Cách 1. Đặt \(\frac{x}{3}=\frac{y}{7}=k\) \(\Rightarrow\begin{cases}x=3k\\y=7k\end{cases}\)

\(\Rightarrow xy=189\Leftrightarrow3k.7k=189\Leftrightarrow k^2=9\Leftrightarrow k=\pm3\)

Nếu k = 3 thì x = 9 , y = 21

Nếu k = -3 thì x = -9 , y = -21

a) \(9x^2+12x+7=\left(9x^2+12x+4\right)+3=\left(3x+2\right)^2+3\ge3\)

Min = 3 <=> x = -2/3

b) \(x^2-26x+180=\left(x^2-26x+169\right)+11=\left(x-13\right)^2+11\ge11\)

Min = 11 <=> x = 13

a/ \(\sqrt{8\left(\sqrt{2}-\sqrt{3}\right)^2}=2\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)=2\sqrt{6}-4\)

b/ \(ab\sqrt{1+\frac{1}{a^2b^2}}=ab.\sqrt{\frac{a^2b^2+1}{a^2b^2}}=\sqrt{a^2b^2.\frac{a^2b^2+1}{a^2b^2}}=\sqrt{a^2b^2+1}\)

c/ \(\sqrt{\frac{a}{b^3}+\frac{a}{b^4}}=\sqrt{\frac{a}{b^3}\left(1+\frac{1}{b}\right)}=\frac{1}{b}.\sqrt{\frac{a}{b}\left(1+\frac{1}{b}\right)}\)

d/ \(\frac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}\)

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

b/ \(B=\left(1+tan^2\alpha\right)\left(1-sin^2\alpha\right)-\left(1+cotg^2\alpha\right)\left(1-cos^2\alpha\right)\)

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

\(=\frac{1}{cos^2\alpha}.cos^2\alpha-\frac{1}{sin^2\alpha}.sin^2\alpha=1-1=0\)

Ta có : \(\begin{cases}x+y=2\\x-y=\frac{3\sqrt{2}}{2}\end{cases}\)

Xét : \(\left(x+y\right)^2=x^2+y^2+2xy=4\left(1\right)\)

\(\left(x-y\right)^2=x^2-2xy+y^2=\frac{9}{2}\left(2\right)\)

Cộng (1) và (2) được : \(2\left(x^2+y^2\right)=4+\frac{9}{2}\Leftrightarrow x^2+y^2=\frac{17}{4}\)

Thừa điểm N ???

Ta có : \(\sin\alpha=\frac{2}{3}\Rightarrow\sin^2\alpha=\frac{4}{9}\) 

Lại có : \(sin^2\alpha+cos^2\alpha=1\Rightarrow cos^2\alpha=1-sin^2\alpha\) thay vào C

\(C=5\left(1-sin^2\alpha\right)+2sin^2\alpha=5-3sin^2\alpha=5-3.\frac{4}{9}=\frac{11}{3}\)

Ta có : \(\left|x-2015\right|\ge0\) \(\Leftrightarrow-\left|x-2015\right|\le0\) \(\Leftrightarrow B=2016-\left|x-2015\right|\le2016\)

=> Max B = 2016 <=> x = 2015

a) \(x^2+7x-8=0\Leftrightarrow\left(x+8\right)\left(x-1\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=-8\\x=1\end{array}\right.\)

b) \(\left(x-3\right)\left(16-4x\right)=0\Leftrightarrow4\left(x-3\right)\left(4-x\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=4\end{array}\right.\)

c) \(5x^2+9x+4=0\Leftrightarrow\left(x+1\right)\left(5x+4\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\x=-\frac{4}{5}\end{array}\right.\)