\(M=2x^2+4x+7\)

\(=2\left(x^2+2x+\dfrac{7}{2}\right)\)

\(=2\left(x^2+2x+1+\dfrac{5}{2}\right)\)

\(=2\left[\left(x+1\right)^2+\dfrac{5}{2}\right]\)

\(=2\left(x+1\right)^2+5\)

Vì \(2\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x+1\right)^2+5\ge5\forall x\)

\(\Rightarrow M_{min}=5\Leftrightarrow2\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Tương tự: \(N=x^2-x+1\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

\(\Rightarrow N_{min}=\dfrac{3}{4}\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

\(E=-4x^2+x-1\)

\(=-4\left(x^2-\dfrac{1}{4}x+\dfrac{1}{4}\right)\)

\(=-4\left[x^2-2.x.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2-\left(\dfrac{1}{8}\right)^2+\dfrac{1}{4}\right]\)

\(=-4\left[\left(x-\dfrac{1}{8}\right)^2+\dfrac{15}{64}\right]\)

\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{15}{16}\)

Vì \(-4\left(x-\dfrac{1}{8}\right)^2\le0\forall x\)

\(\Rightarrow-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{15}{16}\le-\dfrac{15}{16}\forall x\)

\(\Rightarrow E_{max}=-\dfrac{15}{16}\Leftrightarrow-4\left(x-\dfrac{1}{8}\right)^2=0\Leftrightarrow x=\dfrac{1}{8}\)

Tương tự: \(F=5x-3x^2+6\)

\(=-3x^2+5x+6\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)

\(\Rightarrow F_{max}=\dfrac{97}{12}\Leftrightarrow-3\left(x-\dfrac{5}{6}\right)^2=0\Leftrightarrow x=\dfrac{5}{6}\)

Bài 1:

\(a,\)

\(P=2\left(2x-1\right)^2-\left(x+1\right)^2+3\)

\(=2\left(4x^2-4x+1\right)-\left(x^2+2x+1\right)+3\)

\(=8x^2-8x+2-x^2-2x-1+3\)

\(=7x^2-10x+4\)

\(=7\left(x^2-\dfrac{10}{7}x+\dfrac{4}{7}\right)\)

\(=7\left[x^2-2.x.\dfrac{5}{7}+\left(\dfrac{5}{7}\right)^2-\left(\dfrac{5}{7}\right)^2+\dfrac{4}{7}\right]\)

\(=7\left[\left(x-\dfrac{5}{7}\right)^2+\dfrac{3}{49}\right]\)

\(=7\left(x-\dfrac{5}{7}\right)^2+\dfrac{3}{7}\)

Vì \(7\left(x-\dfrac{5}{7}\right)^2\ge0\forall x\)

\(\Rightarrow7\left(x-\dfrac{5}{7}\right)^2+\dfrac{3}{7}\ge\dfrac{3}{7}\forall x\)

\(\Rightarrow P_{min}=\dfrac{3}{7}\Leftrightarrow7\left(x-\dfrac{5}{7}\right)^2=0\Leftrightarrow x=\dfrac{5}{7}\)

\(b,\)

\(Q=3\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)\)

\(=3\left(x^2+4x+4\right)-\left(x^2-4\right)\)

\(=3x^2+12x+12-x^2+4\)

\(=2x^2+12x+16\)

\(=2\left(x^2+6x+8\right)\)

\(=2\left(x^2+6x+9-1\right)\)

\(=2\left[\left(x+3\right)^2-1\right]\)

\(=2\left(x+3\right)^2-2\)

Vì \(2\left(x+3\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x+3\right)^2-2\ge-2\forall x\)

\(\Rightarrow Q_{min}=-2\Leftrightarrow2\left(x+3\right)^2=0\Leftrightarrow x=-3\)

\(c,\)

Tương tự: \(M=\left(x+1\right)^3-\left(x-2\right)^3-5\)

\(=9x^2-9x+4\)

\(=9\left(x^2-x+\dfrac{4}{9}\right)\)

\(=9\left[x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+\dfrac{4}{9}\right]\)

\(=9\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{36}\right]\)

\(=9\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\)

Ta có: \(9\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\)

\(\Rightarrow M_{min}=\dfrac{7}{4}\Leftrightarrow9\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

Bài 2:

\(a,\)

\(M=-3x^2-4x+1\)

\(=-3\left(x^2+\dfrac{4}{3}x-\dfrac{1}{3}\right)\)

\(=-3\left[x^2+2.x.\dfrac{2}{3}+\left(\dfrac{2}{3}\right)^2-\left(\dfrac{2}{3}\right)^2-\dfrac{1}{3}\right]\)

\(=-3\left[\left(x+\dfrac{2}{3}\right)^2-\dfrac{7}{9}\right]\)

\(=-3\left(x+\dfrac{2}{3}\right)^2+\dfrac{7}{3}\)

Ta có: \(-3\left(x+\dfrac{2}{3}\right)^2\le\dfrac{7}{3}\forall x\)

\(\Rightarrow M_{max}=\dfrac{7}{3}\Leftrightarrow-3\left(x+\dfrac{2}{3}\right)^2=0\Leftrightarrow x=-\dfrac{2}{3}\)

\(b,\)

\(Q=\left(x+2\right)^2-3\left(x-1\right)^2+3\)

\(=-2x^2+10x+4\)

\(=-2\left(x^2-5x-2\right)\)

\(=-2\left[\left(x-\dfrac{5}{2}\right)^2-\dfrac{33}{4}\right]\)

\(=-2\left(x-\dfrac{5}{2}\right)^2+\dfrac{33}{2}\)

Tương tự, \(Q_{max}=\dfrac{33}{2}\Leftrightarrow x=\dfrac{5}{2}\)

\(c,\)

\(P=\left(x-1\right)\left(x+2\right)-2\left(x+3\right)^2\)

\(=-x^2-11x-20\)

\(=-\left(x^2+11x+20\right)\)

\(=-\left(x+\dfrac{11}{2}\right)^2+\dfrac{41}{4}\)

Tương tự, \(Q_{max}=\dfrac{41}{4}\Leftrightarrow x=-\dfrac{11}{2}\)

\(d,\)

\(1+2:\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)

\(=1+2:\dfrac{1}{2}\cdot\left(-\dfrac{9}{4}\right)\)

\(=1+2.2.\left(-\dfrac{9}{4}\right)\)

\(=1+4.\left(-\dfrac{9}{4}\right)\)

\(=1-9\)

\(=-8\)

Xét \(\Delta AEB\) vuông tại E:

\(\widehat{BAE}+\widehat{ABE}=90^o\) (2 góc phụ nhau)

\(\Rightarrow\widehat{ABE}=90^o-30^o=60^o\)

Ta có: \(tan\widehat{ABE}=\dfrac{AE}{BE}\)

\(\Rightarrow BE=\dfrac{AE}{tan\widehat{ABE}}=\dfrac{35}{tan60^o}=\dfrac{35\sqrt{3}}{3}m\) (do \(AE=DC=35m\))

\(\Rightarrow BC=BE+EC=\dfrac{35\sqrt{3}}{3}+1,5=\dfrac{9+70\sqrt{3}}{6}m\)

\(\dfrac{5}{7}x^3.\dfrac{2}{3}x^4.\left(-\dfrac{7}{10}x\right)\)

\(=\dfrac{5}{7}.\dfrac{2}{3}.\left(-\dfrac{7}{10}\right).\left(x^3.x^4.x\right)\)

\(=-\dfrac{1}{3}x^8\)

+ Phần hệ số: \(-\dfrac{1}{3}\)

+ Phần biến: \(x^8\)

\(c,\)

\(2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^2+0,125^3.8^3-\left(-12\right)^4:6^4\)

\(=2:\left(-\dfrac{1}{6}\right)^2+\left(\dfrac{1}{8}\right)^3.8^3-\left(-12\right)^4:6^4\)

\(=2:\dfrac{1}{36}+\left(\dfrac{1}{8}.8\right)^3-\left(-12:6\right)^4\)

\(=2.36+1^3-\left(-2\right)^4\)

\(=72+1-16\)

\(=57\)

Ta có: \(\left\{{}\begin{matrix}\dfrac{AB}{AC}=\dfrac{3}{4}\\AB=15cm\end{matrix}\right.\)

\(\Rightarrow AC=\dfrac{4.15}{3}=20cm\)

Áp dụng định lí pitago trong \(\Delta ABC\) vuông tại A:

\(BC^2=AB^2+AC^2\)

\(\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{15^2+20^2}=25cm\)

\(\Rightarrow y=25cm\)

Áp dụng hệ thức lượng trong \(\Delta ABC\) vuông tại A:

\(AB.AC=BC.AH\)

\(\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{15.20}{25}=12cm\)

\(a,\)

\(A=\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)

\(=sin^2\alpha+cos^2\alpha+2.sin\alpha.cos\alpha+sin^2\alpha+cos^2\alpha-2.sin\alpha.cos\alpha\)

\(=1+1=2\)

\(b,\)

\(VT=\dfrac{sin^2\alpha+cos^2\alpha-2.sin\alpha.cos\alpha-1}{tan\alpha-sin\alpha.cos\alpha}\)

\(=\dfrac{-2.sin\alpha.cos\alpha}{tan\alpha.sin^2\alpha}\)

\(=\dfrac{-2.cos\alpha}{tan\alpha.sin\alpha}\)

\(=-2.\dfrac{cos\alpha}{sin\alpha}.\dfrac{1}{tan\alpha}\)

\(=\dfrac{-2.cot\alpha}{tan\alpha}\)

\(=-2.cot^2\alpha=VP\) (điều phải chứng minh)

\(c,\)

\(VT^2=\dfrac{1+sin\alpha}{1-sin\alpha}+\dfrac{1-sin\alpha}{1+sin\alpha}+2\)

\(=\dfrac{\left(1+sin\alpha\right)^2+\left(1-sin\alpha\right)^2+2.\left(1+sin\alpha\right).\left(1-sin\alpha\right)}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}\)

\(=\dfrac{\left(1+sin\alpha+1-sin\alpha\right)^2}{1-sin^2\alpha}\)

\(=\dfrac{4}{cos^2\alpha}\)

\(\Rightarrow VT=\pm\dfrac{2}{cos\alpha}\)

Vì \(0< \alpha< \dfrac{\pi}{2}\Rightarrow cos\alpha>0\Rightarrow VT=\dfrac{2}{cos\alpha}=VP\) (điều phải chứng minh)

Áp dụng định lí pitago trong \(\Delta ABC\) vuông tại A:

\(BC^2=AB^2+AC^2\)

\(\Rightarrow BC=\sqrt{6^2+8^2}=10cm\)

Áp dụng hệ thức lượng trong \(\Delta ABC\) vuông tại A:

\(\left\{{}\begin{matrix}AB^2=BH.BC\\AC^2=CH.BC\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=\dfrac{6^2}{10}=3,6cm\\CH=\dfrac{AC^2}{BC}=\dfrac{8^2}{10}=6,4cm\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3,6cm\\y=6,4cm\end{matrix}\right.\)

\(\dfrac{1}{30^2}=\dfrac{1}{\left(\dfrac{5}{6}AC\right)^2}+\dfrac{1}{AC^2}\)

\(\Leftrightarrow\dfrac{1}{900}=\dfrac{1}{\dfrac{25}{36}AC^2}+\dfrac{1}{AC^2}\left(1\right)\)

Đặt \(AC=x\left(x>0\right)\)

\(\Rightarrow AC^2=x^2\)

\(\left(1\right)\Leftrightarrow\dfrac{1}{900}=\dfrac{1}{\dfrac{25}{36}x^2}+\dfrac{1}{x^2}\)

\(\Leftrightarrow\dfrac{1}{900}=\dfrac{36}{25x^2}+\dfrac{1}{x^2}\)

\(\Leftrightarrow\dfrac{25x^4}{22500x^4}=\dfrac{32400x^2}{22500x^4}+\dfrac{22500x^2}{22500x^4}\)

\(\Rightarrow25x^4=32400x^2+22500x^2\)

\(\Leftrightarrow25x^4=54900x^2\) (chia cả hai vế cho \(25x^2\))

\(\Leftrightarrow x^2=2196\)

\(\Rightarrow AC^2=2196\)

\(\Rightarrow AC=6\sqrt{61}\)

Vậy: \(AC=6\sqrt{61}\)