Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
M=\(\dfrac{1}{x-1}-\dfrac{4}{x+1}+\dfrac{7x-1}{x^2-1}\)
a, Rút gọn M
b, Tính giá trị của M tại x = -3
\(a,M=\dfrac{x+1-4x+4+7x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\\ b,x=-3\Rightarrow M=\dfrac{4}{-3-1}=-1\)
b.\(M=\dfrac{1}{-3-1}-\dfrac{4}{-3+1}+\dfrac{7\left(-3\right)-1}{\left(-3\right)^2-1}\)
\(M=\dfrac{-1}{4}-\left(-2\right)+\dfrac{11}{5}=3,95\)
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
\(1a.A=\dfrac{x}{x+1}-\dfrac{3-3x}{x^2-x+1}+\dfrac{x+4}{x^3+1}=\dfrac{x\left(x^2-x+1\right)-3\left(1-x^2\right)+x+4}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^3+2x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^3+x^2+x^2+x+x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+x+1}{x^2-x+1}\left(x\ne-1\right)\)
\(b.A=\dfrac{x^2+x+1}{x^2-x+1}=\dfrac{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+1-\dfrac{1}{4}}{x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+1-\dfrac{1}{4}}=\dfrac{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}>0\left(x\ne-1\right)\)
\(2a.M=\left(\dfrac{x}{x^2-4}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right]:\dfrac{x^2-4+10-x^2}{x+2}=\dfrac{6}{\left(2-x\right)\left(x+2\right)}.\dfrac{x+2}{6}=\dfrac{1}{2-x}\left(x\ne\pm2\right)\)
\(b.Để:M\in Z\Leftrightarrow\dfrac{1}{2-x}\in Z\Leftrightarrow2-x\in\left\{\pm1\right\}\)
\(\oplus2-x=1\Leftrightarrow x=1\left(TM\right)\)
\(\oplus2-x=-1\Leftrightarrow x=3\left(TM\right)\)
\(c.\circledast x=\dfrac{1}{2}\left(TM\right)\) , ta có :
\(M=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\)
\(\circledast x=2\left(KTM\right)\) , giá trị của M không xác định tại x = 2
a: \(P=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{x-2}{-4\left(x-3\right)}\)
\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)}{4\left(x-3\right)}\)
\(=\dfrac{-4x^2-8x}{x+2}\cdot\dfrac{-1}{4\left(x-3\right)}=\dfrac{-4x\left(x+2\right)}{4\left(x+2\right)}\cdot\dfrac{1}{x-3}\)
\(=\dfrac{-x}{x-3}\)
b: Để P=-1/2 thì \(\dfrac{x}{x-3}=\dfrac{1}{2}\)
=>2x=x-3
=>x=-3
c: |x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3(loại) hoặc x=-1(nhận)
Thay x=-1 vào P, ta được:
\(P=\dfrac{-\left(-1\right)}{-1-3}=\dfrac{1}{-4}=\dfrac{-1}{4}\)
a: \(M=1:\dfrac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=1:\dfrac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x-1\right)}=\dfrac{x^2+x+1}{x}\)
b: \(M-3=\dfrac{x^2-2x+1}{x}=\dfrac{\left(x-1\right)^2}{x}>0\)
=>M>3
e: Khi x=1/4 thì \(M=\dfrac{\dfrac{1}{16}+\dfrac{1}{4}+1}{\dfrac{1}{4}}=\dfrac{21}{4}\)
\(A=\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
\(\Leftrightarrow\dfrac{1}{x^2+2x+x+2}+\dfrac{1}{x^2+2x+3x+6}+\dfrac{1}{x^2+3x+4x+12}+\dfrac{1}{x^2+4x+5x+20}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\left(1\right)\)
a, ĐKXĐ của pt :
\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\\x+3\ne0\\x+4\ne0\end{matrix}\right.\) và \(x+5\ne0\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\\x\ne-3\\x\ne-4\end{matrix}\right.\) và \(x\ne-5\)
b, pt(1) \(=\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+5}\)
\(=\dfrac{x+5-x-1}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4}{x^2+6x+5}\)
c, Thay x = 3 vào bt trên ,có :
\(\dfrac{4}{3^2+6.3+5}=\dfrac{4}{32}=\dfrac{1}{8}\)
Vậy tại ..............
d, Để \(A=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{4}{x^2+6x+5}=\dfrac{1}{3}\)
\(\Leftrightarrow x^2+6x+5=12\)
\(\Leftrightarrow x^2+6x-7=0\)
\(\Leftrightarrow x^2-7x+x-7=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\left(t/m\right)\\x=-1\left(kot/m\right)\end{matrix}\right.\)
Vậy x = 7 thì A = 1/3
Mình giải được rồi, cảm ơn bạn. Nhưng câu d đáp án sai rồi nhé. Do chỗ bạn tách 6x ra -7x + x ấy, đúng ra là 7x - x nhé! Đáp án có 2 nghiệm là 1 và -7 nha bạn.
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...
a: \(M=\dfrac{2x+2+4x-4+7x+15}{\left(x-1\right)\left(x+1\right)}=\dfrac{13x+13}{\left(x-1\right)\left(x+1\right)}=\dfrac{13}{x-1}\)