A = ( \(\sqrt{x}\)+1 -\(\frac{4\sqrt{x}}{\sqrt{x}+1}\)) : (\(\frac{\sqrt{x}}{\sqrt{x}+1}\)- \(\frac{1}{1-\sqrt{x}}\)- \(\frac{2-\sqrt{x}}{x-1}\)
a) Tìm điều kiện để biểu thức A được xác định
b) Rút gọn
c) Tìm x để A < 1
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B =\(\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\) + \(\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)- \(\frac{\sqrt{x}+3}{\sqrt{x}-2}\)( \(x\ge0\); \(x\ne2;3\))
= \(\frac{2\sqrt{x}-9+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b, B = \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)= \(\frac{\sqrt{x}-3+4}{\sqrt{x}-3}\)= \(1+\frac{4}{\sqrt{x}-3}\)
để B có gtri nguyên thì \(\frac{4}{\sqrt{x}-3}\)phải nguyên
\(\Rightarrow\left(\sqrt{x}-3\right)\varepsilonƯ\left(4\right)\)
\(\Rightarrow\left(\sqrt{x}-3\right)\varepsilon\left\{1;-1;2;-2;4;-4\right\}\)
ta có bảng sau
\(\sqrt{x}-3\) 1 -1 2 -2 4 -4
\(\sqrt{x}\) 4 2 5 1 7 -1 (L)
x 16 4 25 1 49
vậy x \(\varepsilon\){ 16 ; 4 ; 25; 1 ; 49 }
#mã mã#
Cách liên hợp
ĐK \(x\ge-2\)
PT <=> \(\sqrt{x+2}+5x+2\ne0\)
\(25x^2+19x+2+2\left(x+1\right)\left(\sqrt{x+2}-5x-2\right)=0\)
Xét \(\sqrt{x+2}+5x+2=0\)=> \(x=\frac{-19-\sqrt{161}}{50}\)
Thay vào ta thấy nó không phải là nghiệm của PT
=> \(\sqrt{x+2}+5x+2\ne0\)
<=> \(25x^2+19x+2+2\left(x+1\right).\frac{x+2-\left(5x+2\right)^2}{\sqrt{x+2}+5x+2}=0\)
<=> \(25x^2+19x+2+2\left(x+1\right).\frac{-25x^2-19x-2}{\sqrt{x+2}+5x+2}=0\)
<=> \(\orbr{\begin{cases}25x^2+19x+2=0\\1-\frac{2\left(x+1\right)}{\sqrt{x+2}+5x+2}=0\left(2\right)\end{cases}}\)
Pt (2)
<=> \(\sqrt{x+2}=-3x\)
<=> \(\hept{\begin{cases}x\le0\\9x^2-x-2=0\end{cases}}\)=> \(x=\frac{1-\sqrt{73}}{18}\)(TM ĐKXĐ)
Pt (1) có nghiệm \(x=\frac{-19+\sqrt{161}}{50}\)(Tm ĐKXĐ)
Vậy Pt có nghiệm \(S=\left\{\frac{1-\sqrt{73}}{18};\frac{-19+\sqrt{161}}{50}\right\}\)
Cách đặt ẩn phụ không hoàn toàn
ĐK\(x\ge-2\)
PT
<=> \(15x^2+6x+2\left(x+1\right)\sqrt{x+2}-\left(x+2\right)=0\)
Đặt \(\sqrt{x+2}=a\left(a\ge0\right)\)
=> \(15x^2+6x+2\left(x+1\right).a-a^2=0\)
<=> \(\left(15x^2+2ax-a^2\right)+\left(6x+2a\right)=0\)
<=> \(\left(5x-a\right)\left(3x+a\right)+2\left(3x+a\right)=0\)
<=> \(\left(3x+a\right)\left(5x-a+2\right)=0\)
<=> \(\orbr{\begin{cases}3x+a=0\\5x-a+2=0\end{cases}}\)
+ 3x+a=0
=> \(3x+\sqrt{2+x}=0\)
=> \(\hept{\begin{cases}x\le0\\9x^2-x-2=0\end{cases}}\)=> \(x=\frac{1-\sqrt{73}}{18}\)(TM ĐKXĐ)
+ 5x-a+2=0
=> \(5x+2=\sqrt{x+2}\)
=> \(\hept{\begin{cases}x\ge-\frac{2}{5}\\25x^2+19x+2=0\end{cases}}\)=> \(x=\frac{-19+\sqrt{161}}{50}\)(TM ĐKXĐ)
vậy \(S=\left\{\frac{-19+\sqrt{161}}{50};\frac{1-\sqrt{73}}{18}\right\}\)
Biến đổi vế trái :
\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}.\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\left(1-a\sqrt{a}+\sqrt{a}-a\right)\frac{1-\sqrt{a}}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a-\sqrt{a}+a.\left(\sqrt{a}\right)^2-\left(\sqrt{a}\right)^2+a\sqrt{a}}{\left(1-a\right)^2}\)
\(=\frac{a^2-2a+1}{\left(1-a\right)^2}\)
\(=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}\)
\(=\left(\frac{a-1}{1-a}\right)^2=\left(-1\right)^2=1=VP\left(đpcm\right)\)
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=\left(\frac{1-\sqrt{a}^3}{1-\sqrt{a}}+\sqrt{a}\right)\left[\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]^2\)
\(=\left[\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right].\left(\frac{1}{1+\sqrt{a}}\right)^2\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right).\left(\frac{1}{1+\sqrt{a}}\right)^2\)
\(=\left(1+2\sqrt{a}+a\right).\frac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=\left(1+\sqrt{a}\right)^2.\frac{1}{\left(1+\sqrt{a}\right)^2}=1\)( đpcm )