\(\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}\)

\(=\sqrt{\dfrac{\left(3\sqrt{3}-4\right)\left(2\sqrt{3}-1\right)}{\left(2\sqrt{3}+1\right)\left(2\sqrt{3}-1\right)}}\)\(-\sqrt{\dfrac{\left(\sqrt{3}+4\right)\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}}\)

\(=\sqrt{\dfrac{18-3\sqrt{3}-8\sqrt{3}+4}{\left(2\sqrt{3}\right)^2-1}}\)\(-\sqrt{\dfrac{5\sqrt{3}+6+20+8\sqrt{3}}{5^2-\left(2\sqrt{3}\right)^2}}\)

\(=\sqrt{\dfrac{22-11\sqrt{3}}{11}}\)\(-\sqrt{\dfrac{26+13\sqrt{3}}{13}}\)

\(=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

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

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

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

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{3}-1-\sqrt{3}-1\right)=-\sqrt{2}\)

Ý còn lại đợi trưa nhá đi nấu cơm đã

2:

BC^2*MB

\(=\dfrac{BH^2}{BA}\cdot BC^2=\left(\dfrac{BA^2}{BC}\right)^2\cdot\dfrac{BC^2}{BA}\)

\(=\dfrac{BA^4}{BA}\cdot\dfrac{BC^2}{BC^2}=BA^3\)

=>\(MB=\dfrac{BA^3}{BC^2}\)

Lời giải:

Cộng 3 PT lại ta có:

$x(a+b+c)+y(a+b+c)=a+b+c$

$\Leftrightarrow (a+b+c)(x+y-1)=0$

$\Rightarrow a+b+c=0$ hoặc $x+y-1=0$

TH1: $a+b+c=0\Leftrightarrow a+b=-c$

Khi đó: $a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3$

$=(-c)^3-3ab(-c)+c^3=3abc$

$\Rightarrow \frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=3$ (đpcm)

TH2: $x+y-1=0\Leftrightarrow y=1-x$

Thay vô hpt \(\left\{\begin{matrix} ax+b(1-x)=c\\ bx+c(1-x)=a\\ cx+a(1-x)=b\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x(a-b)=c-b\\ x(b-c)=a-c\\ x(c-a)=b-a\end{matrix}\right.\)

\(\Rightarrow x^3(a-b)(b-c)(c-a)=(c-b)(a-c)(b-a)=-(a-b)(b-c)(c-a)\)

\(\Leftrightarrow (a-b)(b-c)(c-a)(x^3+1)=0\)

Nếu $a-b=0$ thì kéo theo $b-c=c-a=0$

$\Rightarrow a=b=c$

Nếu $b-c=0; c-a=0$ thì tương tự

Nếu $x^3+1=0\Leftrightarrow x=-1$

$\Rightarrow b-a=c-b=a-c\Rightarrow a=b=c$

Tóm lại $a=b=c$

Do đó: $\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=1+1+1=3$ (đpcm)

Bạn cần bổ sung điều kiện của $a,b$ thì mới giải được nhé.

Thưa thầy,

Điều kiện là a,b thuộc Z

\(P=\left(x^2+2x\right)\left(y^2-4y\right)+5\left(x^2+2x\right)+4\left(y^2-4y\right)+2021\)

\(=\left[\left(x+1\right)^2-1\right]\left[\left(y-2\right)^2-4\right]+5\left(x+1\right)^2+4\left(y-2\right)^2+2000\)

\(=\left(x+1\right)^2\left(y-2\right)^2+\left(x+1\right)^2+3\left(y-2\right)^2+2024\ge2024\)

\(P_{min}=2024\) khi \(\left(x;y\right)=\left(-1;2\right)\)

lm câu mấy ạ

Nếu được thì mong cả 3 câu luôn bạn nhé. 

bài toán lớp 8 hả anh

cú tui

Xét ΔABC vuông tại A ta có:

\(BC=\sqrt{AB^2+AC^2}=\sqrt{AB^2+\left(2AB\right)^2}=AB\sqrt{5}\)

Mà: 

\(\left\{{}\begin{matrix}sinC=\dfrac{AB}{BC}=\dfrac{AB}{AB\sqrt{5}}=\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}\\cosC=\dfrac{AC}{BC}=\dfrac{2AB}{AB\sqrt{5}}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\\tanC=\dfrac{AB}{AC}=\dfrac{AB}{2AB}=\dfrac{1}{2}\\cotC=\dfrac{AC}{AB}=\dfrac{2AB}{AB}=2\end{matrix}\right.\)