Cho biểu thức \(Q=\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right):\left(1-\frac{2\sqrt{x}}{x+1}\right)ĐKXĐ:x\ge0;x\ne1\)
a, Rút gọn Q
b, Tìm x sao cho Q < 0.
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8 tháng 3 2022
a, Với x >= 0 ; x khác 16
\(A=\left(\frac{x+5\sqrt{x}-27+\left(3-\sqrt{x}\right)\left(\sqrt{x}+4\right)}{x-16}\right):\frac{1}{\sqrt{x}+4}\)
\(=\left(\frac{x+5\sqrt{x}-27+3\sqrt{x}+12-x-4\sqrt{x}}{x-16}\right):\frac{1}{\sqrt{x}+4}\)
\(=\left(\frac{4\sqrt{x}-15}{x-16}\right):\frac{1}{\sqrt{x}+4}=\frac{4\sqrt{x}-15}{\sqrt{x}-4}\)
b, Ta có \(B=-2A\Rightarrow\sqrt{x}-4=-\frac{8\sqrt{x}-30}{\sqrt{x}-4}\)
\(\Leftrightarrow x-8\sqrt{x}+16=-8\sqrt{x}+30\Leftrightarrow x-14=0\Leftrightarrow x=14\left(tm\right)\)
16 tháng 5 2021
a, Với \(x\ge0;x\ne1\)
\(Q=\left(\frac{x-1}{\sqrt{x}-1}-\frac{x\sqrt{x}-1}{x-1}\right):\left(\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1-\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x-1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1-\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x+2\sqrt{x}+1-x-\sqrt{x}-1}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
LT
20 tháng 9 2018
Ai trả lời nhanh và chính xác mình k
Lê Thị Tuyết
N
26 tháng 3 2019
\(P=\dfrac{x\sqrt{x}-x-\sqrt{x}-2}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)^2}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)\left(x-1\right)}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)\left(1-x^2\right)}{2\left(x+\sqrt{x}+1\right)}\)
LM
8 tháng 11 2020
A=\(\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
=\(\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x-2}}\)
Vậy A=\(\frac{\sqrt{x}}{\sqrt{x}-2}\)vs x\(\ge0;x\ne4\)
LM
9 tháng 11 2020
C=\(\left(\frac{1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\times\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{1+x}{\sqrt{x}}\)
Vậy C=\(\frac{1+x}{\sqrt{x}}\)vs x>0
16 tháng 7 2020
1: Ta có: \(Q=\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}-\frac{x\sqrt{x}-1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{x+\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\frac{x-2\sqrt{x}+x\sqrt{x}-x\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}:\frac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\frac{x-2\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}:\frac{x-1}{x+\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}\cdot\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
2: Ta có: \(\frac{1}{Q}=4\sqrt{x}-4\)
\(\Leftrightarrow Q=\frac{1}{4\sqrt{x}-4}\)
\(\Leftrightarrow\frac{x+\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{4\sqrt{x}-4}\)
\(\Leftrightarrow\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=\left(x+\sqrt{x}+1\right)\left(4\sqrt{x}-4\right)\)
\(\Leftrightarrow x+x\sqrt{x}-\sqrt{x}-1=4x\sqrt{x}-4\)
\(\Leftrightarrow x+x\sqrt{x}-\sqrt{x}-1-4x\sqrt{x}+4=0\)
\(\Leftrightarrow x-3x\sqrt{x}-\sqrt{x}+3=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-\left(3x\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-3\left(x\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left[\sqrt{x}-3\left(x+\sqrt{x}+1\right)\right]=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-3x-3\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(-3x-2\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)=0\)(vì \(-3x-2\sqrt{x}-3\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\sqrt{x}=1\)
hay x=1(không thỏa mãn ĐKXĐ)
Vậy: Không có giá trị nào của x thỏa mãn \(\frac{1}{Q}=4\sqrt{x}-4\)
31 tháng 7 2020
a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(=\left(\frac{1-x\sqrt{x}+\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\cdot\left(\frac{1}{1+\sqrt{x}}\right)^2\)
\(=\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-\left(x-1\right)\left(-1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{\left(1+\sqrt{x}\right)\cdot\left(-1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-1\cdot\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=-1\)