Rut gon bieu thuc A=\(\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\)
Những câu hỏi liên quan
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
A=\(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
Tim cac gia tri cua x de A co nghia
Rut gon bieu thuc A
A có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)
Ta có:
A = \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\)
= \(\dfrac{-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{-\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-1\)
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ĐK của nó còn là: x ≥ 0 nữa dung doan nhé, mình viết thiếu...
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1. Tim GTNN cua bieu thuc:
\(A=\dfrac{1}{2}\sqrt{x^2}+\sqrt{x^2-2x+1}\)
2. Rut gon bieu thuc:
a) \(A=\sqrt{29-4\sqrt{7}}+\sqrt{23+8\sqrt{7}}\)
b)\(B=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}},voix>=2\)
Bài 2:
a: \(A=2\sqrt{7}-1+\left(\sqrt{7}+4\right)\)
\(=2\sqrt{7}-1+\sqrt{7}+4=3\sqrt{7}+3\)
b: \(B=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
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rut gon bieu thuc: \(\frac{\sqrt{\sqrt{\frac{x-1}{x+1}}+\sqrt{\frac{x+1}{x-1}}-2}\left(2x+\sqrt{x^2+1}\right)}{\sqrt{\left(x+1\right)^3}-\sqrt{\left(x-1\right)^2}}\)
RUT GON BIEU THUC \(\frac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}\)
nhan ca tu va mau voi\(\sqrt{2}\) ta dc
\(\frac{\sqrt{2x-4\sqrt{2x-4}}}{2}=\frac{\sqrt{2x-4-4\sqrt{2x-4}}}{2}=\frac{\sqrt{\left(\sqrt{2x-4}-2\right)^2}}{2}\)(dkx>=2)
=\(\frac{\left|\sqrt{2x-4}-2\right|}{2}\)
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Giup minh nhe!!
1. Rut gon bieu thuc:
a) Asqrt{11-6sqrt{2}}+3+sqrt{2}
b) Bsqrt{12+2sqrt{11}}+sqrt{12-2sqrt{11}}
2. Rut gon bieu thuc:
a) Asqrt{x-2+2sqrt{x-3}}+sqrt{x+6+6sqrt{x-3}},voix3
3. Tim GTNN cua bieu thuc:
a) Asqrt{4x^2-12x+9}+sqrt{x^2-10x+25}+sqrt{9x^2-6x+1}+sqrt{16x^2-72x+81}
b) Bdfrac{1}{2}sqrt{x^2}+sqrt{x^2-2x+1}
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Giup minh nhe!!
1. Rut gon bieu thuc:
a) \(A=\sqrt{11-6\sqrt{2}}+3+\sqrt{2}\)
b) \(B=\sqrt{12+2\sqrt{11}}+\sqrt{12-2\sqrt{11}}\)
2. Rut gon bieu thuc:
a) \(A=\sqrt{x-2+2\sqrt{x-3}}+\sqrt{x+6+6\sqrt{x-3}},voix>=3\)
3. Tim GTNN cua bieu thuc:
a) \(A=\sqrt{4x^2-12x+9}+\sqrt{x^2-10x+25}+\sqrt{9x^2-6x+1}+\sqrt{16x^2-72x+81}\)
b) \(B=\dfrac{1}{2}\sqrt{x^2}+\sqrt{x^2-2x+1}\)
1)
a)
\(\sqrt{11-6\sqrt{2}}=\sqrt{2-2.3.\sqrt{2}+9}=\left|\sqrt{2}-3\right|=3-\sqrt{2}\)
\(A=3-\sqrt{2}+3+\sqrt{2}=6\)
b)
\(B^2=24+2\sqrt{12^2-4.11}=24+2\sqrt{100}=24+20=44\)
\(B=\sqrt{44}=2\sqrt{11}\)
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Giúp mình nhé!!
2. Rut gon bieu thuc:
a) \(A=\sqrt{x-2+2\sqrt{x-3}}+\sqrt{x+6+6\sqrt{x-3}},voix>=3\)
3. Tim GTNN cua bieu thuc:
a) \(A=\sqrt{4x^2-12x+9}+\sqrt{x^2-10x+25}+\sqrt{9x^2-6x+1}+\sqrt{16x^2-72x+81}\)
b) \(B=\dfrac{1}{2}\sqrt{x^2}+\sqrt{x^2-2x+1}\)
Cho bieu thuc E= \(\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right).\dfrac{x-1}{2x+\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
a)Rut gon E
b)Tim GTNN cua E
c) Tìm x để E ≥ \(\dfrac{6}{7}\)
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: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
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