\(\dfrac{AC}{3}=\dfrac{BC}{4}=\dfrac{AC+BC}{3+4}=\dfrac{14}{7}=2\)

\(\Rightarrow\left\{{}\begin{matrix}AC=3.2=6\\BC=4.2=8\end{matrix}\right.\)

Theo định lý Pitago cho Δ vuông ABC:

\(BC^2=AB^2+AC^2\Rightarrow AB^2=BC^2-AC^2=64-36=28\)

\(\Rightarrow AB=\sqrt[]{28}=2\sqrt[]{7}\)

Tam giác ABC vuông tại A

 A = \(\dfrac{6}{3}\) - | 2\(x\) - 4| 

  | 2\(x\) - 4| ≥ 0 ∀ \(x\);  ⇒ - |2\(x\) - 4|  ≤ 0 

⇒ \(\dfrac{6}{3}\) - |2\(x\) - 4| ≤  \(\dfrac{6}{3}\) ∀ \(x\)

A_max = \(\dfrac{6}{3}\) ⇔ \(2x\) - 4 = 0 ⇔ \(x\) = 2

Ta có \(A=\overset{2n}{11...1}+\overset{n}{44...4}+1\)

\(A=\dfrac{1}{9}.\overset{2n}{99...9}+\dfrac{4}{9}.\overset{n}{99...9}+1\)

\(A=\dfrac{1}{9}\left(10^{2n}-1\right)+\dfrac{4}{9}\left(10^n-1\right)+1\)

\(A=\dfrac{10^{2n}-1+4.10^n-4+9}{9}\)

\(A=\dfrac{\left(10^n\right)^2+4.10^n+4}{9}\)

\(A=\left(\dfrac{10^n+2}{3}\right)^2\) 

 Dễ thấy \(10^n+2⋮3\) vì có tổng các chữ số là 3 nên \(\dfrac{10^n+2}{3}\inℕ^∗\). Vậy A là số chính phương (đpcm)

Lời giải:

\(ab+bc+ac=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=\frac{2^2-2}{2}=1\)

Khi đó:

\(\text{VT}=\frac{a}{ab+bc+ac+a^2}+\frac{2b}{ab+bc+ac+b^2}+\frac{3c}{ab+bc+ac+c^2}\)

\(=\frac{a}{(a+b)(a+c)}+\frac{2b}{(b+a)(b+c)}+\frac{3c}{(c+a)(c+b)}\)

\(=\frac{a(b+c)+2b(a+c)+3c(a+b)}{(a+b)(b+c)(c+a)}\)

\(=\frac{3ab+4ac+5bc}{(a+b)(b+c)(c+a)}=\text{VP}\)

Lời giải:

$x^2-y^2+a^2-b^2+2ax+2by=(x^2+a^2+2ax)-(y^2+b^2-2by)$

$=(x+a)^2-(y-b)^2=(x+a-y+b)(x+a+y-b)$

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

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

\(\Rightarrow E=-4\left(x^2-\dfrac{x}{4}+\dfrac{1}{64}\right)+1+\dfrac{1}{16}\)

\(\Rightarrow E=-4\left(x-\dfrac{1}{8}\right)^2+\dfrac{17}{16}\)

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

\(\Rightarrow E=-4\left(x-\dfrac{1}{8}\right)^2+\dfrac{17}{16}\le\dfrac{17}{16}\)

\(\Rightarrow GTLN\left(E\right)=\dfrac{17}{16}\left(tạix=\dfrac{1}{8}\right)\)

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

\(\Rightarrow F=-3\left(x^2-\dfrac{5x}{3}\right)+6\)

\(\Rightarrow F=-3\left(x^2-\dfrac{5x}{3}+\dfrac{25}{36}\right)+6+\dfrac{25}{12}\)

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

mà \(-3\left(x-\dfrac{5}{6}\right)^2\le0,\forall x\)

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

\(\Rightarrow GTLN\left(F\right)=\dfrac{97}{12}\left(tạix=\dfrac{5}{6}\right)\)

\(x\) mũ bao nhiêu thì cô và các bạn mới giúp được chứ em?

7) Chứng minh rằng: x^2 +4y^2 + z^2- 2x -6z +8y + 15 > 0 với mọi x, y, z.