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23 tháng 7 2019
ĐK: \(x-9\ne0\Rightarrow x\ne9\)
\(\sqrt{x}\ge0\Rightarrow x\ge0\)
\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)
\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)
2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)
Y
24 tháng 7 2019
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{7\sqrt{x}-9}{x-9}\\ =\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{7\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\frac{x+3\sqrt{x}-\sqrt{x}-3-7\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\frac{x-5\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)
b) ĐKXĐ: x>0
\(x=\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}+1-\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{2}{\left(\sqrt{2}\right)^2-1}=\frac{2}{2-1}=2\)
\(\Rightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{\sqrt{2}-2}{\sqrt{2}}=\frac{\sqrt{2}\left(1-\sqrt{2}\right)}{\sqrt{2}}=1-\sqrt{2}\)
c)
\(P=\frac{A}{B}=\frac{\frac{\sqrt{x}-2}{\sqrt{x}+3}}{\frac{\sqrt{x}-2}{\sqrt{x}}}=\frac{\sqrt{x}-2}{\sqrt{x}+3}\cdot\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{x}}{\sqrt{x}+3}\)
Còn khúc sau ko hiểu cho lắm ._.
10 tháng 8 2017
Mỗi biểu thức trong dấu căn có dạng:
\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}\) ( Với \(k\ge2\))
Ta có:
\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}=\frac{k^2\left(k+1\right)^2+\left(k+1\right)^2+k^2}{k^2\left(k+1\right)^2}=\frac{k^4+2k^3+k^2+k^2+2k+1+k^2}{k^2\left(k+1\right)^2}\)
\(=\frac{k^4+2k^2\left(k+1\right)+\left(k+1\right)^2}{k^2\left(k+1\right)^2}=\frac{\left(k^2+k+1\right)^2}{\left(k\left(k+1\right)\right)^2}\)
\(\Rightarrow\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\frac{k^2+k+1}{k^2+k}=1+\frac{1}{k\left(k+1\right)}=1+\frac{1}{k}-\frac{1}{k+1}\)
\(\Rightarrow S=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2013}-\frac{1}{2014}=2014-\frac{1}{2014}\)
B
22 tháng 9 2017
Mỗi biểu thức trong dấu căn có dạng:
1+1k2 +1(k+1)2 ( Với k≥2)
Ta có:
1+1k2 +1(k+1)2 =k2(k+1)2+(k+1)2+k2k2(k+1)2 =k4+2k3+k2+k2+2k+1+k2k2(k+1)2
=k4+2k2(k+1)+(k+1)2k2(k+1)2 =(k2+k+1)2(k(k+1))2
⇒√1+1k2 +1(k+1)2 =k2+k+1k2+k =1+1k(k+1) =1+1k −1k+1
⇒S=1+1−12 +1+12 −13 +1+13 −14 +...+1+12013 −12014 =2014−12014
NM
Nguyễn Minh Quang
Giáo viên
9 tháng 9 2021
\(\frac{3}{\sqrt{7}-1}+\frac{3}{\sqrt{7}+1}=\frac{3\left[\sqrt{7}+1+\sqrt{7}-1\right]}{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}=\frac{6\sqrt{7}}{6}=\sqrt{7}\)
\(\frac{3}{\sqrt{X}-1}-\frac{2}{\sqrt{X}+1}+\frac{X-7}{X-1}=\frac{3\left(\sqrt{X}+1\right)-2\left(\sqrt{X}-1\right)+X-7}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{X+\sqrt{X}-2}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{\sqrt{X}+2}{\sqrt{X}+1}\)
VT
9 tháng 9 2021
TÍNH GIÁ TRỊ BIỂU THỨC:
\(\frac{3}{\sqrt{7}-1}\) + \(\frac{3}{\sqrt{7}+1}\)= \(\frac{3\left(\sqrt{7}+1\right)+3\left(\sqrt{7}-1\right)}{\left(\sqrt{7}-1\right)\left(\sqrt{7}+1\right)}\)= \(\frac{3\sqrt{7}+3+3\sqrt{7}-3}{6}\)=\(\frac{6\sqrt{7}}{6}\)=\(\sqrt{7}\)
RÚT GỌN BIỂU THỨC:
\(\frac{3}{\sqrt{X}-1}\)-\(\frac{2}{\sqrt{X}+1}\)+\(\frac{X-7}{X-1}\)
= \(\frac{3\left(\sqrt{X}+1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)-\(\frac{2\left(\sqrt{X}-1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)+\(\frac{X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{3\sqrt{X}+3-2\sqrt{X}+2+X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{X+\sqrt{X}-2}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\left(\sqrt{X}+1\right)\left(\sqrt{X}-2\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\sqrt{X}-2}{\sqrt{X}-1}\)
CHÚC EM HỌC TỐT!