cho tam giác abc cân tại a.đường cao ah.biết BC=a,AH=h.Tính độ dài cạnh bên theo a,h
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\(P=\frac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{1-\sqrt{x}}\)(ĐK: \(x\ge0,x\ne1\))
\(=\frac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(3\sqrt{x}+8\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{3\sqrt{x}+8}{\sqrt{x}+2}\)
\(P< \frac{15}{4}\)
\(\Leftrightarrow\frac{3\sqrt{x}+8}{\sqrt{x}+2}< \frac{15}{4}\)
\(\Leftrightarrow4\left(3\sqrt{x}+8\right)< 15\left(\sqrt{x}+2\right)\)
\(\Leftrightarrow3\sqrt{x}>2\)
\(\Leftrightarrow x>\frac{4}{9}\)
Vậy \(x>\frac{4}{9},x\ne1\)thì \(P< \frac{15}{4}\).
\(P=\frac{3\sqrt{x}+8}{\sqrt{x}+2}=3+\frac{2}{\sqrt{x}+2}\le3+\frac{2}{0+2}=4\)
Dấu \(=\)khi \(x=0\).
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\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}\)
\(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}\)
\(a+\sqrt{a}+1\)
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ĐK: \(x\ge-1\).
\(\sqrt[3]{x-2}+\sqrt{x+1}=3\)
\(\Leftrightarrow\sqrt[3]{x-2}-1+\sqrt{x+1}-2=0\)
\(\Leftrightarrow\frac{x-2-1}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{x+1-4}{\sqrt{x+1}+2}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt[3]{\left(x-2\right)^2+\sqrt[3]{x-2}+1}}+\frac{1}{\sqrt{x+1}+2}\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)(thỏa mãn)
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https://h7.net/hoi-dap/toan-10/giai-phuong-trinh-can-bac-3-x-2-can-x-1-3-faq399261.html
link tham khảo nhé
chúc bạn hok tốt