Cho A= \(\frac{x-2}{x+2\sqrt{x}}-\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x}+2}\)
a) RG
b) tính A khi x= 4 +2\(\sqrt{3}\)
c)tìm x\(\in Z\)để A\(\in Z\)
cho B=\(\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right):\sqrt{x}-2+\frac{10-x}{\sqrt{x+2}}\)
a/ tìm điều kiện xác định của B
b/ rút gọn B
c/ tìm x \(\in\)Z để B \(\in\)Z
d/ tìm x để B=3
e/ Tính B khi x=?
1/ Cho biểu thức \(A=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{3\sqrt{x}-3}{x-5\sqrt{x}+6}\)
a)Tìm các giá trị của x để A<-1
b) Tìm các giá trị của \(x\in Z\) sao cho \(2A\in Z\)
2/ Cho \(A=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)tìm các giá trị của x để A>-6
1. Cho A = \(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
a) Rút gọn A
b) Tìm A khi x = \(\frac{1}{9}\)
c) Tìm x\(\in\)Z để A\(\in\)Z
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
a) \(A=\frac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(A=\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) Thay \(x=\frac{1}{9}\)vào A, ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=\frac{\frac{1}{3}+1}{\frac{1}{3}-3}=\frac{\frac{4}{3}}{\frac{-8}{3}}=-\frac{1}{2}\)
c) \(A\in Z\Rightarrow\frac{\sqrt{x}-3+4}{\sqrt{x}-3}\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\Leftrightarrow\sqrt{x}-3\)là ước số của 4 gồm:\(1;2;4;-1;-2;-4\)
\(\sqrt{x}-3=1\Leftrightarrow x=16\)(chọn)
\(\sqrt{x}-3=2\Leftrightarrow x=25\)(chọn)
\(\sqrt{x}-3=4\Leftrightarrow x=49\)(chọn)
\(\sqrt{x}-3=-1\Leftrightarrow x=4\)(loại)
\(\sqrt{x}-3=-2\Leftrightarrow x=1\)(chọn)
\(\sqrt{x}-3=-4\Leftrightarrow\sqrt{x}=-1\)(vô lý)
Vậy với \(x=1;16;25;49\)thì \(A\in Z\)
Cho C = \(\frac{3\sqrt{x}+2}{2\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+4}-\frac{x-6\sqrt{x}+5}{2x+7\sqrt{x}-4}.\)
a) rút gọn C
b) tìm x\(\in\)Z để C \(\in\)Z
c) tìm x để C > \(\frac{1}{2}\)
Bài 1 : Cho bt M = \(\left(\frac{\sqrt{x}}{x-4}\right)+\left(\frac{1}{\sqrt{x}-2}\right).\frac{\sqrt{x}-2}{2}\)
a) tim dkxd va RG A .
b) tim \(x\in Z\)de \(2M\in Z\)
Bài 2 : cho bt A = \(\left(\frac{\sqrt{x}+1}{x-2\sqrt{x}}-\frac{1}{\sqrt{x}-2}\right).\left(x-3\sqrt{x}+2\right)\)
a) tìm ĐKXĐ và RG A
b) tìm x để \(A< \frac{1}{2}\)
c) tim \(x\in Z\)de \(A\in Z\)
Câu 1) a) ĐKXĐ \(x\ge0,\)\(x\ne4\)A=\(\frac{x+2\sqrt{x}-4}{2\left(x-4\right)}\)b) Mình chưa làm được Câu 2) a) ĐKXĐ \(x>0,\)\(x\ne4\)A=\(\frac{\sqrt{x}-1}{\sqrt{x}}\)b) Để a<\(\frac{1}{2}\)\(\Rightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}< \frac{1}{2}\)\(\Rightarrow x< 1\)\(\Rightarrow0< x< 1\)thỏa mãn bài toán c) Ta có A=\(\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\), để A \(\in Z\)\(\Rightarrow\sqrt{x}\inƯ\left(1\right)\), \(\Rightarrow x=1\)( thỏa mãn ĐK)
Cho biểu thức : \(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
a, Rút gọn A
b, Tìm x để A < 1
c, Tìm \(x\in Z\) để \(A\in Z\)
\(ĐKXĐ:\)
\(\hept{\begin{cases}x-9\ne0\\\sqrt{x}-2\ne0\\\sqrt{x}+3\ne0;x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ne4\\x\ge0\end{cases}}\)
Vậy...................................................
\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}-3}{\left(\sqrt{x}+3\right)}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4-x}\)
\(=\frac{3\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\frac{3}{\left(2+\sqrt{x}\right)}\)
Đế A<1 \(\Rightarrow\frac{3}{2+\sqrt{x}}< 1\)
\(\Leftrightarrow\frac{3}{2+\sqrt{x}}-1< 0\)
\(\Leftrightarrow\frac{3-2-\sqrt{x}}{2+\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{1-\sqrt{x}}{2+\sqrt{x}}< 0\)
Vì \(2+\sqrt{x}>0\forall x\in R\)
\(\Rightarrow1-\sqrt{x}< 0\)
\(\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
Kết hợp ĐKXĐ \(\Rightarrow\hept{\begin{cases}x>1\\x\ne4\\x\ne9\end{cases}}\)
cho A= \(\left(\frac{\sqrt{X}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)
a, rút gọn A
b, tính A khi x= \(5+2\sqrt{6}\)
c, tìm GTNN của A
d, tìm \(x\in Z\)để \(A\in Z\)
Cho M=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\) \(\left(x\ge0;x\ne4;x\ne9\right)\)
a, Rút gọn M
b, Tính M khi x=\(11-6\sqrt{2}\)
c, Tìm x để M<1
d, Tìm \(x\in Z\) để M\(\in Z\)
M = \(\frac{2\sqrt{x}-9x}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)}\)
=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{9-x+2x-3\sqrt{x}}{x-5\sqrt{x}+6}\)
=\(\frac{x-\sqrt{x}}{x-5\sqrt{x}+6}\)
1, cho biểu thức
\(A=\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}\)
a, rút gọn ( tìm đkxd )
b,tính A khi \(x=9-4\sqrt{5}\)
c, tìm x để A > -1
d, Tìm Min của A
e, tìm \(x\in Z\) để A nhận giá trị nguyên
Giúp mình nhé :3 nhanh nhanh tick cho nè <3
\(dkxd\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-2\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}}\)
\(A=\left(\frac{\sqrt{x}}{x-4}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}.\)
\(=\left(\frac{\sqrt{x}}{x-4}-\frac{2\left(\sqrt{x}+2\right)}{x-4}+\frac{\sqrt{x}-2}{x-4}\right):\frac{1}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+2}{1}\)
\(=\frac{-6\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-\frac{6}{\sqrt{x}-2}\)
\(A=\)\(\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\)\(:\frac{1}{\sqrt{x}+2}\)
a,ĐKXĐ:\(\hept{\begin{cases}x\ge0\\2-\sqrt{x}\\x-4\ne0\end{cases}\ne0}\)\(\Rightarrow\)\(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\)\(\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\)\(:\frac{1}{\sqrt{x}+2}\)
\(A=\)\(\left(\frac{\sqrt{x}}{x-4}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right)\)\(.\left(\sqrt{x}+2\right)\)
\(A=\)\(\left(\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)
\(A=\)\(\left(\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)
\(A=\)\(\left(\frac{-6}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)
\(A=\)\(\frac{-6}{\sqrt{x}-2}\)
b,\(x=9-4\sqrt{5}\)\(\Rightarrow\)\(A=\)\(\frac{-6}{\sqrt{9-4\sqrt{5}}-2}\)\(=\frac{-6}{\sqrt{5-2.2\sqrt{5}+4}-2}\)
\(A=\)\(\frac{-6}{\sqrt{\left(\sqrt{5}-2\right)^2}-2}\)\(=\frac{-6}{\sqrt{5}-2-2}\)\(=\frac{-6}{\sqrt{5}-4}\)
c,\(A>-1\)\(\Rightarrow\)\(\frac{-6}{\sqrt{x}-2}\)\(>-1\)\(\Rightarrow\)\(\frac{-6}{\sqrt{x}-2}+1>0\)
\(\Leftrightarrow\)\(\frac{-6+\sqrt{x}-2}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-8}{\sqrt{x}-2}>0\)
\(d,\frac{-6}{\sqrt{x}-2}\)nhỏ nhất \(\Leftrightarrow\frac{6}{\sqrt{x}-2}\)lớn nhất
\(\Rightarrow\sqrt{x}-2\)nhỏ nhất \(\Rightarrow\sqrt{x}=0\Leftrightarrow x=0\)
\(\Rightarrow A_{min}=-2\Leftrightarrow x=0\)
\(e,\)\(A\in Z\Leftrightarrow\frac{6}{\sqrt{x}-2}\in Z\)\(\Leftrightarrow\sqrt{x}-2\inƯ_6\)
Mà \(Ư_6=\left\{\pm1;\pm2;\pm3;\pm6\right\}\Rightarrow...\)