câu 3:Tính
b)
với x khác -1 ;x khác 0; x khác 1
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
1.
$A=(x-2)^2+6x+5=x^2-4x+4+6x+5=x^2+2x+9$
2.
$B=\frac{15x^2y^3}{5x^2y^2}-\frac{10x^3y^2}{5x^2y^2}+\frac{5x^2y^2}{5x^2y^2}$
$=3y-2x+1$
Bài 3:
$f(x)=x+4x^2-5x+3=4x^2-4x+3=4x(x-3)+8(x-3)+27$
$=(x-3)(4x+8)+27=g(x)(4x+8)+27$
Vậy $f(x):g(x)$ có thương là $4x+8$ và dư là $27$
tính
b)
c)
tìm x
b)
c)
d) (với x là số nguyên dương)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
b: \(=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}=\dfrac{1}{2}\)
c: \(=\dfrac{1}{9}\left(\dfrac{2}{9}+\dfrac{7}{9}\right)-\dfrac{8}{9}=\dfrac{1}{9}-\dfrac{8}{9}=-\dfrac{7}{9}\)
Bài 2:
b: \(\Leftrightarrow x+\dfrac{3}{10}=\dfrac{17}{30}\)
hay x=4/15
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(4^5\cdot10\cdot5^6+25^5\cdot2^8\right):\left(2^8\cdot5^4+5^7\cdot2^5\right)\)
\(B=\dfrac{4^5\cdot10\cdot5^6+25^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B=\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B=\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B=\dfrac{2^8\cdot5^7\cdot\left(2^3\cdot1+5^3\cdot1\right)}{2^5\cdot5^4\cdot\left(2^3\cdot1+5^3\cdot1\right)}\)
\(B=\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\)
\(B=\dfrac{2^3\cdot5^3}{1\cdot1}\)
\(B=\left(2\cdot5\right)^3\)
\(B=10^3\)
\(B=1000\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: 3/91*96
\(B=\dfrac{3}{1\cdot6}+\dfrac{3}{6\cdot11}+...+\dfrac{3}{91\cdot96}\)
=3/5(5/1*6+5/6*11+...+5/91*96)
=3/5(1-1/6+1/6-1/11+...+1/91-1/96)
=3/5*95/96=57/96=19/32
![](https://rs.olm.vn/images/avt/0.png?1311)
7 / 4 - 5/8 = 14 / 8 - 5 / 8 = 9/8
9 / 1 . 3 / 11 = 27 / 11
1/2 . 5 / 2 = 5 / 4
\(\dfrac{7}{4}-\dfrac{5}{8}=\dfrac{14}{8}-\dfrac{5}{8}=\dfrac{9}{8}\)
\(9\times\dfrac{3}{11}=\dfrac{9\times3}{11}=\dfrac{27}{11}\)
\(\dfrac{1}{2}:\dfrac{2}{5}=\dfrac{1}{2}\times\dfrac{5}{2}=\dfrac{5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) 2 - 8/9 = 10/9
c) 5/8 x 3 = 15/8
d) 5 : 6/21= 105/6
đ) 3/5 x 4/5 x 6/7= 72/175
\(\dfrac{1}{x^2-x}+\dfrac{x-3}{x^2-1}\left(x\ne\pm1;x\ne0\right)\)
\(=\dfrac{1}{x\left(x-1\right)}+\dfrac{x-3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1}{x\left(x-1\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1+x^2-3x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1}{\left(x-1\right)\left[x\left(x+1\right)\right]}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+x\right)}\)
\(=\dfrac{x-1}{x^2+x}\)