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a) Với \(a\ne0,a\ne2\), hệ phương trình có nghiệm duy nhất:\(\left(x;y\right)=\left(\frac{a+1}{a};\frac{1}{a}\right)\)
Từ \(x=\frac{a+1}{a}=1+\frac{1}{a};y=\frac{1}{a}\Rightarrow x-y=1\)
b) Thay \(x=\frac{a+1}{a};y=\frac{1}{a}\) vào \(6x^2-17y=5\) ta được:
\(a^2-5a+6=0\Leftrightarrow\left(a-2\right)\left(a-3\right)=0\Leftrightarrow\left[{}\begin{matrix}a=2\\a=3\end{matrix}\right.\)
Kết hợp với điều kiện \(a\ne2\Rightarrow a=3\left(tm\right)\)
Ta xét : \(\left(n-1\right).n.\left(n+1\right)\left(n+2\right)+1=\left[\left(n-1\right)\left(n+2\right)\right].\left[n\left(n+1\right)\right]+1\)
\(=\left(n^2+n+2\right)\left(n^2+n\right)+1=\left(n^2+n\right)^2+2\left(n^2+n\right)+1=\left(n^2+n+1\right)^2\)
Suy ra \(A=12\sqrt{\left(n^2+n+1\right)^2}+23=12\left(n^2+n+1\right)+23=\left(2n+1\right)^2+\left(2n-3\right)^2+\left(2n+5\right)^2\)
ĐK: \(x-9\ne0\Rightarrow x\ne9\)
\(\sqrt{x}\ge0\Rightarrow x\ge0\)
\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)
\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)
2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)
Câu 3:
bạn cứ áp dụng cái \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
Câu 4:
từ giả thiết :\(a+b+c+\sqrt{abc}=4\Leftrightarrow\sqrt{abc}=4-a-b-c\Leftrightarrow abc=\left(4-a-b-c\right)^2\)
ta có: \(a\left(4-b\right)\left(4-c\right)=a\left(16-4c-4b+bc\right)=16a-4ac-4ab+abc\)
\(=16a-4ab-4ac+\left[4-\left(a+b+c\right)\right]^2=16a-4ab-4ac+16-8\left(a+b+c\right)+\left(a+b+c\right)^2\)
\(=a^2+b^2+c^2-2ab-2ac+2bc+8a-8b-8c+16\)
\(=\left(a-b-c\right)^2+8\left(a-b-c\right)+16=\left(a-b-c+4\right)^2\)
\(\Rightarrow\sqrt{a\left(4-b\right)\left(4-c\right)}=a-b-c+4\)(vì \(a-b-c+4=a-b-c+a+b+c+\sqrt{abc}=2a+\sqrt{abc}>0\))
các căn thức còn lại tương tự ...
ĐK: a\(\ge0,a\ne1\)
P=\(\left(2+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(2-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=\left[2+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right]\left[2-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right]=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a\)
Ta có \(\sqrt{\dfrac{\sqrt{2}-1}{1+\sqrt{2}}}=\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)
Ta lại có \(P=\sqrt{\dfrac{\sqrt{2}-1}{1+\sqrt{2}}}\Leftrightarrow\)\(4-a=\sqrt{2}-1\Leftrightarrow a=5-\sqrt{2}\)
Vậy a=\(5-\sqrt{2}\) thì \(P=\sqrt{\dfrac{\sqrt{2}-1}{1+\sqrt{2}}}\)