Đề thiếu VP nha b:)

Bạn ơi 

Chứng minh gì vậy bạn

( mà xin lỗi vì mình mới lớp 6 )

\(f\left(x\right)=x^2-2\left(m+5\right)x+m^2+4m-3=0\)

Phương trình cho có 2 nghiệm phân biệt \(\Leftrightarrow\Delta'>0\Leftrightarrow6m+28>0\Leftrightarrow m>-\frac{14}{3}\left(1\right)\)

ycbt\(\Leftrightarrow\hept{\begin{cases}-2< m+5< 4\\f\left(-2\right)>0\\f\left(4\right)>0\end{cases}}\Leftrightarrow\hept{\begin{cases}-7< m< -1\\m^2+8m+21>0\\m^2-4m-27>0\end{cases}}\Leftrightarrow-7< m< 2-\sqrt{31}\left(2\right)\)

Từ (1),(2) suy ra \(-\frac{14}{3}< m< 2-\sqrt{31}.\)

\(\hept{\begin{cases}\frac{2}{x}+\frac{1}{y}=2\\\frac{6}{x}-\frac{2}{y}=1\end{cases}}\)ĐK : \(x;y\ne0\)

Đặt \(\frac{1}{x}=a;\frac{1}{y}=b\)

\(\Leftrightarrow\hept{\begin{cases}2a+b=2\\6a-2b=1\end{cases}\Leftrightarrow\hept{\begin{cases}4a+2b=4\\6a-2b=1\end{cases}\Leftrightarrow}\hept{\begin{cases}10a=5\\2a+b=2\end{cases}\Leftrightarrow}\hept{\begin{cases}a=\frac{1}{2}\\2a+b=2\end{cases}}}\)

Thay a = 1/2 vào pt 2 ta được : 

\(1+b=2\Leftrightarrow b=1\)

Theo cách đặt \(\Leftrightarrow\hept{\begin{cases}a=\frac{1}{2}\\b=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}}\)( tm )

Vậy hệ pt có một nghiệm ( x ; y ) = ( 2 ; 1 ) 

\(a^2+bc\ge2\sqrt{a^2bc}=2a\sqrt{bc}\)

\(\Rightarrow\frac{1}{a^2+bc}\le\frac{1}{2a\sqrt{bc}}=\frac{\sqrt{bc}}{2abc}\).

Tương tự ta có: \(\frac{1}{b^2+ac}\le\frac{\sqrt{ac}}{2abc},\frac{1}{c^2+ba}\le\frac{\sqrt{ba}}{2abc}\).

Ta lại có: \(a+b\ge2\sqrt{ab},b+c\ge2\sqrt{bc},c+a\ge2\sqrt{ca}\Rightarrow a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\).

Do đó ta có: 

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ba}\le\frac{\sqrt{bc}}{2abc}+\frac{\sqrt{ac}}{2abc}+\frac{\sqrt{ba}}{2abc}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\)

\(\le\frac{a+b+c}{2abc}\)

Dấu \(=\)xảy ra khi \(a=b=c>0\).

\(A=\left(\sqrt{22}+7\sqrt{2}\right)\sqrt{30-7\sqrt{11}}\)

\(2A=\left(\sqrt{44}+7\sqrt{4}\right)\sqrt{60-2.7\sqrt{11}}\)

\(2A=\left(2\sqrt{11}+14\right)\sqrt{7^2-2.7\sqrt{11}+\left(\sqrt{11}\right)^2}\)

\(2A=\left(2\sqrt{11}+14\right)\sqrt{\left(7-\sqrt{11}\right)^2}\)

\(2A=\left(2\sqrt{11}+14\right)\left|7-\sqrt{11}\right|\)

\(2A=\left(2\sqrt{11}+14\right)\left(7-\sqrt{11}\right)\)

\(A=\left(7+\sqrt{11}\right)\left(7-\sqrt{11}\right)\)

\(A=49-11=38\)

49, \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}-\sqrt{9-6\sqrt{2}+2}\)

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

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

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

51, \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{5-2\sqrt{5}\sqrt{3}+3}-\sqrt{5-2\sqrt{5}\sqrt{3}+3}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|-\left|\sqrt{5}-\sqrt{3}\right|=-2\sqrt{3}\)

52, \(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\)

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

47659:9

M giải luôn nha

\(\frac{1}{2}=\frac{x^2}{\left(y+1^2\right)}+\)\(\frac{y^2}{\left(x+1\right)^2}\) \(\ge\frac{2xy}{\left(x+1\right)\left(y+1\right)}\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)\ge4xy\)

\(\Leftrightarrow3xy\le x+y+1\)

Dấu " = " xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}\frac{x^2}{\left(y+1\right)^2}=\frac{y^2}{\left(x+1\right)^2}\\3xy=x+y+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=y\\3x^2-2x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=y=1\left(tm\right)\\x=y=-\frac{1}{3}\left(tm\right)\end{cases}}\)

Vậy ( x ; y ) ......

a, tự tìm tự vẽ 

b, Ta có : \(\hept{\begin{cases}y=x^2\\y=-x+2\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+x-2=0\\y=-x+2\end{cases}}\)

\(\left(1\right)\Rightarrow\Delta=1+8=9>0\)

\(x_1=\frac{-1-3}{2}=-2;x_2=\frac{-1+3}{2}=1\)

Với x = -2 => \(y=2+2=4\)

Với x = 1 => \(-1+2=1\)

Vậy giao điểm của 2 đồ thị trên là A ( -2 ; 4 ) ; B ( 1 ; 1 )

đk: \(-3\le x< 3\)

Ta có: 

\(A=\frac{x^2+5x+x\sqrt{9-x^2}+6}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}=\frac{\left(x^2+5x+6\right)+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)

\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(x+3\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)

\(=\frac{\left[\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right]\sqrt{x+3}}{\left(x\sqrt{3-x}+\left(x+2\right)\sqrt{3+x}\right)\sqrt{3-x}}\)

\(=\frac{\sqrt{x+3}}{\sqrt{3-x}}\)

t post lại lên thôi Dung nhá :))