Cho \(A=\frac{x-1}{\sqrt{x}-3},B=\frac{10\sqrt{x}+12}{x-9}+\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{1}{3-\sqrt{x}}\left(x\ge0,x\ne9\right)\)
a. Rút gọn B
b. Biết C = A : B. Tìm minC
Rút gọn biểu thức:
a) \(A=\left(\frac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}\left(x\ge0,x\ne1\right)\)
b) \(B=\frac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\frac{2\left(\sqrt{x-3}\right)}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{3-\sqrt{x}}\left(x>0,x\ne9\right)\)
c) \(C=\frac{2\sqrt{x}-9}{x-5+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,x\ne9\right)\)
cho biểu thức: P=\(\left[1-\frac{x-3\sqrt{x}}{x-9}\right]:\left[\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9x}{x+\sqrt{x}-6}\right]\) \(\left(x\ge0;x\ne9;x\ne4\right)\)
a) Rút gọn P
b) Tìm giá trị của x để P=1
a/ \(P=\left[1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left(\frac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{9-x+x-4\sqrt{x}+4-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\frac{3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{13-4\sqrt{x}-9x}\)
\(=\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}\)
b/ \(P=1\Rightarrow\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}=1\Rightarrow3\sqrt{x}-6=13-4\sqrt{x}-9x\)
\(\Rightarrow9x+7\sqrt{x}-19=0\)
Mình k biết mình sai chỗ nào nữa, bạn xem giúp mình với
Cho biểu thức: B = \(\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}+\frac{3x-9}{x-9}\right):\left(\frac{\sqrt{x}-2}{3}+1\right)\)với \(x\ge0;x\ne9\)
Rút gọn B
Cho biểu thức C=\(\frac{x\sqrt{x}}{x-2\sqrt{x}-3}+\frac{\sqrt{x}+3}{3-\sqrt{x}}-\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)với \(x\ge0,x\ne9.\)
a/Rút gọn biểu thức C
b/Tìm x để biểu thức C đạt giá trị nhỏ nhất.
Giúp tớ nhanh nhanh nha!Cảm ơn rất rất nhiều.
Giúp mình với:
Cho b/t A=\(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(x+3\right)}{x-9}\) (\(x\ge0\) ; \(x\ne9\) )
Rút gọn b/t A rồi tính giá trị tại x= \(2\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
Mai Kt rồi :( :( :(
a. \(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(x+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\sqrt{x}+3}\)
. \(x=2.\left(4+\sqrt{15}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
\(\Rightarrow x=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^3\)\(=4\left(\sqrt{5}-\sqrt{3}\right)\)
Thay \(x=4\left(\sqrt{5}-\sqrt{3}\right)\Rightarrow A=\frac{3}{\sqrt{4\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
\(=\frac{3}{2\sqrt{\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
\(A=\left(\frac{\sqrt{x}-2}{\sqrt{x}-3}+\frac{\sqrt{x}+1}{\sqrt{x}+3}+\frac{x-4\sqrt{x}-9}{9-x}\right):\frac{\sqrt{x}+5}{3-\sqrt{x}}\)với \(x\ge0,x\ne9\)
a, Rút gọn
b, Tìm x để A = 1
c, Tìm x để \(\left|A\right|< \frac{1}{2}\)
P/s: Không chắc lắm nha!
Cho biểu thức:\(A=\left(1-\frac{\sqrt{x}}{1+\sqrt{x}}\right)\)\(:\)\(\left(\frac{\sqrt{x}+3}{\sqrt{x-2}}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)Với:\(x\ge0,x\ne4,x\ne9\)
a/Rút gọn A
b/Tìm các giá trị nguyên của x để A nhận giá trị nguyên
Cho biểu thức
A = \(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\left(x\ge0\right),x\ne9\)
a) Rứt gọn biểu thức
a) \(A=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
\(A=\left(\frac{1}{\sqrt{x}+3}-\frac{4}{9-x}\right).\frac{2\sqrt{x}-6}{\sqrt{x}+1}\) với \(x\ge0;x\ne9\)
a) rút gọn và CMR \(A=\frac{2}{\sqrt{x}+3}\)
a) \(A=\left(\frac{1}{\sqrt{x}+3}-\frac{4}{9-x}\right).\frac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(A=\left[\frac{\sqrt{x}-3}{x-9}+\frac{4}{x-9}\right].\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}-3+4}{x-9}.\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}+1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(A=\frac{2}{\sqrt{x}+3}\)
vậy \(A=\frac{2}{\sqrt{x}+3}\)