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a) DK : x > 0; x khác 1
\(P=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}+1\)
c ) \(Q=\frac{2\sqrt{x}}{P}=\frac{2\sqrt{x}}{x-\sqrt{x}+1}\)
<=> \(xQ-\left(Q+2\right)\sqrt{x}+Q=0\)(1)
TH1: Q = 0 => x = 0 loại
TH2: Q khác 0
(1) là phương trình bậc 2 với tham số Q ẩn x.
(1) có nghiệm <=> \(\left(Q+2\right)^2-4Q^2\ge0\)
<=> \(-3Q^2+4Q+4\ge0\)
<=> \(-\frac{2}{3}\le Q\le2\)
Vì Q nguyên và khác 0 nên Q = 1 hoặc Q = 2
Với Q = 1 => \(x-3\sqrt{x}+1=0\)
<=> \(\sqrt{x}=\frac{3}{2}\pm\frac{\sqrt{5}}{2}\)----> Tìm được x
Với Q = 2 => \(2x-4\sqrt{x}+1=0\Leftrightarrow\sqrt{x}=1\pm\frac{1}{\sqrt{2}}\)-----> tìm đc x.
Tự làm tiếp nhé! Kiểm tra lại đề bài câu b.
\(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{x+\sqrt{x}+1-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-\left(x+\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{1}{x+\sqrt{x}+1}\cdot\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{x-1}{\left(x+\sqrt{x}+1\right)^2}\)
a) Ta có: \(P=\left(\frac{2-\sqrt{x}}{1-x}-\frac{\sqrt{x}-2}{x+2\sqrt{x}+1}\right):\frac{2}{x^2-2x+1}\)
\(=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\frac{\left(x-1\right)^2}{2}\)
\(=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\right)\cdot\frac{\left(x-1\right)^2}{2}\)
\(=\frac{x-\sqrt{x}-2-\left(x-3\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{x-\sqrt{x}-2-x+3\sqrt{x}-2}{2}\cdot\frac{\left(\sqrt{x}-1\right)}{1}\)
\(=\frac{2\sqrt{x}-4}{2}\cdot\left(\sqrt{x}-1\right)\)
\(=\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-1\right)\)
\(=x-3\sqrt{x}+2\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(P+x\le2\)
\(\Leftrightarrow x-3\sqrt{x}+2+x-2\le0\)
\(\Leftrightarrow2x-3\sqrt{x}\le0\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-3\right)\le0\)
Trường hợp 1: \(\sqrt{x}\left(2\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\2\sqrt{x}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=\frac{9}{4}\left(nhận\right)\end{matrix}\right.\)
Trường hợp 2: \(\sqrt{x}\left(2\sqrt{x}-3\right)< 0\)
mà \(\sqrt{x}\ge0\forall x\) thỏa mãn ĐKXĐ
nên \(2\sqrt{x}-3< 0\)
\(\Leftrightarrow2\sqrt{x}< 3\)
\(\Leftrightarrow\sqrt{x}< \frac{3}{2}\)
\(\Leftrightarrow x< \frac{9}{4}\)
mà \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
nên \(\left\{{}\begin{matrix}0< x< \frac{9}{4}\\x\ne1\end{matrix}\right.\)
Vây: Để \(P+x\le2\) thì \(\left\{{}\begin{matrix}0\le x\le\frac{9}{4}\\x\ne1\end{matrix}\right.\)
đkxđ: x≥0; x≠4
\(A=\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)
\(=\dfrac{2-\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{2+\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}-\dfrac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(=\dfrac{4-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2}{2+\sqrt{x}}\)
+) A = 1/4 <=> \(\dfrac{2}{2+\sqrt{x}}=\dfrac{1}{4}\Leftrightarrow2+\sqrt{x}=8\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)(tm)
Vậy x = 36
đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(A=\dfrac{2+\sqrt{x}+2-\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\)
\(A=\dfrac{4-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\)
\(A=\dfrac{2}{\sqrt{x}+2}\)
để \(A=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x}+2=8\)
\(\Leftrightarrow x=36\left(tm\right)\)
vậy tại x=36 thì A=1/4