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\(B=\frac{a+1}{a^2-a+1}-\frac{1}{a+1}-\frac{a-2}{a^3+1}=\frac{\left(a+1\right)^2}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a-2}{a^3+1}\\ \)
\(=\frac{a^2+2a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a-2}{\left(a+1\right).\left(a^2-a+1\right)}\)
\(=\frac{a^2+2a+1-\left(a^2-a+1\right)-\left(a-2\right)}{\left(a+1\right).\left(a^2-a+1\right)}=\frac{2a+2}{\left(a+1\right).\left(a^2-a+1\right)}=\frac{2}{a^2-a+1}\)
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
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:
\(A=\left(\dfrac{1}{3}-1\right):\left(\dfrac{2}{3}-1\right)=\dfrac{-2}{3}:\dfrac{-1}{3}=2\)
a. A=\(1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left(\frac{x+1+x+1-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right).\frac{x\left(x^2-x+1\right)}{x^2\left(x-2\right)}\)
\(=1+\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\frac{-2x\left(x-2\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1-\frac{2}{x+1}=\frac{x-1}{x+1}\)
b.\(\left|x-\frac{3}{4}\right|=\frac{5}{4}\Rightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)
Với \(x=2\Rightarrow A=\frac{2-1}{2+1}=\frac{1}{3}\)
Với \(x=-\frac{1}{2}\Rightarrow A=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
\(\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right)=\frac{a^2-1}{a^2-a}=\frac{a+1}{a}\)
ở phàn a+/a thiếu số 1 nhé
\(\frac{1}{a+1}+\frac{2}{a^2-1}=\frac{a-1+2}{a^2-1}=\frac{1}{a-1}\)
=> K =\(\frac{a^2-1}{a}\)
đkxđ: a khác +-1
b, thay vào mà tình
a/ \(K=\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right):\left(\frac{1}{a+1}+\frac{2}{a^2-1}\right)\)
\(=\left(\frac{a}{a-1}-\frac{1}{a\left(a-1\right)}\right):\left(\frac{1}{a+1}+\frac{2}{\left(a-1\right)\left(a+1\right)}\right)\)
\(=\frac{a^2-1}{a\left(a-1\right)}:\frac{a-1+2}{\left(a-1\right)\left(a+1\right)}\)
\(=\frac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)}.\frac{\left(a-1\right)\left(a+1\right)}{a-1}\)
\(=\frac{a+1}{a}.a+1\)
\(=\frac{\left(a+1\right)^2}{a}\)
b, Thay a=1/2
\(\Rightarrow\frac{\left(\frac{1}{2}+1\right)^2}{\frac{1}{2}}=\frac{\frac{9}{4}}{\frac{1}{2}}=\frac{9}{2}\)
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 ??
\(B=\frac{a+1}{a^2-a+1}-\frac{1}{a+1}+\)\(\frac{a-2}{a^3+1}\)
\(B=\frac{\left(a+1\right)^2}{a^3+1}-\frac{a^2-a+1}{a^3+1}+\)\(\frac{a-2}{a^3+1}\)
\(B=\frac{a^2+2a+1-a^2+a-1-a+2}{a^3+1}\)
\(B=\frac{2a+2}{a^3+1}\)
\(B=\frac{2\left(a+1\right)}{\left(a+1\right)\left(a^2+a+1\right)}\)
\(B=\frac{2}{a+1}\)
\(B=\frac{a+1}{a^2-a+1}-\frac{1}{a+1}-\frac{a-2}{a^3+1}\) ĐKXĐ : \(x\ne-1\)
\(=\frac{\left(a+1\right)^2}{\left(a+1\right)\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right)\left(a^2-a+1\right)}\)\(-\frac{a-2}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{a^2+2a+1-a^2+a-1-a+2}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{\left(a^2-a^2\right)+\left(2a+a-a\right)+\left(1-1+2\right)}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{2a+2}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{2\left(a+1\right)}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{2}{a^2-a+1}\)