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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.
Dat \(a=\sqrt[3]{65+x},b=\sqrt[3]{65-x}\)
Bien doi PT thanh \(a^2+4b^2=5ab\)
\(\Leftrightarrow a^2-5ab+4b^2=0\)
\(\Leftrightarrow\left(a^2-ab\right)-\left(4ab-4b^2\right)=0\)
\(\Leftrightarrow a\left(a-b\right)-4b\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\left(1\right)\\a=4b\left(2\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow\sqrt[3]{65+x}=\sqrt[3]{65-x}\)
\(\Leftrightarrow65+x=65-x\)
\(\Leftrightarrow x=0\left(n\right)\)
\(\left(2\right)\Leftrightarrow\sqrt[3]{65+x}=4\sqrt[3]{65-x}\)
\(\Leftrightarrow65+x=64.65-64x\)
\(\Leftrightarrow65x=64.65-65\)
\(\Leftrightarrow x=63\left(n\right)\)
Vay nghiem cua PT la \(x=0,x=63\)
`a)(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4,x ne 9)`
`=(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9+(sqrtx-3)(sqrtx+3)+(2sqrtx+1)(sqrtx-2))/(x-5sqrtx+6)`
`=(2sqrtx-9+x-9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(3x-sqrtx-20)/
\(a,ĐK:\hept{\begin{cases}x\ge0\\\sqrt{x}+2\ne0\\\sqrt{x}-2\ne0;4-x\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
Rút gọn :
\(A=\frac{4}{\sqrt{x}+2}+\frac{2}{\sqrt{x}-2}+\frac{5\sqrt{x}-6}{4-x}\)
\(A=\frac{4}{\sqrt{x}+2}+\frac{2}{\sqrt{x}-2}-\frac{5\sqrt{x}-6}{x-4}\)
\(A=\frac{4\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{4\sqrt{x}-8+2\sqrt{x}+4-5\sqrt{x}+6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{1}{\sqrt{x}-2}\)
\(b,\)Để A nhận giá tri nguyên \(\Leftrightarrow\frac{1}{\sqrt{x}-2}\) nguyên
\(\Leftrightarrow\sqrt{x}-2\inƯ\left(1\right)\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=1\\\sqrt{x}-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=3\\\sqrt{x}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\x=1\end{cases}}}\)
Vậy A có giá tri nguyên \(\Leftrightarrow x\in\left\{1;9\right\}\)
\(A=\frac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}×\frac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\frac{1}{\sqrt{x}+2}\)
A đạt GTLN khi \(2+\sqrt{x}\)đạt GTNN hay x là nhỏ nhất. Vậy A đạt GTLN là \(\frac{1}{2}\)khi x = 0
a) \(G=\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{1-\frac{2}{x}+\frac{1}{x^2}}}\)
Tử : \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1\)
Mẫu : \(\sqrt{1-\frac{2}{x}+\frac{1}{x^2}}=\sqrt{\left(\frac{1}{x}-1\right)^2}=\left|\frac{1}{x}-1\right|\)
\(\Rightarrow G=\frac{\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1}{\left|\frac{1}{x}-1\right|}\)
b) \(x>2\Leftrightarrow\left\{{}\begin{matrix}x-1>1\Leftrightarrow\sqrt{x-1}>1\\\frac{1}{x}< 1\end{matrix}\right.\)
\(\Rightarrow G=\frac{\sqrt{x-1}-1+\sqrt{x-1}+1}{1-\frac{1}{x}}\)
\(G=\frac{2\sqrt{x-1}}{\frac{x-1}{x}}=\frac{2x\sqrt{x-1}}{x-1}=\frac{2x}{\sqrt{x-1}}\)
Để G nguyên thì \(2x⋮\sqrt{x-1}\)
\(\Leftrightarrow2x-2+2⋮\sqrt{x-1}\)
\(\Leftrightarrow2\left(x-1\right)+2⋮\sqrt{x-1}\)
Ta có \(2\left(x-1\right)⋮\sqrt{x-1}\)
\(\Rightarrow2⋮\sqrt{x-1}\)
\(\Rightarrow\sqrt{x-1}\inƯ\left(2\right)=\left\{1;2\right\}\)
\(\Leftrightarrow x-1\in\left\{1;4\right\}\)
\(\Leftrightarrow x\in\left\{2;5\right\}\)( thỏa )
Vậy....