a) \(A=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4-2\sqrt{3}}}\)

\(=\frac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\frac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}\)

\(=\frac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-\sqrt{18}+6\sqrt{2}+2\sqrt{6}-3\sqrt{6}-\sqrt{18}}{6}\)

\(=\frac{12\sqrt{2}-2\sqrt{18}}{6}=\frac{6\sqrt{2}}{6}=\sqrt{2}\)

Drichle^^

a/ \(1+tan^2a=1+\frac{sin^2a}{cos^2a}=\frac{sin^2a+cos^2a}{cos^2a}=\frac{1}{cos^2a}\)

b/ \(1+cot^2a=1+\frac{cos^2a}{sin^2a}=\frac{sin^2a+cos^2a}{sin^2a}=\frac{1}{sin^2a}\)

cảm ơn nhìu

Để \(\sqrt{x^2+x+3}\) nguyên thì

\(\Rightarrow x^2+x+3=a^2\left(a\in Z\right)\)

\(\Leftrightarrow4x^2+4x+12=4a^2\)

\(\Leftrightarrow4a^2-\left(2x+1\right)^2=11\)

\(\Leftrightarrow\left(2a+2x+1\right)\left(2a-2x-1\right)=11\)

\(\Leftrightarrow\left(2a+2x+1,2a-2x-1\right)=\left(1,11;11,1;-1,-11;-11,-1\right)\)

\(\Leftrightarrow\left(a,x\right)=\left(3,-3;3,2;-3,-3;-3,2\right)\)

Vậy ....

\(BĐT\Leftrightarrow\left(\frac{a+1}{a}\right)\left(\frac{b+1}{b}\right)\left(\frac{c+1}{c}\right)\ge64\)(*)

Mà \(\frac{a+1}{a}=\frac{\left(a+a\right)+\left(b+c\right)}{a}\ge\frac{2a+2\sqrt{bc}}{a}\ge\frac{2\sqrt{2a.2\sqrt{bc}}}{a}=\frac{4\sqrt{a\sqrt{bc}}}{a}\) (1)

Tương tự \(\frac{b+1}{b}\ge\frac{4\sqrt{b\sqrt{ac}}}{b}\) (2)  ;           \(\frac{c+1}{c}\ge\frac{4\sqrt{c\sqrt{ab}}}{c}\) (3)

Từ (1), (2) và (3) nhân vế theo vế ta được   (*) \(\ge\frac{4\sqrt{a\sqrt{bc}}.4\sqrt{b\sqrt{ac}}.4\sqrt{c\sqrt{ab}}}{abc}=\frac{64abc}{abc}=64\)

Dấu ''='' xảy ra khi \(\hept{\begin{cases}a+b+c=1\\1+\frac{1}{a}=1+\frac{1}{b}=1+\frac{1}{c}=4\end{cases}\Leftrightarrow a=b=c=\frac{1}{3}}\)

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Ta có:

\(4A=\frac{\left(x+y+z+t\right)^2\left(x+y+z\right)\left(x+y\right)}{xyzt}\)

\(\ge\frac{4\left(x+y+z\right)t\left(x+y+z\right)\left(x+y\right)}{xyzt}\)

\(=\frac{4\left(x+y+z\right)^2\left(x+y\right)}{xyz}\ge\frac{16\left(x+y\right)z\left(x+y\right)}{xyz}\)

\(=\frac{16\left(x+y\right)^2}{xy}\ge\frac{64xy}{xy}=64\)

\(\Rightarrow A\ge16\)

Đấu = xảy ra khi \(t=2z=4x=4y=1\)

x;y;z;t >0 áp dụng bất đẳng thức Cô-si cho 2 số dương ta có :

=\(x+y\ge2\sqrt{xy}\)

=\(\left(x+y\right)+z\ge2\sqrt{\left(x+y\right)z}\)

=\(\left(x+y+z\right)+t\ge2\sqrt{\left(x+y+z\right)t}\)

nhân các vế tương ứng ta có:

\(\left(x+y\right)\left(x+y+z\right)\left(x+y+z+t\right)\ge8\sqrt{xyzt\left(x+y\right)\left(x+y+z\right)}\)

mà x+y+z+t=2

\(\left(x+y\right)\left(x+y+z\right)2\ge8\sqrt{xyzt\left(x+y\right)\left(x+y+z\right)}\)

=\(\sqrt{\left(x+y\right)\left(x+y+z\right)}\ge4\sqrt{xyzt}\)

=\(\left(x+y\right)\left(x+y+z\right)\ge16xyzt\)

\(\Rightarrow B=\frac{\left(x+y\right)\left(x+y+z\right)}{xyzt}\ge\frac{16xyzt}{xyzt}=16\)

vậy minB=16 khi\(\hept{\begin{cases}x=y\\x+y=z\\x+y+z=t\end{cases}};x+y+z+t=2\Rightarrow x=y=0.25;z=0.5;t=1\)

\(\hept{\begin{cases}8x^3y^3+27=18y^3\\4x^2y+6x=y^2\end{cases}}\)

Dễ thấy y = 0 không phải là nghiệm của hệ.

Xét \(y\ne0\)

\(\Rightarrow\hept{\begin{cases}8x^3y^3+27=18y^3\left(1\right)\\4x^2y^2+6xy=y^3\left(2\right)\end{cases}}\)

Lấy (1) - 18.(2) ta được

\(8x^3y^3-72x^2y^2-108xy+27=0\)

\(\Leftrightarrow\left(2xy+3\right)\left(4x^2y^2-42xy+9\right)=0\)

Đặt \(xy=a\)

\(\Rightarrow\left(2a+3\right)\left(4a^2-42a+9\right)=0\)

Tới đây thì bạn làm tiếp nhé.

\(\hept{\begin{cases}\left(x+2y-2\right)\left(2x+y\right)=2x\left(5y-2\right)-2y\\x^2-7y=-3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x^2-5xy+2y^2=0\left(1\right)\\x^2-7y=-3\left(2\right)\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}y=2x\\x=2y\end{cases}}\)

Thế ngược lại (2) giải tiếp sẽ được nghiệm nhé.

1+2-2+444=?

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

\(\Rightarrow y^2=2x+2\sqrt{x+2\sqrt{x-1}}.\sqrt{x-2\sqrt{x-1}}\)

\(\Leftrightarrow y^2=2x+2\sqrt{\left(2-x\right)^2}=2x+4-2x=4\)

\(\Rightarrow y=2\)