cho A= \(\frac{1-\sqrt{x}}{1+\sqrt{x}}\) ,B=\((\frac{15-\sqrt{x}}{x-25}\)+\(\frac{2}{\sqrt{x+5}}\)) : \(\frac{\sqrt{x}+1}{\sqrt{x}-5}\)
a) tinh gia tri bieu thuc A khi x = 16
b) rut gon B
c) tim x để ( B-A) nguyen
cho bieu thuc A = \(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
a. tim x de bieu thuc A co nghia ?rut gon A ?
b. tinh gia tri cua bieu thuc A tai x=7+4√3
a. A có nghĩa khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne\\\frac{x+\sqrt{x}}{\sqrt{x}+1}\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
A\(=\frac{x-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{x+\sqrt{x}}\)\(=\frac{x-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b. \(x=7+4\sqrt{3}\Rightarrow\)A = \(\frac{\sqrt{7+4\sqrt{3}}+1}{\sqrt{7+4\sqrt{3}}}=\frac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}}=\frac{3+\sqrt{3}}{2+\sqrt{3}}\)
cho bieu thuc A =\(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
(x≥0;x≠1)
a. tim x de bieu thuc A co nghia ?rut gon A ?
b. tinh gia tri bieu thuc A tai x=7+4√3
a/ Ta có: A=\(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+1\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1\right):\left(\sqrt{x}\right)=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b/ Ta có :\(x=7+4\sqrt{3}=3+4\sqrt{3}+4=\left(\sqrt{3}+2\right)^2
\)
\(\Rightarrow\sqrt{x}=|\sqrt{3}+2|=\sqrt{3}+2\)
Thay x vào A ta có:
A\(=\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}+2+1}{\sqrt{3}+2}=\frac{\sqrt{3}+3}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+3\right)\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{3-\sqrt{3}}{1}=3-\sqrt{3}\)
a\Rut gon bieu thuc
b/Tim gia tri nho nhat cua 4A
A=\(\frac{5\sqrt{x}+4}{x+\sqrt{x}-2}+\frac{\sqrt{x}-1}{\sqrt{x}+2}-\)\(\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(a.A=\frac{5\sqrt{x}+4}{x+\sqrt{x}-2}+\frac{\sqrt{x}-1}{\sqrt{x}+2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}.\)
\(=\frac{5\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)\(+\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)\(-\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{5\sqrt{x}+4+x-2\sqrt{x}+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=-\frac{1}{\sqrt{x}+2}\)
\(b,4A_{min}\Leftrightarrow A_{min}\Rightarrow\frac{-1}{\sqrt{x}+2}\)nhỏ nhất
\(\frac{\Rightarrow1}{\sqrt{x}+2}\)lớn nhất \(\Leftrightarrow\sqrt{x}+2\)nhỏ nhất
\(\sqrt{x}+2\ge2\Leftrightarrow\sqrt{x}=0\Rightarrow x=0\)
\(\Rightarrow A_{min}=\frac{-1}{0+2}=-\frac{1}{2}\Rightarrow4A_{min}=-1\Leftrightarrow x=0\)
\(M=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)N=\frac{\sqrt{x}+1}{\sqrt{x}}\) voi x>0 va x khac 1
a) tim gia tri bieu thuc cua N khi x = 25
b) rut gon S = M.N
c) tim m de S<-1
a) Với x = 25 thì \(N=\frac{\sqrt{25}+1}{\sqrt{25}}=\frac{6}{5}\)
b) Ta có \(M=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}\)
\(M=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\)
Suy ra \(S=M.N=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
cho bieu thuc A=\(\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
(x≥0;x≠1)
a.Tim x de bieu thuc A co nghia ? rut gon A ?
b. Tinh gia tri cua bieu thuc A tai x=7=4√3
cho bieu thuc \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x+2}}{x-2\sqrt{x}+1}\right)\cdot\frac{\left(1-x\right)^2}{2}\)
a) rut gon P
b) tim gia tri lon nhat cua P
\(ĐKXĐ:0\le x\ne x\)
a) \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}\)
\(P=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(1-x\right)^2}{2}\)
\(P=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(P=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(P=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b) \(P=-x+\sqrt{x}=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}.\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
\(\Rightarrow MAX_P=\frac{1}{4}\text{ khi }x=\frac{1}{4}\)
Cho bieu thuc P = \(\frac{x-\sqrt{x}}{x+\sqrt{x}+1}\) --\(\frac{2x+\sqrt{x}}{\sqrt{x}}\)+\(\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
a) rut gon P
b) tim x de Q=\(\frac{2\sqrt{x}}{P}\)nhan gia tri nguyen
Cho bieu thuc: \(p=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a) Tim DKXD cua bieu thuc p
b) Rut gon bieu thuc p
Cho bieu thuc:
P=\(\frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}\)
a. Rut gon bieu thuc P
b.Tim GTLN cua P sau khi rut gon
đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)