Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1:
a: \(A=\dfrac{x+1-x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x^2+1-2x}{2}\)
\(=\dfrac{2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}=\dfrac{x-1}{x+1}\)
b: Để A=x/6 thì \(\dfrac{x-1}{x+1}=\dfrac{x}{6}\)
\(\Leftrightarrow x^2+x-6x+6=0\)
=>x=3 hoặc x=2
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
B3;a,ĐKXĐ:\(x\ne\pm4\)
A=\(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right)\dfrac{x^2+8x+16}{32}=\left(\dfrac{4x+16}{x^2-16}-\dfrac{4x-16}{x^2-16}\right)\dfrac{x^2+2.4x+4^2}{32}=\left(\dfrac{4x+16-4x+16}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\left(\dfrac{32}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\dfrac{32\left(x+4\right)^2}{32.\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\\ \\ \\ \\ \\ \\ b,Tacó\dfrac{x+4}{x-4}=\dfrac{1}{3}\Leftrightarrow3x+12=x-4\Leftrightarrow x=-8\left(TM\right)c,TAcó\dfrac{x+4}{x-4}=3\Leftrightarrow x+4=3x-12\Leftrightarrow x=8\left(TM\right)\)
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
a: \(M=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)^2+4x^2}{2\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-4x^2+4x+4x^2}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
b: Thay x=1/2 vào M, ta được:
\(M=\left(\dfrac{1}{2}+1\right):\left(2\cdot\dfrac{1}{2}\right)=\dfrac{3}{2}\)
\(M=\left(\dfrac{1}{x-1}-\dfrac{1}{1-x^3}.\dfrac{x^2+x+1}{x+1}\right):\dfrac{1}{x^2-1}\)
\(M=\left(\dfrac{1}{x-1}+\dfrac{1}{\left(x-1\right)\left(x^2+x+1\right)}.\dfrac{x^2+x+1}{x+1}\right).\dfrac{\left(x+1\right)\left(x-1\right)}{1}\)
\(M=\left(\dfrac{1}{x-1}+\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right).\left(x+1\right)\left(x-1\right)\)
\(M=\dfrac{x+2}{\left(x-1\right)\left(x+1\right)}.\left(x+1\right)\left(x-1\right)\)
\(M=x+2\)
Với \(x=\dfrac{1}{2}\)
ta có: \(M=\dfrac{1}{2}+2=\dfrac{5}{2}\)
Để M có giá trị dương \(\Rightarrow M>0\)
\(\Leftrightarrow x+2>0\)
\(\Rightarrow x>-2\)
a,\(P=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x^2-x}\right)\)\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\dfrac{x}{x\left(x-1\right)}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2}{x-1}\)
\(b,\) Để \(P=-\dfrac{1}{2}\) hay \(\dfrac{x^2}{x-1}=-\dfrac{1}{2}\)
\(\Leftrightarrow2x^2=-\left(x-1\right)\)
\(\Leftrightarrow2x^2=-x+1\)
\(\Leftrightarrow2x^2+x-1=0\)
\(\Leftrightarrow2x^2+2x-x-1=0\)
\(\Leftrightarrow2x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
\(a,A=\left(\dfrac{x-2}{x+2}-\dfrac{x+2}{x-2}\right).\left(\dfrac{2}{x}-1\right)\)
\(=\left(\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}\right).\left(\dfrac{2}{x}-\dfrac{x}{x}\right)\)
\(=\left(\dfrac{x^2-4x+4-x^2-4x-4}{\left(x+2\right)\left(x-2\right)}\right).\left(\dfrac{2-x}{x}\right)\)
\(=\dfrac{-8x}{\left(x+2\right)\left(x-2\right)}.\dfrac{x-2}{-x}\)
\(=\dfrac{8}{x+2}\)
b, Thay \(x=-\dfrac{2}{5}\) vào biểu thức A ,có :
\(\dfrac{8}{-\dfrac{2}{5}+2}=\dfrac{8}{\dfrac{8}{5}}=5\)
Vậy tại x = -2/5 giá trị của A là 5
c, Để \(A=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{8}{x+2}=\dfrac{1}{2}\)
\(\Leftrightarrow x+2=16\)
\(\Leftrightarrow x=14\)
Vậy x = 14 thì A = 1/2
Mạn phép không chép đề , tớ làm luôn
a) M = \(\left[\dfrac{x-1-2\left(x+1\right)+x}{x^2-1}\right].\dfrac{x+1}{1}\) ( x # 1 ; x # -1)
M = \(\dfrac{3}{1-x}\)
b) Với : x = \(\dfrac{-1}{2}\) ( thỏa mãn ĐKXĐ ), ta có :
M = \(\dfrac{3}{1+\dfrac{1}{2}}=\dfrac{3}{\dfrac{3}{2}}=\dfrac{6}{3}=2\)
KL...
bạn ơi hình như sai rồi ý