a/\(\frac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{18}{5}\)\(\frac{2^4\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}=\frac{2\cdot11}{5\cdot7}=\frac{22}{35}\)

b/\(\frac{121\cdot75\cdot130\cdot169}{39\cdot60\cdot11\cdot198}=\frac{11^2\cdot5^3\cdot13^3\cdot2\cdot3}{2^3\cdot3^4\cdot5\cdot11^2\cdot13}=\frac{5^2\cdot13^2}{2^2\cdot3^3}=\frac{4225}{108}\)

c/\(\frac{1998\cdot1990+3978}{1992\cdot1991-3984}=\frac{2^2\cdot3^3\cdot37\cdot5\cdot199+2\cdot3^2\cdot13\cdot17}{2^3\cdot3\cdot83\cdot11\cdot181-2^4\cdot3\cdot83}=\frac{2\cdot3^2\cdot11\cdot20101}{2^3\cdot3^3\cdot13\cdot17\cdot83}=\frac{11\cdot20101}{2^2\cdot3\cdot13\cdot17\cdot83}\)

Đề lỗi quá. Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.

1. a) \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=14x^5+21x^7\)

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

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

2: \(x^2-8x+7=0\)

=>\(x^2-x-7x+7=0\)

=>\(x\left(x-1\right)-7\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x-7\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

1:

a: \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=21x^7+14x^5\)

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

\(=x^2+1\)

1)

\(a,7x^2\cdot(2x^3+3x^5)\\=7x^2\cdot2x^3+7x^2\cdot3x^5\\=14x^5+21x^7\\---\\b,(x^3-x^2+x-1):(x-1)(dkxd:x\ne 1)\\=[x^2(x-1)+(x-1)]:(x-1)\\=(x-1)(x^2+1):(x-1)\\=x^2+1\)

2)

\(x^2-8x+7=0\\\Leftrightarrow x^2-x-7x+7=0\\\Leftrightarrow x(x-1)-7(x-1)=0\\\Leftrightarrow (x-1)(x-7)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

\(\text{#}Toru\)

a, Thay x=2 vào A, ta được:

\(A\left(2\right)=3.2^3+5-6.2+5.2^2=37\)

Vậy A= 37 khi x=2.

b,

 \(A\left(x\right)+B\left(x\right)=\left(3x^3+5-6x+5x^2\right)+\left(4x^2+6x-2x^7-9\right)\\ =-2x^7+3x^3+9x^2-4\)

ê ku dạng nào z

tui cg l 6

2x4 ,4 là mũ hay số vậy

a) 3 x 5² + 15 x 2² - 26 : 2

= 3 x 25 + 15 x 4 - 13

= 75 + 60 - 13

= 135 - 13

= 122

b) 5³ x 2 - 100 : 4 + 2³ x 3

= 125 x 2 - 25 + 8 x 3

= 250 - 25 + 24

= 225 + 24

= 249

c) 3² : 5 + 2³ x 10 - 81 : 3

9 : 5 + 8 x 10 - 27

= 1,8 + 80 - 27

= 81,8 - 27

= 54,8

a) \(\left(2x^3-x^2+5x\right):x\)

\(=\dfrac{2x^3-x^2+5x}{x}\)

\(=\dfrac{x\left(2x^2-x+5\right)}{x}\)

\(=2x^2-x+5\)

b) \(\left(3x^4-2x^3+x^2\right):\left(-2x\right)\)

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

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

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

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

c) \(\left(-2x^5+3x^2-4x^3\right):2x^2\)

\(=\dfrac{-2x^5+3x^2-4x^3}{2x^2}\)

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

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

d) \(\left(x^3-2x^2y+3xy^2\right):\left(-\dfrac{1}{2}x\right)\)

\(=\dfrac{x^3-2x^2y+3xy^2}{-\dfrac{1}{2}x}\)

\(=\dfrac{\dfrac{1}{2}x\left(2x^2-4xy+6y^2\right)}{-\dfrac{1}{2}x}\)

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

\(=-2x^2+4xy-6y^2\)

e) \(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:5\left(x-y\right)^2\)

\(=\dfrac{3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2}{5\left(x-y\right)^2}\)

\(=\dfrac{5\left(x-y\right)^2\left[\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\right]}{5\left(x-y\right)^2}\)

\(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)

f) \(\left(3x^5y^2+4x^3y^3-5x^2y^4\right):2x^2y^2\)

\(=\dfrac{3x^5y^2+4x^3y^3-5x^2y^4}{2x^2y^2}\)

\(=\dfrac{2x^2y^2\left(\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\right)}{2x^2y^2}\)

\(=\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\)