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bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
a, \(B=\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right)\frac{\sqrt{a}+1}{\sqrt{a}}\)ĐKXĐ : \(a>0;a\ne1\)
\(=\left(\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right)\frac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\left(\frac{a+\sqrt{a}-2-a+\sqrt{a}+2}{a-1}\right)\frac{1}{\sqrt{a}}\)
\(=\frac{2\sqrt{a}}{\left(a-1\right)\sqrt{a}}=\frac{2}{a-1}\)
b, quá rõ ràng rồi nhé
a) ĐKXĐ:x>0ĐKXĐ:x>0
A=x2+√xx−√x+1−2x+√x√x+1A=x2+xx−x+1−2x+xx+1
⇔A=√x(√x+1)(x−√x+1)x−√x+1−√x(2√x+1)√x+1⇔A=x(x+1)(x−x+1)x−x+1−x(2x+1)x+1
⇔A=x+√x−2√x−1+1⇔A=x+x−2x−1+1
⇔A=x−√x⇔A=x−x
b) Để A = 0
⇔x−√x=0⇔x−x=0
⇔√x(√x−1)=0⇔x(x−1)=0
⇔[√x=0√x=1⇔[x=0x=1
⇔[x=0(ktm)x=1(tm)⇔[x=0(ktm)x=1(tm)
vậy ...
a. ĐKXĐ: \(x>0,x\ne1\)
A=Đề\(=\left[\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{-1}{\sqrt{x}}\cdot\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)}\)\(=\frac{-\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
Đề sai hả bạn ?
a) ĐK: \(a\ge0;a\ne1\)
b) \(B=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1+\frac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\frac{\sqrt{a}+1+a+\sqrt{a}}{\sqrt{a}+1}.\frac{1-\sqrt{a}+a-\sqrt{a}}{1-\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.\frac{\left(1-\sqrt{a}\right)^2}{1-\sqrt{a}}\)
\(=\left(\sqrt{a}+1\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
Tự làm đi easy quá mà :)))) không biết quy đồng mà rút gọn hay sao
\(a,ĐKXĐ:\hept{\begin{cases}a\ge0,\sqrt{a}\ne0\\\sqrt{a}-1\ne0\\\sqrt{a}-2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}a>0\\a\ne1\\a\ne4\end{cases}}}\)
\(b,\)Rút gọn : \(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(Q=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\)
\(Q=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a^2-1-a^2+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\)
\(Q=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
c, bn thay vào rồi tính nha
a, \(P=\frac{a}{\sqrt{a}-1}-\frac{2a-\sqrt{a}}{a-\sqrt{a}}=\frac{a}{\sqrt{a}-1}-\frac{2\sqrt{a}-1}{\sqrt{a}-1}\)với a > 0, a khác 1
\(=\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}=\sqrt{a}-1\)