tìm số nguyên lớn nhất không vượt quá ((3+căn5)/2)^7
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\(x+y+z=1\Rightarrow z=1-x-y\)Thay vào A ta được:
\(A=2xy+3y\left(1-x-y\right)+4\left(1-x-y\right)x\)
\(\Leftrightarrow2xy+3y-3xy-3y^2+4x-4x^2-4xy-A=0\)
\(\Leftrightarrow3y-3y^2+4x-4x^2-5xy-A=0\)
\(\Leftrightarrow-4x^2-\left(5y-4\right)x-3y^2+3y-A=0\)
\(\Leftrightarrow4x^2+\left(5y-4\right)x+3y^2-3y+A=0\)
\(\Delta=\left(5y-4\right)^2-16\left(3y^2-3y+A\right)\)
Để pt có nghiệm \(\Leftrightarrow\Delta\ge0\)
\(\Leftrightarrow\left(5y-4\right)^2-16\left(3y^2-3y+A\right)\ge0\)
\(\Leftrightarrow25y^2-40y+16-48y^2+48y-16A\ge0\)
\(\Leftrightarrow-23y^2+8y+16\ge16A\)
\(\Leftrightarrow16A\le-23\left(y^2-\frac{8}{23}y-\frac{12}{23}\right)=-23\left(y-\frac{4}{23}\right)^2+\frac{384}{23}\le\frac{384}{23}\)
\(\Rightarrow A\le\frac{24}{23}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2xy+3y\left(1-x-y\right)+4\left(1-x-y\right)x=\frac{24}{23}\\\left(y-\frac{4}{23}\right)^2=0\\x+y+z=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{9}{23}\\y=\frac{4}{23}\\z=\frac{10}{23}\end{cases}}\)
Vậy Max A = \(\frac{24}{23}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{9}{23}\\y=\frac{4}{23}\\z=\frac{10}{23}\end{cases}}\)
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{99.101}\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(=\frac{3}{2}.\frac{100}{101}\)
\(=\frac{150}{101}\)
Đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)
\(\frac{1}{2}A=\frac{1}{2}\left(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{99.101}\right)\)
\(\frac{1}{2}A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(\frac{1}{2}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(\frac{1}{2}A=1-\frac{1}{101}\)
\(\frac{1}{2}A=\frac{100}{101}\)
\(A=\frac{100}{101}:\frac{1}{2}\)
\(A=\frac{200}{101}\)
\(\left(5-2\sqrt{6}\right)^{\frac{x}{2}}+\left(5+2\sqrt{6}\right)^{\frac{x}{2}}=10\)
\(pt\Leftrightarrow\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^{2x}}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^{2x}}=10\)
\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)^x+\left(\sqrt{3}+\sqrt{2}\right)^x=10\)
\(\Leftrightarrow\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^x}+\left(\sqrt{3}+\sqrt{2}\right)^x=10\)
\(\Leftrightarrow\frac{1}{t}+t=10\left(t=\left(\sqrt{3}+\sqrt{2}\right)^x\right)\)
\(\Leftrightarrow t^2-10t+1=0\)\(\Leftrightarrow t=5\pm2\sqrt{6}\)
\(\Rightarrow5\pm2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^x\)
\(\Leftrightarrow\left(\sqrt{3}+\sqrt{2}\right)^{\pm2}=\left(\sqrt{3}+\sqrt{2}\right)^x\)
\(\Rightarrow x=\pm2\). Vậy...
câu này nghĩa là : anh 1 đi không trở lại để lại mình em với thằng bé
\(5x^2-22x+23=0\)
\(\Leftrightarrow25x^2-110x+115=0\)
\(\Leftrightarrow\left(5x\right)^2-2.5x.11+121=6\)
\(\Leftrightarrow\left(5x-11\right)^2=6\)
\(\Leftrightarrow5x-11=\pm\sqrt{6}\)
\(\Leftrightarrow x=\frac{11\pm\sqrt{6}}{5}\)
Vậy...
842 nha bn
là 842 nha bn