Cho biểu thức sau : \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
a.Rút gọn P
b.tìm GTNN của P
c. tìm x để \(Q=\dfrac{2\sqrt{x}}{P}\) nhận giá trị nguyên
cho biểu thức
Q=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
a.rút gọn Q
b.tìm số nguyên x để Q có gtri nguyên
\(\left(đk:x\ne\pm1\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\dfrac{x-\sqrt{x}+2\sqrt{x}-2-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}}{x-1}\)
Cho biểu thức P=\(\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x-3}\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
a.Rút gọn P
b.Tính giái trị của P khi x=\(\dfrac{3-2\sqrt{2}}{4}\)
Sửa đề: \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\dfrac{9}{4}\end{matrix}\right.\)
a) Ta có: \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{2\cdot\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}:\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\cdot\dfrac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{2x+2\sqrt{x}+\sqrt{x}+1}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
b) Ta có: \(x=\dfrac{3-2\sqrt{2}}{4}\)
\(\Leftrightarrow x=\dfrac{2-2\cdot\sqrt{2}\cdot1+1}{4}\)
\(\Leftrightarrow x=\dfrac{\left(\sqrt{2}-1\right)^2}{4}\)(thỏa ĐK)
Thay \(x=\dfrac{\left(\sqrt{2}-1\right)^2}{4}\) vào biểu thức \(P=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\), ta được:
\(P=\left(3\cdot\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{4}}-5\right):\left(2\cdot\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{4}}+1\right)\)
\(\Leftrightarrow P=\left(3\cdot\dfrac{\sqrt{2}-1}{2}-5\right):\left(2\cdot\dfrac{\sqrt{2}-1}{2}+1\right)\)
\(\Leftrightarrow P=\left(\dfrac{3\cdot\left(\sqrt{2}-1\right)}{2}-\dfrac{10}{2}\right):\left(\sqrt{2}-1+1\right)\)
\(\Leftrightarrow P=\dfrac{3\sqrt{2}-3-10}{2}:\sqrt{2}\)
\(\Leftrightarrow P=\dfrac{3\sqrt{2}-13}{2}\cdot\sqrt{2}\)
\(\Leftrightarrow P=\dfrac{6-13\sqrt{2}}{2}\)
Vậy: Khi \(x=\dfrac{3-2\sqrt{2}}{4}\) thì \(P=\dfrac{6-13\sqrt{2}}{2}\)
Cho biểu thức P = \(\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{2-\sqrt{x}}\) (với x>0; x\(\ne\)0)
a,Rút gọn biểu thức P và tìm x để P = \(\dfrac{-3}{5}\)
b,Tìm GTNN của biểu thức A=P . \(\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)
cho biểu thức
Q=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
a.rút gọn Q
b.tìm số nguyên x để Q có gtri nguyên
đáp an: a.Q=\(\dfrac{2}{x-1}\)
b.x=-1;x=0;x=2;x=3 thì Q\(\in Z\)
mk chỉ bt đáp án chứ ko biết cách giải
ĐKXĐ: \(x>0;x\ne1\)
\(Q=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{1}{\sqrt{x}}\)
\(=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right).\dfrac{1}{\sqrt{x}}\)
\(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{1}{\sqrt{x}}=\dfrac{2}{x-1}\)
b.
Để \(Q\in Z\Rightarrow2⋮\left(x-1\right)\Rightarrow x-1=Ư\left(2\right)\)
\(\Rightarrow x-1=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow x=\left\{-1;0;2;3\right\}\)
Kết hợp ĐKXĐ: \(\Rightarrow x=\left\{2;3\right\}\)
(Đáp án của đề bài đã quên mất ĐKXĐ ban đầu nên ko loại 2 giá trị \(x=-1;x=0\))
cho biểu thức
A=\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
a.rút gọn biểu thức A
b. chứng minh 0<A<2
Điều kiện: \(x\ge0,x\ne1\)
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}+1}{x\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)
Ta có \(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0,\forall x\Rightarrow A>0\)
Lại có: \(A-2=\dfrac{2}{x+\sqrt{x}+1}-2=\dfrac{-2\left(x+\sqrt{x}\right)}{x+\sqrt{x}+1}\)
Mà \(x+\sqrt{x}+1>0;x+\sqrt{x}>0\) với mọi \(x\in TXĐ\)
\(\Rightarrow A-2< 0\Rightarrow A< 2\)
Vậy \(0< A< 2\)
A.rút gọn biểu thức a
b.tính giá trị của x khi a>1/6
A=\(\left(\dfrac{1}{\sqrt{x}-1}\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)vớix>0,xkhac1\)
Ngoặc thứ nhất dấu giữa 2 phân số là gì vậy bạn?
Cho biểu thức:
A=\(\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{4-x}+\dfrac{1}{2+\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}-1\right)\)(với x>0;x\(\ne\)4)
a.Rút gọn A
b.Tìm x để A<-1
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{4-x}+\dfrac{1}{2+\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}-1\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\left(\dfrac{2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}}\right)\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}}\)
\(=\dfrac{-4}{\sqrt{x}+2}\)
Lời giải:
a)
\(A=\left[\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\right].\frac{2-\sqrt{x}}{\sqrt{x}}\)
\(=\frac{\sqrt{x}+2+2\sqrt{x}+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{2-\sqrt{x}}{\sqrt{x}}=\frac{4\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{2-\sqrt{x}}{\sqrt{x}}=\frac{-4}{\sqrt{x}+2}\)
b)
$A< -1\Leftrightarrow \frac{-4}{\sqrt{x}+2}+1< 0$
$\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+2}< 0$
$\Leftrightarrow \sqrt{x}-2< 0\Leftrightarrow 0\leq x< 4$
Kết hợp với ĐKXĐ suy ra $0< x< 4$
ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
\(A=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{4-x}+\dfrac{1}{2+\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}-1\right)\)
\(=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{2+\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}+2+2\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{2-\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{2-\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{-2}{\sqrt{x}+2}\)
b/ \(A< -1\)
\(\Leftrightarrow\dfrac{-2}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{\sqrt{x}+2}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}< 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-2< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\)
Vậy..
Câu 6: Cho biểu thức Q = \(\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)\)
a) Tìm x để biểu thức Q xác định và rút gọn biểu thức Q
b) Tìm các giá trị của x để biểu thức Q có giá trị âm
a) ĐKXĐ: \(x>0;x\ne4\)
\(Q=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
b) Để biểu thức \(Q\) có giá trị âm thì \(\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)
\(\Rightarrow\sqrt{x}-2< 0\) (vì \(3\sqrt{x}>0\forall x>0;x\ne4\))
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\)
Kết hợp với điều kiện xác định của \(x\), ta được: \(0< x< 4\)
\(\text{#}\mathit{Toru}\)
Cho biểu thức sau: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
1, Rút gọn P
2, Tính giá trị nhỏ nhất của P
3, Tìm \(x\in Z\) sao cho \(Q=\dfrac{2\sqrt{x}}{P}\in Z\)
1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)