Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm1\\x\ne-\frac{1}{2}\end{cases}}\)
a) \(A=\left(\frac{1}{x-1}+\frac{x}{x^3-1}\cdot\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(\Leftrightarrow A=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(\Leftrightarrow A=\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{2x+1}\)
\(\Leftrightarrow A=\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow A=\frac{x+1}{x-1}\)
b) Thay \(x=\frac{1}{2}\)vào A, ta được :
\(A=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3\)
Bài 1 :
a) \(ĐKXĐ:x\ne1\)
\(A=\left(\frac{3}{x^2-1}+\frac{1}{x+1}\right):\frac{1}{x+1}\)
\(\Leftrightarrow A=\frac{3+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\left(x+1\right)\)
\(\Leftrightarrow A=\frac{x+2}{x-1}\)
b) Thay x = \(\frac{2}{5}\)vào A ta được :
\(A=\frac{\frac{2}{5}+2}{\frac{2}{5}-1}=\frac{\frac{12}{5}}{-\frac{3}{5}}=-4\)
c) Để \(A=\frac{5}{4}\)
\(\Leftrightarrow\frac{x+2}{x-1}=\frac{5}{4}\)
\(\Leftrightarrow4x+8=5x-5\)
\(\Leftrightarrow x=13\)
d) Để \(A>\frac{1}{2}\)
\(\Leftrightarrow\frac{x+2}{x-1}>\frac{1}{2}\)
\(\Leftrightarrow\frac{x+2}{x-1}-\frac{1}{2}>0\)
\(\Leftrightarrow2x+4-x+1>0\)
\(\Leftrightarrow x+5>0\)
\(\Leftrightarrow x>-5\)
Bài 2 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne0\end{cases}}\)
\(A=\frac{x^2}{x^2+x}-\frac{1-x}{x+1}\)
\(A=\frac{x}{x+1}+\frac{x-1}{x+1}\)
\(\Leftrightarrow A=\frac{2x-1}{x+1}\)
b) Để \(A=1\)
\(\Leftrightarrow\frac{2x-1}{x+1}=1\)
\(\Leftrightarrow2x-1=x+1\)
\(\Leftrightarrow x=2\)
b) Để \(A< 2\)
\(\Leftrightarrow\frac{2x-1}{x+1}< 2\)
\(\Leftrightarrow\frac{2x-1}{x+1}-2< 0\)
\(\Leftrightarrow2x-1-2x-1< 0\)
\(\Leftrightarrow-2< 0\)(luôn đúng)
Vậy A < 2 <=> mọi x
cho biểu thức A = (x+y) (x-1)+x(2-x-y)+1
a, rút gọn biểu thức
b, tính giá trị biểu thức khi x=1 y=1/2
Bài 1:
a: \(A=\dfrac{x+1+x}{x+1}:\dfrac{3x^2+x^2-1}{x^2-1}\)
\(=\dfrac{2x+1}{x+1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{x-1}{2x-1}\)
b: Thay x=1/3 vào A, ta được:
\(A=\left(\dfrac{1}{3}-1\right):\left(\dfrac{2}{3}-1\right)=\dfrac{-2}{3}:\dfrac{-1}{3}=2\)
\(ĐKXĐ:x\ne\pm1\)
a) \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{4x^2}{1-x^2}\right):\frac{2x^2-2}{x^2-2x+1}\)
\(\Leftrightarrow A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x^2}{x^2-1}\right):\frac{2\left(x^2-1\right)}{\left(x-1\right)^2}\)
\(\Leftrightarrow A=\frac{\left(x+1\right)^2-\left(x-1\right)^2-4x^2}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)
\(\Leftrightarrow A=\frac{x^2+2x+1-x^2+2x-1}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)
\(\Leftrightarrow A=\frac{4x-4x^2}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)
\(\Leftrightarrow A=\frac{-4x\left(x-1\right)^3}{2\left(x-1\right)^2\left(x+1\right)^2}\)
\(\Leftrightarrow A=\frac{-2x\left(x-1\right)}{\left(x+1\right)^2}\)
b) Thay x = -3 vào A, ta được :
\(A=\frac{\left(-2\right)\left(-3\right)\left(-3-1\right)}{\left(-3+1\right)^2}\)
\(\Leftrightarrow A=\frac{6.\left(-4\right)}{2^2}\)
\(\Leftrightarrow A=-6\)
c) Để A > -1
\(\Leftrightarrow-2x\left(x-1\right)>-\left(x+1\right)^2\)
\(\Leftrightarrow2x\left(x-1\right)< \left(x+1\right)^2\)
\(\Leftrightarrow2x^2-2x< x^2+2x+1\)
\(\Leftrightarrow x^2-4x-1< 0\)
\(\Leftrightarrow\left(x-2\right)^2-5< 0\)
\(\Leftrightarrow\left(x-2\right)^2< 5\)
Đoạn này bạn tự tìm giá trị x thỏa mãn là xong (Chú ý ĐKXĐ)
1,
\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)
\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)
2.
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
3.
Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)
4.
\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)
\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)
5.
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)
https://books.google.com.vn/books?id=tQlmDwAAQBAJ&pg=PA198&lpg=PA198&dq=cho+A+%3D+x%5E2%2B2x%5E3-1+%2B+x%2B1/x%5E2%2Bx%2B1+-+1/x-1+r%C3%BAt+g%E1%BB%8Dn+bi%E1%BB%83u+th%E1%BB%A9c+t%C3%ADnh+a+khi+x+%3D1/2&source=bl&ots=ALIjuS9TGW&sig=ACfU3U2G9ueMTMh3ldwfDCxD-PBbGQ3l2Q&hl=vi&sa=X&ved=2ahUKEwjbjZfGisfmAhXpyIsBHS9VB6oQ6AEwAHoECAgQAQ#v=onepage&q=cho%20A%20%3D%20x%5E2%2B2x%5E3-1%20%2B%20x%2B1%2Fx%5E2%2Bx%2B1%20-%201%2Fx-1%20r%C3%BAt%20g%E1%BB%8Dn%20bi%E1%BB%83u%20th%E1%BB%A9c%20t%C3%ADnh%20a%20khi%20x%20%3D1%2F2&f=false
why sai hả bn ???????????????????????????????????????