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Bài 1 :
+) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a) Ta có :
\(x=4-2\sqrt{3}\)
\(\Leftrightarrow x=3-2\sqrt{3}+1\)
\(\Leftrightarrow x=\left(\sqrt{3}-1\right)^2\)( Thỏa mãn ĐKXĐ )
Vậy tại \(x=\left(\sqrt{3}-1\right)^2\)thì giá trị của biểu thức A là :
\(A=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}+1}{\sqrt{\left(\sqrt{3}-1\right)^2}-3}=\frac{\sqrt{3}-1+1}{\sqrt{3}-1-3}=\frac{\sqrt{3}}{\sqrt{3}-4}=\frac{-\sqrt{3}\left(\sqrt{3}+4\right)}{7}\)
b)
\(B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\)
\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\frac{-3-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
Ta có :
\(P=A:B\)
\(\Leftrightarrow P=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{-3\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\frac{-\sqrt{x}-3}{3}\)
c) \(P=\frac{-\sqrt{x}-3}{3}\ge0\)
Dấu bằng xảy ra
\(\Leftrightarrow-\sqrt{x}-3=0\)
\(\Leftrightarrow\sqrt{x}=-3\)( vô lí )
Vậy không tìm được giá trị nào của x để P đạt GTNN
mình giúp bài 3 cho
\(\sqrt{25x-125}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=6\left(ĐKXĐ:x\ge5\right)\)
\(< =>\sqrt{25\left(x-5\right)}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=6\)
\(< =>\sqrt{25}.\sqrt{x-5}-3\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=6\)
\(< =>5.\sqrt{x-5}-3.\frac{\sqrt{x-5}}{3}-\frac{1}{3}.3.\sqrt{x-5}=6\)
\(< =>5.\sqrt{x-5}-\sqrt{x-5}-\sqrt{x-5}=6\)
\(< =>3\sqrt{x-5}=6< =>\sqrt{x-5}=2\)
\(< =>x-5=4< =>x=4+5=9\left(tmđk\right)\)
giải giúp mình bài này ới ạ mình đng cần gấp
Cho biểu thức
c=(căng x-2/căng x+2+căng x+2/căng x-2)nhân căng x+2/2 - 4 căng x/căng x-2
a)
\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{a-9}\)
\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\frac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{a-3\sqrt{a}+3+3\sqrt{a}-3a-9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{-2a-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{-2a-3}{a-9}\)
b) Để \(P=\frac{1}{3}\Rightarrow\frac{-2a-3}{a-9}=\frac{1}{3}\)
\(\Rightarrow3\left(-2a-3\right)=a-9\)
\(\Rightarrow-6a-9=a-9\)
\(\Rightarrow-6a-a=-9+9\)
\(\Rightarrow-7a=0\left(L\right)\)
Vậy ko có gt của a để P=1/3 ( mk ko chắc.....)
2. b,
\(\sqrt{3x^2-2x}+3=2x\)
ĐKXĐ: \(\left[{}\begin{matrix}x\ge\frac{2}{3}\\x\le0\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{3x^2-2x}=2x-3\\ \Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\3x^2-2x=\left(2x-3\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\3x^2-2x=4x^2-12x+9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\4x^2-3x^2-12x+2x+9=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\x^2-10x+9=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\\left(x-1\right)\left(x-9\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow x=9\)
Vậy phương trình có 1 nghiệm duy nhất là x = 9.
2.a,
\(9\sqrt{\frac{4x-8}{9}}-5\sqrt{\frac{16x-32}{25}}+18\sqrt{\frac{25x^2-100}{81}}=15\sqrt{x^2-4}\)
ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow9\sqrt{\frac{4\left(x-2\right)}{9}}-5\sqrt{\frac{16\left(x-2\right)}{25}}+18\sqrt{\frac{25\left(x^2-4\right)}{81}}=15\sqrt{x^2-4}\)
\(\Leftrightarrow9.\frac{2}{3}\sqrt{\left(x-2\right)}-5.\frac{4}{5}\sqrt{\left(x-2\right)}+18.\frac{5}{9}\sqrt{\left(x-2\right)\left(x+2\right)}=15\sqrt{\left(x-2\right)\left(x+2\right)}\)\(\Leftrightarrow6\sqrt{\left(x-2\right)}-4\sqrt{\left(x-2\right)}+10\sqrt{\left(x-2\right)\left(x+2\right)}=15\sqrt{\left(x-2\right)\left(x+2\right)}\)\(\Leftrightarrow6\sqrt{\left(x-2\right)}-4\sqrt{\left(x-2\right)}+10\sqrt{\left(x-2\right)\left(x+2\right)}-15\sqrt{\left(x-2\right)\left(x+2\right)}=0\)\(\Leftrightarrow\sqrt{x-2}\left(6-4+10\sqrt{x+2}-15\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\sqrt{x-2}\left(2-5\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\2-5\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=\frac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=\frac{4}{25}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tmDKXD\right)\\x=-\frac{11}{6}\left(khongtmDKXD\right)\end{matrix}\right.\)
Vậy pt có 1 nghiệm duy nhất là x = 2.