rút gọn phân thức
x^3-7x-6/x^2*(x-3)^2+4x*(3-x)^2+4*(x-3)^2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
JW
15 tháng 6 2016
Nhìn ko hiểu dâu "|" là dấu ngoặc hay dấu giá trị tuyệt đối
Bạn ghi rõ đề bài ra nha 
15 tháng 6 2016
cái này || ko phải là dấu tuyệt đối đâu mà là phép nhân, chia giua 2 phân số
PN
1 tháng 8 2020
\(F=-3\left(x-8\right)\left(2x+1\right)-\left(x+5\right)\left(2-3x\right)-4x\left(x-6\right)\)
\(=-3\left(-3-8\right)\left(-6+1\right)-\left(5-3\right)\left(2+9\right)+12\left(-9\right)\)
\(=-3\left(-11\right)\left(-5\right)-\left(-2\right)11-12.9\)
\(=-165+22-108=22-273=-251\)
\(G=\left(5x-4\right)\left(5-2x\right)-7x\left(x^2-4x+3\right)+\left(x^2-4x\right)\left(7x-2\right)\)
\(=\left(5-4\right)\left(5-2\right)-7\left(1-4+3\right)+\left(1-4\right)\left(7-2\right)\)
\(=3-7.0+5.\left(-3\right)=3-15=-12\)
\(H=\left(-3x+5\right)\left(x-6\right)-\left(x-1\right)\left(x^2-2x+3\right)+\left(x+2\right)\left(x^2-3\right)\)
\(=\left(3+5\right)\left(-1-6\right)-\left(-1-1\right)\left(1+2+3\right)+\left(-1+2\right)\left(1-3\right)\)
\(=8\left(-7\right)-\left(-2\right)6+1\left(-2\right)=-56+12-2=-46\)
PN
1 tháng 8 2020
\(L=5x\left(x-1\right)\left(2x+3\right)-10x\left(x^2-4x+5\right)-\left(x-1\right)\left(x-4\right)\)
\(=-\frac{5}{3}\left(-\frac{4}{3}\right)\left(-\frac{2}{3}+3\right)+\frac{10}{3}\left(\frac{1}{9}+\frac{4}{3}+5\right)-\left(-\frac{4}{3}\right)\left(-\frac{1}{3}-4\right)\)
\(=\frac{20}{9}\left(\frac{7}{3}\right)+\frac{10}{3}\left(\frac{13}{9}+5\right)+\frac{4}{3}\left(-\frac{13}{3}\right)\)
\(=\frac{140}{27}+\frac{10}{3}.\frac{58}{9}-\frac{52}{9}\)
\(=\frac{140}{27}+\frac{580}{27}-\frac{156}{27}=\frac{140+580-156}{27}=\frac{720-156}{27}=\frac{564}{27}\)
\(M=-7x\left(x-5\right)-\left(x-1\right)\left(x^2-x-2\right)+x^2\left(x-3\right)-5x\left(x-8\right)\)
\(=\frac{-7}{2}\left(\frac{1}{2}-5\right)+\frac{\left(\frac{1}{4}-\frac{1}{2}-2\right)}{2}+\frac{1}{4}\left(\frac{1}{2}-3\right)-\frac{5}{2}\left(\frac{1}{2}-8\right)\)
\(=\frac{7}{2}.\frac{9}{2}-\frac{9}{8}-\frac{1}{4}.\frac{5}{2}+\frac{5}{2}.\frac{15}{2}\)
\(=\frac{63}{4}-\frac{9}{8}-\frac{5}{8}+\frac{75}{4}=\frac{138}{4}-\frac{7}{4}=\frac{131}{4}\)
BC
1
14 tháng 12 2020
ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)
Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)
14 tháng 4 2023
Bài 1:
a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)
\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)
\(\Rightarrow16x-5=x-2\)
\(\Rightarrow16x-x=5-2\)
\(\Rightarrow15x=3\)
\(\Rightarrow x=\dfrac{15}{3}=5\)
b) \(12x^2-4x\left(3x+5\right)=10x-17\)
\(\Rightarrow12x^2-12x^2-20x=10x-17\)
\(\Rightarrow-20x=10x-17\)
\(\Rightarrow-20x-10x=-17\)
\(\Rightarrow-30x=-17\)
\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)
c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)
\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)
\(\Rightarrow-8x=12\)
\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)
Bài 2:
a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)
\(=x^2-7x+5x-35-7x^2+21x\)
\(=-6x^2+19x-35\)
b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)
\(=x^3-x^2-2x-x^2+x-5x-5\)
\(=x^3-2x^2-6x-5\)
c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)
\(=x^2-7x-5x+35-x^2-3x+4x-12\)
\(=11x+23\)
d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)
\(=x^2-2x-x+2-x^2+2x+5x+10\)
\(=4x+12\)
M
1
23 tháng 12 2021
\(a,\dfrac{x^3-x}{3x+3}=\dfrac{x\left(x^2-1\right)}{3\left(x+1\right)}=\dfrac{x\left(x-1\right)\left(x+1\right)}{3\left(x+1\right)}=\dfrac{x\left(x-1\right)}{3}\\ b,\dfrac{x^2+3xy}{x^2-9y^2}=\dfrac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}=\dfrac{x}{x-3y}\\ c,\dfrac{x^2+4x+4}{3x+6}=\dfrac{\left(x+2\right)^2}{3\left(x+2\right)}=\dfrac{x+2}{3}\)
24 tháng 11 2018
\(a)\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(3x-1)(x-3)^2}(ĐK:x\ne3,x\ne\frac{1}{3})\)
\(=\frac{2x+5}{3x-1}\)
Còn bài b bạn tự làm nhé
24 tháng 11 2018
Điều kiện: \(x\ne\left\{-1;-2;-5\right\}\)
\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x^2-4\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)
\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]}\)
\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+5\right)}=\frac{x-2}{x+5}\)
Điều kiện: \(x\ne\left\{3;\frac{1}{3}\right\}\)
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)
\(=\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)
\(=\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)
\(=\frac{2x^2-x-15}{3x^2-10x+3}=\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)
\(=\frac{\left(2x+5\right)\left(x-3\right)}{\left(3x-1\right)\left(x-3\right)}=\frac{2x+5}{3x-1}\)
30 tháng 6 2019
\(3x^4-4x^3+2x\left(x^3-2x^2+7x\right)\)
\(=3x^4-4x^3+2x^4-4x^3+14x^2\)
\(=5x^4-8x^3+14x^2\)
30 tháng 6 2019
3x4 - 4x3 + 2x(x3 - 2x2 + 7x )
= 3x4 - 4x3 + 2x4 _ 4x3 + 14x2
= 5x4 - 8x3 + 14x2