Để căn thức có nghĩa\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{2}{x+1}\ge0\\x+1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+1\le0\\x+1\ne0\end{matrix}\right.\)\(\Leftrightarrow x+1< 0\Leftrightarrow x< -1\)
Vậy...
cho bieu thuc P= (\(\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}-3}\) ): \(\frac{1}{x-1}\)
a) Tim dieu kien de P co nghia, rut gon bieu thuc P.
b) Tim cac so tu nhien x de \(\frac{1}{P}\)la so tu nhien
c) Tinh gia tri cua P voi x= 4-\(2\sqrt{3}\)
Giup mk vs mk dang can gap
Cho bieu thuc A=\(\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\div\dfrac{1}{\sqrt{x}-1}\)
a/ Tim dieu kien cua x de bieu thuc A co gia tri xac dinh
b/ Rut gon A
c/ Tinh gia tri cua A khi x = \(4-2\sqrt{3}\)
d/ Tim gia tri nho nhat cua A
Cho cac bieu thuc :
\(A=\dfrac{\sqrt{x}+4}{\sqrt{x}+2},B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)
a) Rut gon B ?
b) Tim cac gia tri nguyen cua x de cac gia tri cua bieu thuc B(A-1) la so nguyen.
moi nguoi giup em bai 4 5 6 voi a em cam on mng
1,thu gon bieu thuc
a A=\(\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\)
b,B=\(\dfrac{12\sqrt{6}}{\sqrt{7+2\sqrt{6}-\sqrt{7-2\sqrt{6}}}}\)
c, C=\(\dfrac{\sqrt{c^2+2c+1}}{\left|c\right|-1}\)
2,giai cac phuong trinh
a,\(x^2-9\sqrt{x}+14=0\)
b, \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=-5-x^2+6\)
GIUP MINH VOI MINH CAN GAP
Tinh gt cua da thuc :
\(f\left(x\right)=\left(x^4-3x+1\right)^{2016}\)
voi \(x=9-\dfrac{1}{\sqrt{\dfrac{9}{4}-\sqrt{5}}}+\dfrac{1}{\sqrt{\dfrac{9}{4}+\sqrt{5}}}\)
giai chi tiet nhat co the nha mn ^^!
giup minh cau nay voi
Chung minh bat dang thuc sau \(\dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n+1}\right)\) voi n>=2,n thuoc N
Cho biểu thức:
P= \(\left(\dfrac{2-x}{x-\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\) :\(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
a)Rút gọn biểu thức
b)Tính giá trị của P với x=\(\dfrac{9}{16}\)
c)Tìm giá trị của x để P <\(\dfrac{1}{2}\)
-Giup minh voi a-
