\(b=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)
Những câu hỏi liên quan
\(B=x-4\sqrt{x}+\frac{x+16}{\sqrt{x}+3}+10=x-4\sqrt{x}+4+\frac{4\left(\sqrt{x}+3\right)+x-4\sqrt{x}+4}{\sqrt{x}+3}+6\)
\(=\left(\sqrt{x}-2\right)^2+\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+3}+4+6\ge10\)Dấu = xảy ra tại x=4
\(\left[\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right]:\frac{x+16}{\sqrt{x}+2}\)
\(\left[\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right]:\frac{x+16}{\sqrt{x}+2}\)
\(=\left[\frac{\sqrt{x}\left(\sqrt{x}-4\right)}{x-16}+\frac{4\left(\sqrt{x}+4\right)}{x-16}\right]:\frac{x+16}{\sqrt{x}+2}\)
\(=\left[\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\right].\frac{\sqrt{x}+2}{x+16}\)
\(=\frac{x+16}{x-16}.\frac{\sqrt{x}+2}{x+16}\)
\(=\frac{\sqrt{x}+2}{x-16}\)
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b, MA-Bfrac{sqrt{x}+2}{sqrt{x}+3}-left(frac{5}{x+sqrt{x}-6}+frac{1}{sqrt{x}-2}right)frac{sqrt{x}+2}{sqrt{x}+3}-frac{5}{x+sqrt{x}-6}-frac{1}{sqrt{x}-2}frac{left(sqrt{x}+2right)left(sqrt{x}-2right)}{x+sqrt{x}-6}-frac{5}{x+sqrt{x}-6}-frac{1left(sqrt{x}+3right)}{x+sqrt{x}-6}frac{x-4-5-sqrt{x}-3}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{x-sqrt{x}-12}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{x-4sqrt{x}+3sqrt{x}-12}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{left(sqrt{x}-4right)left(sqrt{x...
Đọc tiếp
b, \(M=A-B=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\left(\frac{5}{x+\sqrt{x}-6}+\frac{1}{\sqrt{x}-2}\right)\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{x+\sqrt{x}-6}-\frac{5}{x+\sqrt{x}-6}-\frac{1\left(\sqrt{x}+3\right)}{x+\sqrt{x}-6}\)
\(=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+3\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)\(=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
bạn trung học hay tiểu học vậy
\(A=\frac{\sqrt{x}+4}{\sqrt{x}+2};B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)
a. rut gon B
b. Tim x nguyen de P = B(A-1) nguyen
giup minh voi a
A=\(\left(\frac{\sqrt{x}-1}{x-4}-\frac{\sqrt{x+1}}{x+4\sqrt{x+4}}\right)\):\(\frac{x\sqrt{x}}{\left(4-x\right)^2}\)
B=\(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\right):\frac{x\sqrt{x}}{\left(4-x\right)^2}\)
rút gọn các biểu thức
\(B=\left(\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\right):\frac{x\sqrt{x}}{\left(4-x\right)^2}\)
\(=\frac{2x+18\sqrt{x}-x-9\sqrt{x}}{x-9}\cdot\frac{\left(4-x\right)^2}{x\sqrt{x}}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+9\right)\left(4-x\right)^2}{x\sqrt{x}\left(x-9\right)}\)
\(=\frac{\left(\sqrt{x}+9\right)\left(4-x\right)^2}{\sqrt{x}\left(x-9\right)}\)
Bài 1 :1) TÍNH A=\(\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)
2) rút gọn biểu thức B=\(\frac{\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right)}{\frac{x+16}{\sqrt{x}+2}}\)
LÀM ƠN GIẢI GIÚP
Rút gọn
a) \(\left(\frac{2+\sqrt{a}}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right)\left(\frac{a\sqrt{a}-\sqrt{a}-1}{\sqrt{a}}\right)\)
b) \(\left(\frac{\sqrt{x}+1}{x-4}-\frac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\left(\frac{x\sqrt{x}+2x+4\sqrt{x}-8}{\sqrt{x}}\right)\)
Bài 1. Tìm điều kiện các BPT sau
a, sqrt{20-x}sqrt{3x-6}+1
b, frac{sqrt{9-x^2}}{x-1}frac{1}{sqrt{x}}+1
c, x+frac{x+1}{sqrt{x-4}}2-frac{2}{x^2-25}
d, sqrt{x}sqrt{-x}
e, 3x+frac{4}{sqrt{x-5}}le9+frac{x}{x-6}
f, frac{x+2}{10+3x^2}ge7+frac{4}{left(3x+9right)^2}
g, frac{sqrt{x+2}}{sqrt{x-2}}+frac{1}{left(x-4right)left(x+6right)}lefrac{3}{sqrt{8-x}}
h, frac{sqrt{x+6}}{left|xright|-sqrt{x+6}}gesqrt{16-2x}
Đọc tiếp
Bài 1. Tìm điều kiện các BPT sau
a, \(\sqrt{20-x}>\sqrt{3x-6}+1\)
b, \(\frac{\sqrt{9-x^2}}{x-1}>\frac{1}{\sqrt{x}}+1\)
c, \(x+\frac{x+1}{\sqrt{x-4}}>2-\frac{2}{x^2-25}\)
d, \(\sqrt{x}>\sqrt{-x}\)
e, \(3x+\frac{4}{\sqrt{x-5}}\le9+\frac{x}{x-6}\)
f, \(\frac{x+2}{10+3x^2}\ge7+\frac{4}{\left(3x+9\right)^2}\)
g, \(\frac{\sqrt{x+2}}{\sqrt{x-2}}+\frac{1}{\left(x-4\right)\left(x+6\right)}\le\frac{3}{\sqrt{8-x}}\)
h, \(\frac{\sqrt{x+6}}{\left|x\right|-\sqrt{x+6}}\ge\sqrt{16-2x}\)
\(\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8\sqrt{x}}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)