(\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\\ =\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\\ =\dfrac{6.2}{2x\left(x+4\right)}+\dfrac{3x}{2x\left(x+4\right)}\\ =\dfrac{12+3x}{2x\left(x+4\right)}\\ =\dfrac{3\left(4+x\right)}{2x\left(x+4\right)}\\ =\dfrac{3}{2x}\)
________
\(\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{x^2-4}\\ \left(\text{đ}k\text{x}\text{đ}:x\ne\pm2\right)\\ =\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x+x+2+x^2-4x+4+x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2x^2-8}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2\left(x^2-4\right)}{x^2-4}\\ =2\)
2 tháng 8 2023
a: \(=\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\)
\(=\dfrac{12+3x}{2x\left(x+4\right)}=\dfrac{3\left(x+4\right)}{2x\left(x+4\right)}=\dfrac{3}{2x}\)
b: \(=\dfrac{\left(x+1\right)\left(x+2\right)+\left(x-2\right)^2+x-14}{x^2-4}\)
\(=\dfrac{x^2+3x+2+x^2-4x+4+x-14}{x^2-4}=\dfrac{2x^2-8}{x^2-4}=2\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9
Bài 1:
a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)
\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{3}\)
\(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)
\(\dfrac{1}{4}\) \(x\)= \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)
\(\dfrac{1}{4}x=\dfrac{11}{24}\)
\(x=\dfrac{11}{24}:\dfrac{1}{4}\)
\(x=\dfrac{11}{24}\times4\)
\(x=\dfrac{11}{6}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9
b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)
\(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)
\(x\) = \(\dfrac{9}{10}\)
x 3 - 8 5 x + 20 . x 2 + 4 x x 2 + 2 x + 4