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C1 :
\(B=\frac{4\left(x^2+x+1\right)}{4\left(x^2+2x+1\right)}=\frac{3\left(x^2+2x+1\right)}{4\left(x^2+2x+1\right)}+\frac{x^2-2x+1}{4\left(x^2+2x+1\right)}=\frac{3}{4}+\frac{\left(x-1\right)^2}{4\left(x^2+2x+1\right)}\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=1\)
C2 :
\(B=\frac{x^2+x+1}{x^2+2x+1}\)\(\Leftrightarrow\)\(Bx^2-x^2+2Bx-x+B-1=0\)
\(\Leftrightarrow\)\(\left(B-1\right)x^2+\left(2B-1\right)x+\left(B-1\right)=0\)
+) Nếu \(B=1\) thì \(x=0\)
+) Nếu \(B\ne1\) thì pt có nghiệm \(\Leftrightarrow\)\(\Delta\ge0\)
\(\Leftrightarrow\)\(\left(2B-1\right)^2-4\left(B-1\right)\left(B-1\right)\ge0\)
\(\Leftrightarrow\)\(4B^2-4B+1-4B^2+8B-4\ge0\)
\(\Leftrightarrow\)\(4B-3\ge0\)
\(\Leftrightarrow\)\(B\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=1\)
a) \(ĐKXĐ:x\ne\pm1\)
\(Q=\frac{1}{2x-2}+\frac{1}{2x+2}+\frac{x^2}{1-x^2}\)
\(\Leftrightarrow Q=\frac{1}{2\left(x-1\right)}+\frac{1}{2\left(x+1\right)}-\frac{x^2}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow Q=\frac{x+1+x-1-2x^2}{2\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow Q=\frac{-2x^2+2x}{2\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow Q=\frac{-1}{x+1}\)
b) Khi \(\left|x+1\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=-3\left(tm\right)\end{cases}}\)
Thay \(x=-3\)vào Q ta được :
\(Q=\frac{-1}{-3+1}=\frac{1}{2}\)
c) Để \(Q\)có giá trị nguyên \(\Leftrightarrow-1⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(-1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{-2;0\right\}\)
Vậy để Q có giá trị nguyên \(\Leftrightarrow x\in\left\{-2;0\right\}\)
c) Bạn lấy mỗi giá trị nguyên nhỏ nhất của x = -2 thôi nhé !
Xin lỗi vì đọc nhầm đề
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 \(M< 1\)
a) M= \(\frac{2x}{1-x^2}\)( \(\frac{1}{x^2+2x+1}\)- \(\frac{1}{x^2-1}\))
= \(\frac{2x}{1-x^2}\)(\(\frac{1}{x^2+2x+1}\)+ \(\frac{1}{1-x^2}\))
= \(\frac{2x}{1-x^2}\)(\(\frac{1}{\left(x+1\right)^2}\)+ \(\frac{1}{\left(1+x\right)\left(1-x\right)}\))
= \(\frac{2x}{1-x^2}\)(\(\frac{1-x}{\left(1-x\right)\cdot\left(x+1\right)^2}\)+ \(\frac{1+x}{\left(1-x\right)\cdot\left(x+1\right)^2}\))
= \(\frac{2x}{1-x^2}\)(\(\frac{1-x^2}{\left(1-x\right)\cdot\left(x+1\right)^2}\))
= \(\frac{2x}{\left(1-x\right)\cdot\left(x+1\right)^2}\)
a: Khi x=1 thì \(A=\dfrac{1+1}{2}=1\)
b: Để A=2 thì x+1=4
=>x=3
c: \(B=\dfrac{2+x-2+x}{x\left(x-2\right)}=\dfrac{2x}{x\left(x-2\right)}=\dfrac{2}{x-2}\)
d: C=A*B=2/(x-2)*x+1/2=x+1/x-2
Để C la số nguyên thì x-2+3 chia hết cho x-2
=>\(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;5;-1;1\right\}\)
gợi ý :Đề cho \(x^2-x-1\ne0\), mà mẫu cũng khác 0
nên mẫu có hạng tử \(x^2-x-1\) chia \(x^4-x^2-2x-1/x^2-x-1\)
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 \(M=\frac{2x+1}{x^2-1}.\frac{x^2-x}{2x+1}=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x}{x+1}\)