Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ Đkxđ: \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Vậy phân thức được xác định khi \(\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
b/ \(A=\left[1+\frac{1}{x}+\frac{2}{x+1}\left(1+\frac{1}{x}\right)\right]:\frac{x^3+27}{2x}\)
\(=\left[1+\frac{1}{x}+\frac{2}{x+1}+\frac{2}{\left(x+1\right)x}\right]:\frac{\left(x+3\right)\left(x^2-3x+9\right)}{2x}\)
\(=\left[\frac{x\left(x+1\right)+\left(x+1\right)+2x+2}{\left(x+1\right)x}\right].\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\frac{x^2+4x+3}{\left(x+1\right)x}.\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}=\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)x}.\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\frac{2}{x^2-3x+9}\)
d, Ta có : \(\frac{x^3+4x^2-x-4}{x+4}\)
\(=\frac{x^2\left(x+4\right)-\left(x+4\right)}{x+4}=\frac{\left(x^2-1\right)\left(x+4\right)}{x+4}=x^2-1\)
- Thay \(x=-2\frac{1}{3}\) vào biểu thức trên ta được :
\(\left(-2\frac{1}{3}\right)^2-1=\frac{58}{9}\)
Vậy biểu thức có giá trị là \(\frac{58}{9}\) tại \(x=-2\frac{1}{3}\)
ĐKXĐ: \(x\ne\left\{-\frac{1}{2};\frac{1}{2};-1\right\}\)
\(B=\left(\frac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\frac{4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\left(\frac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)
\(=\frac{\left(2x^2+3x+1\right)}{\left(2x+1\right)\left(2x-1\right)}.\frac{\left(2x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{\left(x+1\right)\left(2x+1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)\left(x+1\right)\left(x^2-x+1\right)}=\frac{1}{x^2-x+1}\)
ĐKXĐ: \(\left\{{}\begin{matrix}a-1\ne0\\a^2-1\ne0\\a-a^3\ne0\\a+a^3\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ne1\\a\ne\left\{-1;1\right\}\\a\left(1-a^2\right)\ne0\\a\left(1+a^2\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ne1\\a\ne\left\{1;-1\right\}\\a\ne\left\{-1;0;1\right\}\\a\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ne0\\a\ne-1\\a\ne1\end{matrix}\right.\)
\(M=\frac{a^2}{a-1}+\left(\frac{a}{a^2-1}+\frac{1}{a-a^3}\right):\frac{1-a}{a+a^3}\)
\(=\frac{a^2}{a-1}+\left(\frac{a}{\left(a-1\right)\left(a+1\right)}+\frac{1}{a\left(1-a^2\right)}\right):\frac{1-a}{a\left(1+a^2\right)}\)
\(=\frac{a^2}{a-1}+\left(\frac{a^2}{a\left(a-1\right)\left(a+1\right)}-\frac{1}{a\left(a+1\right)\left(a-1\right)}\right):\frac{1-a}{a\left(1+a^2\right)}\)
\(=\frac{a^2}{a-1}+\frac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)\left(a+1\right)}.\frac{a\left(1+a^2\right)}{1-a}\)
\(=\frac{a^2}{a-1}-\frac{1+a^2}{a-1}=\frac{a^2-1-a^2}{a-1}=-\frac{1}{a-1}\)
b/ Thay $a=\frac{1}{2}$ vào M ta được \(M=-\frac{1}{-\frac{1}{2}-1}=-\frac{1}{-\frac{3}{2}}=\frac{1}{\frac{3}{2}}=\frac{2}{3}\)
a,ĐKXĐ:\(x\ne2,x\ne-3\)
\(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
\(=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x-4}{x-2}\)
c,Để A = - 3/4
thì: \(\frac{x-4}{x-2}=-\frac{3}{4}\)
\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)
\(4x-16=-3x+6\)
\(4x+3x=6+16\)
\(7x=22\)
\(x=\frac{22}{7}\)
d,\(A=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=\frac{x-2}{x-2}-\frac{2}{x-2}=1-\frac{2}{x-2}\)
Để A nguyên thì: \(x-2\inƯ\left(2\right)\)
Ta có: \(Ư\left(2\right)=\left\{\pm1,\pm2\right\}\)
Xét từng TH:
_ x - 2 = -1 => x = 1
_ x - 2 = 1 => x = 3
_ x - 2 = -2 => x = 0
_ x- 2 = 2 => x= 4
Vậy: \(x\in\left\{0,1,3,4\right\}\)
=.= hok tốt!!
\(5,\)\(\frac{1}{5}x\left(10x-15\right)-2x\left(x-5\right)+7x\)
\(=2x^2-3x+-2x^2+10x-7x\)
\(=0\)
\(\Rightarrow\)Giá trị biểu thức không phụ thuộc vào biến x
\(6,\)\(F=5\left(x^2-3x\right)-x\left(3-5x\right)+18x+3\)
\(=5x^2-15x-3x-5x^2+18x+3\)
\(=3\)
Vậy giá trị biểu thức không phụ thuộc vào biến x
( À có một số chỗ mình phải sửa đề mới đúng đó. Cậu coi lại giùm mình nha )
a) A có nghĩa <=> \(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}x-1\ne0\\\left(1-x\right)\left(x+1\right)\ne0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\end{matrix}\right.\)
b) Ta có:
A = \(\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\)
A = \(\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}\)
A = \(\frac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
A = \(\frac{x^2+x-x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
A = \(\frac{x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x-1\right)}\)
c) A = -1/2
<=> \(\frac{1}{2\left(x-1\right)}=-\frac{1}{2}\)
<=> 2(x - 1) = -2
<=> x - 1 = -1
<=> x = 0 (tmđk)
Vậy x = 0