Zinyami
Giới thiệu về bản thân
a) \(A=-\left(\dfrac{x+18}{1203}\right)^2-\dfrac{183}{121}\le-\dfrac{183}{121}\)
Dấu bằng xảy ra
\(\Leftrightarrow x+18=0\Leftrightarrow x=-18\)
b) \(B=\dfrac{4}{\left(x+\dfrac{1}{3}\right)^2+5}\)
Ta có : \(\left(x+\dfrac{1}{3}\right)^2+5\ge5\)
\(\Leftrightarrow\dfrac{4}{\left(x+\dfrac{1}{3}\right)^2+5}\le\dfrac{4}{5}\)
Dấu bằng xảy ra
\(\Leftrightarrow x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\)
\(\sqrt{\dfrac{9}{7-4\sqrt{3}}}-\sqrt{\dfrac{4}{7+4\sqrt{3}}}\)
\(=\sqrt{\dfrac{9\left(7+4\sqrt{3}\right)}{7^2-\left(4\sqrt{3}\right)^2}}-\sqrt{\dfrac{4\left(7-4\sqrt{3}\right)}{7^2-\left(4\sqrt{3}\right)^2}}\)
\(=\dfrac{3\sqrt{7+4\sqrt{3}}-2\sqrt{7-4\sqrt{3}}}{\sqrt{49-48}}\)
\(=3\sqrt{4+2.2.\sqrt{3}+3}-2\sqrt{4-2.2.\sqrt{3}+3}\)
\(=3\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=3.\left|2+\sqrt{3}\right|-2.\left|2-\sqrt{3}\right|\)
\(=6+3\sqrt{3}-2+2\sqrt{3}\)
\(=4+5\sqrt{3}\)
a) \(A\left(x\right)=-5x^2-4x+1\)
\(\Leftrightarrow A\left(x\right)=-\left(5x^2+4x-1\right)\)
\(\Leftrightarrow A\left(x\right)=-\left[\left(\sqrt{5}x\right)^2+2.\sqrt{5}x.\dfrac{2\sqrt{5}}{5}+\dfrac{4}{5}-\dfrac{9}{5}\right]\)
\(\Leftrightarrow A\left(x\right)=-\left(\sqrt{5}x+\dfrac{4}{5}\right)^2+\dfrac{9}{5}\le\dfrac{9}{5}\)
Dấu bằng xảy ra
\(\Leftrightarrow\sqrt{5}x+\dfrac{4}{5}=0\Leftrightarrow x==-\dfrac{4\sqrt{5}}{25}\)
b) \(B\left(x\right)=-3x^2+x+1\)
\(\Leftrightarrow B\left(x\right)=-\left[\left(\sqrt{3}x\right)^2-2.\sqrt{3}x.\dfrac{2\sqrt{3}}{3}+\left(\dfrac{2\sqrt{3}}{3}\right)^2-\dfrac{1}{3}\right]\)
\(\Leftrightarrow B\left(x\right)=-\left(\sqrt{3}x-\dfrac{2\sqrt{3}}{3}\right)^2+\dfrac{1}{3}\le\dfrac{1}{3}\)
Dấu bằng xảy ra
\(\Leftrightarrow\sqrt{3}x-\dfrac{2\sqrt{3}}{3}=0\Leftrightarrow x=\dfrac{2}{3}\)
Sơ đồ :
Tuổi bố : |.....|.....|.....|
Tuổi con : |.....|
Tổng số phần bằng nhau là : 3 + 1 = 4 ( phần )
Tuổi con hiện nay là : 48 : 4 x 1 = 12 ( tuổi )
Số cây tháng này đội A trồng được là :
1400 + 1400 x 12 % = 1568 ( cây )
Đáp số : .........
Phần trăm số sảm phẩm đạt chuẩn là :
\(\dfrac{80}{120}\times100\approx66,7\%\)
a) \(x^{15}=x\)
\(\Leftrightarrow x\left(x^{14}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^{14}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
b) \(\left(x-5\right)^6=\left(x-5\right)^4\)
\(\Leftrightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Leftrightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\\x-5=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=1\\2x-15=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Điều kiện : \(x\ge-\dfrac{3}{2}\)
\(x^2+4x+5=2\sqrt{2x+3}\)
\(\Leftrightarrow x^2+4x+4+1=2\sqrt{2x+4-1}\)
\(\Leftrightarrow\left(x+2\right)^2+1=2\sqrt{2\left(x+2\right)-1}\)
Đặt \(x+2=t\left(t\ge\dfrac{1}{2}\right)\)
Ta có phương trình :
\(t^2+1=2\sqrt{2t-1}\)
\(\Leftrightarrow\left(t^2+1\right)^2=4\left(2t-1\right)\)
\(\Leftrightarrow t^4+2t^2+1-8t+4=0\)
\(\Leftrightarrow t^4+2t^2-8t+5=0\)
\(\Leftrightarrow t=1\)
\(\Leftrightarrow x+2=1\)
\(\Leftrightarrow x=-1\left(tm\right)\)
Vậy \(\Leftrightarrow x=-1\left(tm\right)\)
Điều kiện : \(x\le5\)
+) Với \(2x-7< 0\Leftrightarrow x< \dfrac{7}{2}\Rightarrow VP< 0\)
=> Phương trình vô nghiệm
+) Với \(2x-7>0\Leftrightarrow x>\dfrac{7}{2}\Rightarrow VP>0\)
Ta có phương trình :
\(5-x=\left(2x-7\right)^2\)
\(\Leftrightarrow5-x=4x^2-28x+49\)
\(\Leftrightarrow4x^2-27x+44=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{11}{4}\left(ktm\right)\end{matrix}\right.\)
Vậy x = 4
Điều kiện : \(x\ne1\)
\(\dfrac{\sqrt{x^2+2x+3}}{x-1}=x+3\)
\(\Rightarrow\sqrt{x^2+2x+3}=\left(x+3\right)\left(x-1\right)\)
\(\Leftrightarrow\sqrt{x^2+2x+3}=x^2-x+3x-3\)
\(\Leftrightarrow\sqrt{x^2+2x+3}=x^2+2x-3\)
+) Với \(-3< x< 1\) thì VP < 0 => Phương trình vô nghiệm
+) Với \(\left[{}\begin{matrix}x\le-3\\x>1\end{matrix}\right.\) thì VP > 0 , lúc này ta có phương trình :
\(x^2+2x+3=\left(x^2+2x-3\right)^2\)
\(\Leftrightarrow x^2+2x+3=\left(x^2+2x+1-4\right)^2\)
\(\Leftrightarrow x^2+2x+3=\left(x+1\right)^4-8\left(x+1\right)^2+16\)
\(\Leftrightarrow\left(x+1\right)^2+2=\left(x+1\right)^4-8\left(x+1\right)^2+16\)
\(\Leftrightarrow\left(x+1\right)^4-9\left(x+1\right)^2+14=0\)
Đặt \(\left(x+1\right)^2=t\left(t>0\right)\) , ta có :
\(t^2-9t+14=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=7\\t=2\end{matrix}\right.\)
+) \(t=7\Rightarrow\left(x+1\right)^2=7\Leftrightarrow x=\sqrt{7}-1\left(tm\right)\)
+) \(t=2\Leftrightarrow\left(x+1\right)^2=2\Leftrightarrow x=\sqrt{2}-1\left(ktm\right)\)
Vậy \(x=\sqrt{7}-1\)