Violympic toán 9
Các câu hỏi tương tự
\(A=\frac{\sqrt{x}+4}{\sqrt{x}+2};B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)
a. rut gon B
b. Tim x nguyen de P = B(A-1) nguyen
giup minh voi a
Tim x,y,z biet: \(\dfrac{1}{2}\left(x+y+z\right)-3=\sqrt{x-2}+\sqrt{y-3}+\sqrt{z-4}\)
1) Cho bieu thuc: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\left(x\ge0,x\ne16\right)\)
a) Cho bieu thuc A= \(\frac{\sqrt{x}+4}{\sqrt{x}+2}\) ; voi cac cua bieu thuc A va B da cho, hay tim cac gia tri cua x nguyen de gia tri cua bieu thuc B(A;-1) la so nguyen
1.Chm cac so sau la so vo ti: \(\sqrt{3}-\sqrt{2};2\sqrt{2}+\sqrt{3}\)
2. Chm \(\sqrt{x-2}+\sqrt{4-x}\le2\)
Cho cac so thuc duong x,y,z thoa man :\(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2=2015}\)
Tim ja tri nho nhat cua bieu thuc :\(T=\dfrac{x^2}{y+x}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)
Cho P = \(\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right)\): \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{x-1}\right)\)
a/ Tim DKXD va rut gon P
b/ Tim cac gia tri nguyen cua x de P co gia tri nguyen
cho 3 so thuc x,y,z khac khong va thoa man hai dieu kien \(ax^3=by^3=cz^3\) va \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
chung minh rang : \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)
Tim x, y, z
1/ sqrt{x-2}+sqrt{y-2008}+sqrt{z-2009}dfrac{1}{2}left(x+y+zright)
2/ x+y+z+42sqrt{x-2}+4sqrt{y-3}+6sqrt{x-5}
3/ Tinh T x^2+y^2+z^2-7 biet x-y-z 2sqrt{x-34}+4sqrt{y-21}+6sqrt{z-4}+45
4/ 2x^2+9y^2-6xy-12y-6x+290
5/4x^2+3y-4x+4xy-10y+90
Đọc tiếp
Tim x, y, z
1/ \(\sqrt{x-2}+\sqrt{y-2008}+\sqrt{z-2009}=\dfrac{1}{2}\left(x+y+z\right)\)
2/ \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{x-5}\)
3/ Tinh T = \(x^2+y^2+z^2-7\) biet x-y-z = \(2\sqrt{x-34}+4\sqrt{y-21}+6\sqrt{z-4}+45\)
4/ \(2x^2+9y^2-6xy-12y-6x+29=0\)
5/\(4x^2+3y-4x+4xy-10y+9=0\)
Làm hộ mình câu c nha
Cho \(H=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x^3}-\sqrt{y^3}}{x-y}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\).
a) Rút gọn H
b) Chứng minh \(H\ge0\)
c) So sánh H với \(\sqrt{H}\)