Đk: \(\left\{{}\begin{matrix}x\ge0\\y\ge1\end{matrix}\right.\)
đặt \(\left\{{}\begin{matrix}a=\sqrt{x}\\b=\sqrt{y-1}\end{matrix}\right.\) (a,b >/ 0)
được: \(a^2+b^2+13=4a-6b\Leftrightarrow\left(a-2\right)^2+\left(b+3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=-3\left(L\right)\end{matrix}\right.\)
ptvn
hơi kỳ!
giải phương trình
1. y2 - 2y + 3 = \(\dfrac{6}{x^2+2x+4}\)
2. \(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\) biết rằng a + b + c = 3
Rút gọn:
a,\(\frac{3+\sqrt{3}}{1+\sqrt{3}}\)
b,\(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-\sqrt{2}}\)
c,\(\frac{y-2\sqrt{y}}{\sqrt{y}-2}\)
d,\(\frac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
e,\(\frac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
g,\(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
Tính GTBT: \(M=\left(x-y\right)^3+3\left(x-y\right)\left(xy+1\right)\) biết
\(x=\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\)
\(y=\sqrt[3]{17+12\sqrt{2}}-\sqrt[3]{17-12\sqrt{2}}\)
bài tập 1 rút gọn
a) \(\dfrac{\sqrt{7}-5}{2}-\dfrac{6-\sqrt{7}}{4}+\dfrac{6}{4-\sqrt{7}}-\dfrac{5}{4+\sqrt{7}}\)
b) \(\dfrac{x\sqrt{y}+y\sqrt{y}}{\sqrt{xy}}:\dfrac{x+y}{\sqrt{x}-\sqrt{y}}\left(x,y>0\right)\)
c) (\(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}+\dfrac{1}{3\sqrt{x}+1}\)) : \(\dfrac{3\sqrt{x}-5}{3\sqrt{x}-1}\)
Giải:
1)a) \(17\sqrt{3x-1}=3x\)
b) \(\sqrt{2+\sqrt{3x-5}}=\sqrt{x+1}\)
c)\(\sqrt{\dfrac{5x+7}{x+3}}=4\)
2)Giai pt :
a) x+y+12=\(4\sqrt{x}+6\sqrt{y}-1\)
b) \(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\)
c)\(\sqrt{x+\sqrt{14x-4y}}+\sqrt{x-\sqrt{14x-4y}}=\sqrt{14}\)
d)x-\(4\sqrt{2x+2}-2\sqrt{2-x+9}=0\)
Cho E= \(\dfrac{1+xy}{x+y} - \dfrac{1-xy}{x-y} \)
Biết x= \(\sqrt{4+\sqrt{8}} . \sqrt{2+\sqrt{2 + \sqrt{2}}} . \sqrt{2 -\sqrt{2 +\sqrt{2}}}\)
y =\(\dfrac{ 3 \sqrt{8} -2 \sqrt{12}+ \sqrt{20}}{ 3\sqrt{18} -2\sqrt{27} + \sqrt{45}}\)
rút gọn
\(\dfrac{9-x}{\sqrt{x}+3}-\dfrac{9-6\sqrt{x}+x}{\sqrt{x}-3}-6\) (với x>_9)
\(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)/\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\) (với x>=0, x#1)
\(\sqrt{x+12+6\sqrt{x+3}}-\sqrt{x+12-6\sqrt{x+3}}\) ( với x>_6)
\(\sqrt{m^2+6m+9}+\sqrt{m^2-6m+9}\) (m bát kì)
\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\dfrac{x+1}{\sqrt{x}}\)
\(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}/\dfrac{\sqrt{x}-\sqrt{y}}{x-y}\)
\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(\left(\dfrac{\sqrt{x}+2}{3\sqrt{x}}+\dfrac{2}{\sqrt{x}+1}-3\right)/\dfrac{2-4\sqrt{x}}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1-x}{3\sqrt{x}}\)
Cho biểu thức:
\(A=\left[\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\times\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right]:\dfrac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)
a, Rút gọn A
b, Biết xy=6 Tìm x, y để A đạt GTNN
Với mọi x, y, z >= 0 . Chứng minh rằng
\(\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{6\left(x+y+z\right)}\)
