A=2004+\(\sqrt{2003-x}\)
a)Tim x de A co nghia
b)Tim x de A=2005
c)Tim GTNN cua A
Cho A \(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}+1}{X+\sqrt{X}}\right):\frac{\sqrt{X}+1}{X}\)
a)Tìm dk cua x de A co nghia
b)Rut gon A
c)Tim gia tri nho nhat cua A
tim dieu kien cua x de a co nghia va rut gon a
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 = \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\)
= \(\dfrac{-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{-\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-1\)
Kết luận: ...
ĐK của nó còn là: x ≥ 0 nữa dung doan nhé, mình viết thiếu...
Bai 1)Cho bieu thuc A=\(\frac{x+y-2\sqrt{xy}}{x-y}\)
a)Tim dieu kien de A co nghia
b)Rut gon A
c)Tinh A biet x=\(3+2\sqrt{2}\)va y=\(3-2\sqrt{2}\)
Bai 2) Cho bieu thuc B=\(\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\)
a)Tim dieu kien de B co nghia
b)Rut gon B
c) Tinh B voi x=\(4\left(2-\sqrt{3}\right)\)
d)Tim x de B co gia tri nho nhat
a) A có nghĩa\(\Leftrightarrow x-y\ne0\Leftrightarrow x\ne y\)
b) \(A=\frac{x+y-2\sqrt{xy}}{x-y}=\frac{\left(\sqrt{x-\sqrt{y}}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
giai dum minh bai nay voi :c=x/2x-2+x^2+1/2-2x^2
a)tim x de a co nghia
b)rut gon bieu thuc c
c)tim gia tri cua x de bieu thuc kia =1/2
giúp mình cảm ơn nhìu nha !
Cho bthuc: \(A=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}\)
a) Rut gon A
b) Tim gia tri cua x de /A/ =\(\frac{1}{2}\)
c) Tim x nguyen de A co gia tri 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 phuong trinh:\(\dfrac{2+\sqrt{x}}{\sqrt{2}+\sqrt{2+\sqrt{x}}}+\dfrac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)
a/tim dieu kien cua x de phuong trinh co nghia
b/giai phuong trinh
a: ĐKXĐ: x>=0
b: \(\Leftrightarrow\dfrac{2\sqrt{2}-2\sqrt{2-\sqrt{x}}+\sqrt{2x}-\sqrt{x\left(2-\sqrt{x}\right)}+2\sqrt{2}+2\sqrt{2+\sqrt{x}}-\sqrt{2x}-\sqrt{x\left(2+\sqrt{x}\right)}}{2-2+\sqrt{x}}=\sqrt{2}\)
\(\Leftrightarrow4\sqrt{2}-2\sqrt{x\left(\sqrt{x}+2\right)}=\sqrt{2x}\)
\(\Leftrightarrow\sqrt{4x\left(\sqrt{x}+2\right)}=4\sqrt{2}-\sqrt{2x}\)
\(\Leftrightarrow4x\left(\sqrt{x}+2\right)=32-16\sqrt{x}+2x\)
\(\Leftrightarrow4x\sqrt{x}+8x-32+16\sqrt{x}-2x=0\)
=>\(x\in\left\{0;1.2996\right\}\)
cho bieu thuc C =\(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
a. tim dieu kien de C co nghia ?
b.rut gon C ?
c. tinh C tai x=\(\frac{4}{9}\)
d. tim x de C= 5
e.tim gia tri x nguyen de C co gia tri nguyen