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Do : \(4x^2=1\)
\(< =>\orbr{\begin{cases}2x=1\\2x=-1\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
Ta thấy điều kiện xác định của B là \(x\ne-\frac{1}{2}\)
Suy ra \(x=\frac{1}{2}\)
Ta có : \(B=\frac{x^2-x}{2x+1}=\frac{\frac{1}{4}-\frac{1}{2}}{\frac{1}{2}.2+1}=\frac{\frac{-1}{4}}{2}=-\frac{1}{8}\)
Vậy ......
Ta có : \(A=\frac{1}{x-1}+\frac{x}{x^2-1}=\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x+1}{x^2-1}\)
Suy ra \(M=\frac{2x+1}{x^2-1}.\frac{x^2-x}{2x+1}=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x}{x+1}\)
a)Ta có : \(4x^2=1\)
\(\Rightarrow\orbr{\begin{cases}2x=1\\2x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
mà \(x\ne-\frac{1}{2}\Rightarrow x=\frac{1}{2}\)
Thay \(x=\frac{1}{2}\)vào B , ta được:
\(B=\frac{\left(\frac{1}{2}\right)^2-\frac{1}{2}}{2.\frac{1}{2}+1}=\frac{\frac{1}{4}-\frac{1}{2}}{1+1}=\frac{-\frac{1}{4}}{2}=-\frac{1}{8}\)
Vậy \(B=-\frac{1}{8}\)khi \(4x^2=1\)
b)Ta có : \(A=\frac{1}{x-1}-\frac{x}{1-x^2}\)
\(=\frac{1}{x-1}+\frac{x}{x^2-1}\)
\(=\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow M=A.B=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x^2-x}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x\left(x-1\right)}{2x+1}\)
\(=\frac{x}{x+1}\)
Vậy \(M=\frac{x}{x+1}\)
c)Ta có: \(x< x+1\forall x\)
\(\Rightarrow M=\frac{x}{x+1}< \frac{x+1}{x+1}=1\forall x\ne-1\)
Vậy với mọi \(x\ne-1\)thì \(M< 1\)
ĐKXĐ: \(x\ne\pm2\)
a)\(A=\frac{1}{x-2}+\frac{1}{x+2}+\frac{x^2+4}{x^2-4}=\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{x^2+4}{x^2-4}\)
\(=\frac{x+2}{x^2-4}+\frac{x-2}{x^2-4}+\frac{x^2+4}{x^2-4}=\frac{x+2+x-2+x^2+4}{x^2-4}=\frac{x^2+2x+4}{x^2-4}=\frac{\left(x+1\right)^2+3}{x^2-4}\)
b) \(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+3\ge3>0\)
=> A<0 khi \(x^2-4< 0\Leftrightarrow x^2< 4\)
Vì \(x^2\ge0\Rightarrow0\le x^2< 4\Leftrightarrow-2< x< 2\)
Tại sao lại x khác -1 thì A<0 vì khi x=-1 thì A=-1<0 mà!
a) Ta thấy x=-2 thỏa mãn ĐKXĐ của B.
Thay x=-2 và B ta có :
\(B=\frac{2\cdot\left(-2\right)+1}{\left(-2\right)^2-1}=\frac{-3}{3}=-1\)
b) Rút gọn :
\(A=\frac{3x+1}{x^2-1}-\frac{x}{x-1}\)
\(=\frac{3x+1-x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
Xấu nhỉ ??
2) a) Ta có B = \(\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{16}{4-x^2}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{8}{x-2}\)
Khi |x - 1| = 2
=> \(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Khi x = 3 (thỏa mãn) => A = \(\frac{3^2-2.3}{3+1}=\frac{3}{4}\)
Khi x = - 1 (không thỏa mãn) => Không tìm được A
b) Ta có P = \(A.B=\frac{x^2-2x}{x+1}.\frac{8}{x-2}=\frac{8x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{8x}{x+1}\)
Đẻ P < 8
=> \(\frac{8x}{x+1}< 8\Leftrightarrow\frac{x}{x+1}< 1\)
=> \(\orbr{\begin{cases}x< x+1\left(x>-1\right)\\x>x+1\left(x< -1\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x< 1\left(tm\right)\\0x>1\left(\text{loại}\right)\end{cases}}\)
Vậy x > - 1 thì P < 8
Thay x = 1/2 vào
a) Thay x = 1/2 (tm) vào A ta được A = \(\frac{\frac{1}{2}+2}{\frac{1}{2}-5}=\frac{\frac{5}{2}}{-\frac{9}{2}}=-\frac{5}{9}\)
b) B = \(\frac{3}{x+5}+\frac{20-2x}{x^2-25}=\frac{3\left(x-5\right)+20-2x}{\left(x-5\right)\left(x+5\right)}=\frac{x+5}{\left(x-5\right)\left(x+5\right)}=\frac{1}{x-5}\)
c) A = B|x2 - 4|
=> \(\left|x^2-4\right|=\frac{A}{B}\)
=> \(\left|x^2-4\right|=x+2\)(x \(\ge-2\))
<=> \(\orbr{\begin{cases}x^2-4=x+2\\x^2-4=-\left(x+2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)\left(x+2\right)=0\\\left(x-1\right)\left(x+2\right)=0\end{cases}}\)
Khi (x - 3)(x + 2) = 0
<=> \(\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)(tm)
Khi (x - 1)(x + 2) = 0
<=> \(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)(tm)
Vậy x \(\in\left\{1;-2;3\right\}\)thì A = B|x2 - 4|
Bài 2.
a.Ta có:\(\left|x-1\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\left(tmđkxđ\right)\\x=-1\left(ktmđkxđ\right)\end{cases}}}\)
Với \(x=3\)thì giá trị biểu thức là:
\(A=\frac{3^2-2.3}{3+1}=\frac{3}{4}\)
b.P có phải rút gọn không vậy hay chỉ đặt thôi :?Chắc rút gọn nữa nhỉ:?//
\(P=A.B\)
\(=\frac{x^2-2x}{x+1}.\left(\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{16}{4-x^2}\right)\)
\(=\frac{x^2-2x}{x+1}.\left(\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{16}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\frac{x^2-2x}{x+1}.\left(\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\frac{x\left(x-2\right)}{x+1}.\left(\frac{x^2+4x+4-x^2+4x-4+16}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\frac{x\left(x-2\right)}{x+1}.\left(\frac{8x+16}{\left(x-2\right)\left(x+2\right)}\right)=\frac{x\left(x-2\right)}{x+1}.\left(\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\frac{x\left(x-2\right)}{x+1}.\frac{8}{x-2}=\frac{8x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{8x}{x+1}\)
c.Để \(P< 8\)thì \(\frac{8x}{x+1}< 8\)
\(\Leftrightarrow\frac{x}{x+1}< 1\)
Bài 3.
a.Với \(x=\frac{1}{2}\)thì giá trị của A là:
\(A=\frac{\frac{1}{2}+2}{\frac{1}{2}-5}=-\frac{5}{9}\)
b.\(B=\frac{3}{x+5}+\frac{20-2x}{x^2-25}\)
\(=\frac{3}{x+5}+\frac{20-2x}{\left(x-5\right)\left(x+5\right)}=\frac{3\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\frac{20-2x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{3x-15+20-2x}{\left(x-5\right)\left(x+5\right)}=\frac{x+5}{\left(x-5\right)\left(x+5\right)}=\frac{1}{x-5}\)
Bài 3.c/
\(A=B\left|x^2-4\right|\)
\(\Leftrightarrow\left|x^2-4\right|=\frac{A}{B}\)
\(\Leftrightarrow\left|x^2-4\right|=\frac{x+2}{x-5}\div\frac{1}{x-5}\)
\(\Leftrightarrow\left|x^2-4\right|=\frac{x+2}{x-5}.\frac{x-5}{1}\)
\(\Leftrightarrow\left|x^2-4\right|=\frac{\left(x-2\right)\left(x-5\right)}{\left(x-5\right)}\)
\(\Leftrightarrow\left|x^2-4\right|=x+2\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4=x+2\\x^2-4=-x-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2-4-x-2=0\\x^2-4+x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-x-6=0\\x^2+x-2=0\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)\left(x+2\right)=0\\\left(x-1\right)\left(x+2\right)=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3;x=-2\\x=1;x=-2\end{cases}\left(tm\right)}}\)
Vậy \(x\in\left\{-2;1;3\right\}\)thì \(A=B.\left|x^2-4\right|\)