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a) \(\frac{1}{x-1+\sqrt{x^2-2x+3}}+\frac{1}{x-1-\sqrt{x^2-2x+3}}=1\)
ĐKXĐ : \(x\inℝ\)
\(\Leftrightarrow\frac{x-1-\sqrt{x^2-2x+3}}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}+\frac{x-1+\sqrt{x^2-2x+3}}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}=\frac{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}\)
\(\Rightarrow2x-2=\left[\left(x-1\right)+\left(\sqrt{x^2-2x+3}\right)\right]\left[\left(x-1\right)-\left(\sqrt{x^2-2x+3}\right)\right]\)
\(\Leftrightarrow2x-2=\left(x-1\right)^2-\left(\sqrt{x^2-2x+3}\right)^2\)
\(\Leftrightarrow2x-2=x^2-2x+1-\left(x^2-2x+3\right)\)
\(\Leftrightarrow2x-2=x^2-2x+1-x^2+2x-3\)
\(\Leftrightarrow2x-2=-2\)
\(\Leftrightarrow2x=0\)
\(\Leftrightarrow x=0\)
Vậy phương trình có nghiệm duy nhất x = 0
Sửa đề: \(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{5\sqrt{x}}{\sqrt{x}+3}=\dfrac{22}{x-9}\)
\(\Leftrightarrow x+5\sqrt{x}+6-5x+15\sqrt{x}=22\)
\(\Leftrightarrow-4x+20\sqrt{x}-16=0\)
\(\Leftrightarrow x-5\sqrt{x}+4=0\)
=>x=1 hoặc x=16
\(< =>\sqrt[3]{x+5}=-2\)
<=> \(\left(\sqrt[3]{x+5}\right)^3=-8\)
<=> \(x+5=-8\)
<=> x=-13
2. ĐK: \(x\ge0\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x}\ge0\\b=\sqrt{x^2+4}\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=2a^2\\x^2+4=b^2\\3\sqrt{x^3+4x}=3ab\end{matrix}\right.\)
pt trên được viết lại thành
\(2a^2+b^2-3ab=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=\dfrac{1}{2}b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{x^2+4}\\\sqrt{x}=\dfrac{1}{2}\sqrt{x^2+4}\end{matrix}\right.\)
Đến đây dễ rồi nhé ^^
b: \(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
=>x-3=0 hoặc x+3=9
=>x=3 hoặc x=6
c: \(\Leftrightarrow\sqrt{x^3-6x^2+9}=2x-6=\sqrt{4x^2-24x+36}\)
\(\Leftrightarrow x^3-6x^2+9-4x^2+24x-36=0\)
=>\(x^3-10x^2+24x-27=0\)
hay \(x\in\left\{7.18\right\}\)
3) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{4x-20}=4\)
\(\Leftrightarrow4x-20=16\)
\(\Leftrightarrow4x=36\)
\(\Leftrightarrow x=9\)
vậy ...
1)
\(A=\dfrac{\sqrt{x}-2}{x-4}=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}\right)^2-2^2}\\ A=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}\)
\(B=\dfrac{x^2-2x\sqrt{2}+2}{x^2-2}=\dfrac{x^2-2x\sqrt{2}+\left(\sqrt{2}\right)^2}{x^2-\sqrt{2}}\\ B=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\dfrac{\left(x-\sqrt{2}\right)}{\left(x+\sqrt{2}\right)}\)
\(C=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+\left(\sqrt{5}\right)^2}\\ C=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)
\(D=\dfrac{\sqrt{a}-2a}{2\sqrt{a}-1}=\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)}{2\sqrt{a}-1}=\sqrt{a}\)
\(E=\dfrac{x^2-2}{x-\sqrt{2}}=\dfrac{x^2-\left(\sqrt{2}\right)^2}{x-\sqrt{2}}\\ E=\dfrac{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}{x-\sqrt{2}}=x+\sqrt{2}\)
\(F=\dfrac{\sqrt{x}-3}{x-9}=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}\right)^2-3^2}\\ F=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ F=\dfrac{1}{\sqrt{x}+3}\)
Lời giải:
ĐK: \(x\geq 0; x\neq 9\)
PT tương đương:
\(\frac{(\sqrt{x}+2)(\sqrt{x}+3)-5\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{22}{x-9}\)
\(\Leftrightarrow \frac{-4x+20\sqrt{x}+6}{x-9}=\frac{22}{x-9}\)
\(\Rightarrow -4x+20\sqrt{x}+6=22\)
\(\Leftrightarrow -x+5\sqrt{x}-4=0\)
\(\Leftrightarrow x-5\sqrt{x}+4=0\)
\(\Leftrightarrow (\sqrt{x}-1)(\sqrt{x}-4)=0\)
\(\Rightarrow \left[\begin{matrix} \sqrt{x}=1\rightarrow x=1\\ \sqrt{x}=4\rightarrow x=16\end{matrix}\right.\) (đều thỏa mãn)