\(\left(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1}\right).Rut\:gon\:bieu\:thuc\:nay\)
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cho biểu thức p=\(\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{1-\sqrt{a^3}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
a)rut gon p
b) xet dau cua bieu thuc M = a. \(\left(P-\frac{1}{2}\right)\)
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 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 = \(\frac{\sqrt{7+4\sqrt{3}}+1}{\sqrt{7+4\sqrt{3}}}=\frac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}}=\frac{3+\sqrt{3}}{2+\sqrt{3}}\)
rut gon cac bieu thuc sau:\(\left(1+\frac{a=\sqrt{a}}{\sqrt{a+1}}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a-1}}\right)\)
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 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\(=\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}+2+1}{\sqrt{3}+2}=\frac{\sqrt{3}+3}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+3\right)\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{3-\sqrt{3}}{1}=3-\sqrt{3}\)
rut gon bieu thuc
a) A= \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
b) B=\(\frac{\sqrt{a-1-2\sqrt{a-2}}}{\sqrt{a-2}-1}\)
a) \(A=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{6}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(A=\frac{\left(-\sqrt{a}+1\right)^2}{\left(-a+1\right)^2}.\left(\sqrt{a}+\frac{-a\sqrt{a}+1}{-\sqrt{a}+1}\right)\)
\(A=\frac{\left(1-\sqrt{a}\right)^2\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)}{\left(1-a\right)^2}\)
\(A=\frac{\frac{-a\sqrt{a}+\sqrt{a}.\left(-\sqrt{a}+1\right)+1}{-\sqrt{a}+1}.\left(-\sqrt{a}+1\right)^2}{\left(1-a\right)^2}\)
\(A=\frac{a^2-2a+1}{\left(1-a\right)^2}\)
\(A=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}\)
\(A=1\)
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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
rut gon bieu thuc: \(\frac{\sqrt{\sqrt{\frac{x-1}{x+1}}+\sqrt{\frac{x+1}{x-1}}-2}\left(2x+\sqrt{x^2+1}\right)}{\sqrt{\left(x+1\right)^3}-\sqrt{\left(x-1\right)^2}}\)
Cho bieu thuc p=\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2.\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
a/tim dkxd va rut gon p
b/ tìm giá trị của a để p<0
a) ĐKXĐ: \(x\ge0;x\ne1\)
P=\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2.\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
=\(\left(\frac{a-1}{2\sqrt{a}}\right)^2.\left(\frac{-1-3\sqrt{a}}{a-1}\right)\)
=\(\frac{\left(a-1\right)^2}{4a}.\frac{-1-3\sqrt{a}}{a-1}\)
=\(\frac{\left(a-1\right)\left(-1-3\sqrt{a}\right)}{4a}\)
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