B=\((\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}):\frac{\sqrt{x}}{\sqrt{x}-3}\)) (vs x>0; x \(\ne\)9)
a,Rut gon B
b,Tinh gt cua B khi x=\(\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
c,CM: B>\(\frac{1}{3}\)
(giup mk vs..)
cho A=\(\frac{x+7}{\sqrt{x}}\); B=\(\frac{\sqrt{x}}{\sqrt{x+3}}\)+\(\frac{2\sqrt{x-1}}{\sqrt{x-3}}\)-\(\frac{2x-\sqrt{x-3}}{x-9}\)
a, tinh A khi x=16
b, rut gon B
c,tim Min cua p=A+1/B
P=\(\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
a,Rút gọn P
b,Tính P khi x=\(\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
c,CM: B>\(\frac{1}{3}\)
ĐKXĐ: ...
\(P=\left(\frac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(P=\left(\frac{x+\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{\left(\sqrt{x}-3\right)}{\sqrt{x}}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)
\(x=5+\sqrt{2}-4-\sqrt{2}=1\)
\(\Rightarrow P=\frac{1+1}{1+3}=\frac{1}{2}\)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}+3}=1-\frac{2}{\sqrt{x}+3}\)
Do \(\sqrt{x}>0\Rightarrow\sqrt{x}+3>3\Rightarrow\frac{2}{\sqrt{x}+3}< \frac{2}{3}\)
\(\Rightarrow P>1-\frac{2}{3}=\frac{1}{3}\) (đpcm)
P=\(\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
a,Rút gọn P
b,Tính P khi x=\(\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
c,CM: B>\(\frac{1}{3}\)
\(P=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
ĐKXĐ:\(x\ge0;x\ne9\)
\(=\left(\frac{x+3}{x-9}+\frac{1\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x+3}\right)}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\left(\frac{x+3+\sqrt{x}-3}{x-9}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{x+\sqrt{x}}{x-9}.\frac{\sqrt{x-3}}{\sqrt{x}}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b)
\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(=\sqrt{5^2+2.5\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{4^2+2.4\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)
\(=5+\sqrt{2}-4-\sqrt{2}\)
\(=1\)
Thay x=1 vào P ta có:
\(P=\frac{\sqrt{1}+1}{\sqrt{1}-3}\)
\(=\frac{2}{-2}=-1\)
cho bieu thuc:P=\(\frac{\sqrt{x}}{\sqrt{x}-3}\)+\(\frac{2\sqrt{x}}{\sqrt{x}-3}\)--\(\frac{3x+9}{x-9}\) voi x>= 0;x#9 .a; Rut gon bieu thuc P . b; Tinh gia tri cua bieu thuc voi \(x=4-2\sqrt{3}\)
\(B=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x+3}}\right)\div\frac{\sqrt{x}}{\sqrt{x-3}}\)
\(x>0,x\ne9\)
a) rút gọn B
b) tính B khi \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
c) chứng minh khi B>\(\frac{1}{2}\)
ai nhanh mình tick nhé cảm ơn các bạn
Có bị sai đề không vậy bạn ? Mình nghĩ nó là \(\sqrt{x}+3\) với \(\sqrt{x}-3\)chứ không phải là \(\sqrt{x+3}\) với \(\sqrt{x-3}\)?
Rút gọn:
a, A = \(\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\) (đk: x ≥ 0 và x ≠ 36)
b, B = \(\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\) (đk: x ≥ 0 và x ≠ 9)
c, C = \(\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2\) (đk: a > 0, b > 0 và a ≠ b)
d, D = \(\left(\frac{2-a\sqrt{a}}{2-\sqrt{a}}+\sqrt{a}\right)\left(\frac{2-\sqrt{a}}{2-a}\right)\) (đk: a ≥ 0, a ≠ 2, a ≠ 4)
\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)
\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)
\(B=3-\sqrt{x}-\sqrt{x}+3-6\)
\(B=-2\sqrt{x}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3}{\sqrt{x}-6}\)
\(P=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a)Rut gon P?
b)Tinh gia tri cua P voi \(x=3-2\sqrt{2}\)?
Rut gon \(B=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{\sqrt{2}+3}-\frac{9-x}{x+\sqrt{x-6}}\right)\)
RÚT GỌN BIỂU THỨC
A=\(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)(với a>_ 0, b>_ 0, a#b)
B=\(\left(\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right).\left(\frac{\sqrt{x}+\sqrt{y}}{x-y}\right)\)(với x>_ 0, y>_ 0, x#y)
C=\(x-4-\sqrt{16-8x^2+x^4}\)(với x>4)
D=\(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}\)(với a>0, b>0, a#b)
E=\(\left(2+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right).\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\)(với a>0, a#1)
F=\(\frac{a-3\sqrt{a}}{\sqrt{a}-3}-\frac{a+4\sqrt{a}+3}{\sqrt{a}+3}\)( với a>_ 9)
G=\(\frac{9-x}{\sqrt{x}+3}-\frac{9-6\sqrt{x}+x}{\sqrt{x}-3}-6\)( với x>_ 9 )