Bài 8: Rút gọn biểu thức chứa căn bậc hai
Các câu hỏi tương tự
Adfrac{sqrt{x}+2}{sqrt{x}-2}-dfrac{3}{sqrt{x}+2}+dfrac{12}{x-4}Bdfrac{sqrt{x}}{sqrt{x}-3}-dfrac{sqrt{x}-21}{9-x}dfrac{1}{sqrt{x}+3}Cdfrac{sqrt{x}}{sqrt{x}+3}+dfrac{2sqrt{x}}{sqrt{x}-3}-dfrac{3x+9}{x-9}Ddfrac{1}{sqrt{x}+3}-dfrac{sqrt{x}}{3-sqrt{x}}+dfrac{2sqrt{x}+12}{x-9}Ndfrac{sqrt{x}}{sqrt{x}-1}+dfrac{2sqrt{x}-1}{sqrt{x}+1}-dfrac{6}{x-1}Mdfrac{3}{sqrt{x}-3}+dfrac{2}{sqrt{x}+3}+dfrac{x-5sqrt{x}-3}{x-9}
Đọc tiếp
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{x-4}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{\sqrt{x}-21}{9-x}\dfrac{1}{\sqrt{x}+3}\)
\(C=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(D=\dfrac{1}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}+\dfrac{2\sqrt{x}+12}{x-9}\)
\(N=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{6}{x-1}\)
\(M=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)
\(B=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9\sqrt{x}}{9-x};\left(x\ge0;x\ne9;x\ne16\right)\)
\(B=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)
\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1};\left(x>0;x\ne1\right)\)
\(A= (\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}):(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1)\)
a) Rút gọn A
b) Tìm x khi A=1
Rút gọn biểu thức:
\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
A= \(\left[\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\right)+\dfrac{\sqrt{x}}{\sqrt{x}+3}+3\left(\dfrac{\sqrt{x}}{x-9}\right)\right]:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{1}{1}\right)\)với x>= 0 , x #9
Rút gọn biểu thức
A=\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}\)- \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)+\(\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
16 tháng 7 2021 lúc 17:15
Cho \(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
a, Rút gọn Q
b, Tính Q biết \(x=6+4\sqrt{2}\)
c, Tìm xϵZ để QϵZ
Rút gọn
A=\(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right)\div\left(\dfrac{\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
B=\(\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)\div\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}+1\)
Rút gọn:
A=\(\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right)\div\dfrac{1-\sqrt{x}}{2-\sqrt{x}}vớix>0,x\ne1\)
B=\(\left(\dfrac{x}{3+\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right)\div\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Lm nhanh giúp mk nhé!
Cho biểu thức D = \(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
với \(x\ne9,x\ge0\)
a) Rút gọn D
b)Tìm x để \(D< \dfrac{-1}{4}\)