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ĐKXĐ: \(x\ge0;x\ne1\)
\(\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11+3x+7\sqrt{x}-6-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3x+19\sqrt{x}-14}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+7\right)\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\left(\frac{2x+1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(x+\sqrt{x}+1\right)}\right).\left(\frac{\sqrt{x}.\left(3+x\right)}{-2x}-\sqrt{x}\right) \)
\(A=\left(\frac{2x+1-\sqrt{x}.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\right).\left(\frac{3+x}{-2\sqrt{x}}-\sqrt{x}\right)\)
\(A=\left(\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\right).\left(\frac{3+x+2x}{-2\sqrt{x}}\right)\)
\(A=\left(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\right).\left(\frac{3x+3}{-2\sqrt{x}}\right)\)
\(A=\frac{1}{\sqrt{x}-1}.\frac{3.\left(x+1\right)}{-2\sqrt{x}}\)
\(A=\frac{3x+3}{-2\sqrt{x}.\left(\sqrt{x}+1\right)}\)
P/s: hình như đề sai hay sao á, thường thì người ta không cho mẫu là 2 số trừ được như ( x - 3x ) đâu
Dk: x\(\ge0\)
lien hop
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x+3}=2\Rightarrow x=1\)
c,Có x=\(\frac{1}{2}\left(\sqrt{\frac{1-a}{a}}-\sqrt{\frac{a}{1-a}}\right)\left(0< a< 1\right)\)
<=> \(x=\frac{1}{2}\left(\frac{\sqrt{1-a}}{\sqrt{a}}-\frac{\sqrt{a}}{\sqrt{1-a}}\right)\) (vì 0<a<1)
<=>\(x=\frac{1}{2}.\frac{\sqrt{1-a}^2-\sqrt{a}^2}{\sqrt{a}.\sqrt{1-a}}=\frac{1}{2}.\frac{1-a-a}{\sqrt{a\left(1-a\right)}}=\frac{1}{2}.\frac{1-2a}{\sqrt{a\left(1-a\right)}}=\frac{1-2a}{2\sqrt{a\left(1-a\right)}}\)(1)
<=> 1+x2=1+\(\frac{1}{4}.\frac{\left(1-2a\right)^2}{a\left(1-a\right)}\)= \(\frac{4a\left(1-a\right)+\left(1-2a\right)^2}{4a\left(1-a\right)}\)
<=> 1+x2=\(\frac{4a-4a^2+1-4a+4a^2}{4a\left(1-a\right)}=\frac{1}{4a\left(1-a\right)}\)>0
<=> \(\sqrt{1+x^2}=\frac{1}{2\sqrt{a\left(1-a\right)}}\) (2)
Thay (1),(2) vào C có:
C= \(\frac{2a.\frac{1}{2\sqrt{a\left(1-a\right)}}}{\frac{1}{2\sqrt{a\left(1-a\right)}}-\frac{1-2a}{2\sqrt{a\left(1-a\right)}}}=\frac{\frac{a}{\sqrt{a\left(1-a\right)}}}{\frac{1-1+2a}{2\sqrt{a\left(1-a\right)}}}=\frac{\frac{a}{\sqrt{a\left(1-a\right)}}}{\frac{2a}{2\sqrt{a\left(1-a\right)}}}=1\)
Vậy C=1
ĐK: \(x\ge-7\)
PT \(\Leftrightarrow\left(\sqrt[3]{x-8}-\left(x-8\right)\right)+\left[\sqrt{x+7}-4\right]+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\frac{-\left(x-9\right)\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}+\frac{x-9}{\sqrt{x+7}+4}+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left[x^2+x+2+\frac{1}{\sqrt{x+7}+4}-\frac{\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}\right]=0\)
\(\Leftrightarrow x=9\)
P/s:em chả biết đánh giá cái ngoặc to thế nào nữa:((((
