Kẻ DH vuông góc với DN tại D

Xét ΔADM và ΔCDH có:

  ^DAM=^DCH=90(gt)

   AD=DC(gt)

  ^ADM=^CDH (cùng phụ với ^NDC)

=>ΔADM=ΔCDH(g.c.g)

=>DM=DH

Xét ΔDHN vuông tại D(gt).Có:

 \(\frac{1}{DH^2}+\frac{1}{DN^2}=\frac{1}{DC^2}=\frac{1}{a^2}\)

hay \(\frac{1}{DM^2}+\frac{1}{DN^2}=\frac{1}{a^2}\)

1) \(P=4x\left(x-1\right)+10=4x^2-4x+10=\left(4x^2-4x+1\right)+9=\left(2x-1\right)^2+9\)

Vì: \(\left(2x-1\right)^2\ge0\)

=> \(\left(2x-1\right)^2+9\ge9\)

Vậy GTNN của P là 9 khi \(x=\frac{1}{2}\)

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

Vì: \(\left(x-\frac{1}{2}\right)^2\ge0\)

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

Vậy GTNN của P là \(\frac{3}{4}\) khi \(x=\frac{1}{2}\)

3) \(P=x^2+5y^2-4xy+2y+5=\left(x^2-4xy+4y^2\right)+\left(y^2+2y+1\right)+4\\ =\left(x-2y\right)^2+\left(y+1\right)^2+4\)

Vì; \(\left(x-2y\right)^2+\left(y+1\right)^2\ge0\)

=> \(\left(x-2y\right)^2+\left(y+1\right)^2+4\ge4\)

Vậy GTNN của P là 4 khi x=-2;y=-1

\(\left(x+2\right)^3-x\left(x-3\right)\left(x+3\right)-6x^2=29\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3+9x-6x^2=29\)

\(\Leftrightarrow21x=21\)

\(\Leftrightarrow x=1\)

\(\left(x-3\right)^5=\left(x-3\right)^3\)

\(\Leftrightarrow x-3=1\)

\(\Leftrightarrow x=4\)

Có; \(x=-2y\)

=> \(\frac{x}{-2}=\frac{y}{1}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x}{-2}=\frac{y}{1}=\frac{x+y}{-2+1}=\frac{10}{-1}=-10\)

=>\(\begin{cases}\frac{x}{-2}=-10\\y=-10\end{cases}\)\(\Leftrightarrow\begin{cases}x=20\\y=-10\end{cases}\)

Có: \(3x=7y\)

=> \(\frac{x}{7}=\frac{y}{3}\)

Áp dụng tính chất cuae dãy tie số bằng nhau ta có:

\(\frac{x}{7}=\frac{y}{3}=\frac{x-y}{7-3}=\frac{-16}{4}=-4\)

=> \(\begin{cases}\frac{x}{7}=-4\\\frac{y}{3}=-4\end{cases}\)\(\Leftrightarrow\begin{cases}x=-28\\y=-12\end{cases}\)

2) M = (x²⁵ + 1 + 1 + 1 + 1) - 5x⁵ + 2

Áp dụng BĐT Cô - si cho 5 số dương x²⁵; 1;1;1;1 ta có: x²⁵ + 1 + 1 + 1 + 1 \(\ge\)5.\(\sqrt[5]{x^{25}.1.1.1.1}=x^5\) = 5x⁵

=> M \(\ge\) 5x⁵ - 5x⁵ + 2 = 2

Vậy M nhỏ nhất = 2 khi x²⁵ = 1 => x = 1

1) \(x^2+5x+6=0\)

\(\Leftrightarrow x^2+2x+3x+6=0\)

\(\Leftrightarrow x\left(x+2\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\end{array}\right.\)

2) \(2\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2-x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\\2-x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=2\end{array}\right.\)

3) \(x^2+4x+3=0\)

\(\Leftrightarrow x^2+x+3x+3=0\)

\(\Leftrightarrow x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\x=-3\end{array}\right.\)

4) \(2x^2-3x-5=0\)

\(\Leftrightarrow2x^2+2x-5x-5=0\)

\(\Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-5\right)=0\)

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