\(x+y+13=2\left(2\sqrt{x}+3\sqrt{y}\right)\)

\(\Leftrightarrow x+y+13-4\sqrt{x}-6\sqrt{y}=0\)

\(\Leftrightarrow x-4\sqrt{x}+4+y-6\sqrt{y}+9=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2=0\)

Xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x}=2\\\sqrt{y}=3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=9\end{matrix}\right.\)

Điều kiện xác định: \(x,y>0\)

\(x+y+13=2\left(2\sqrt{x}+3\sqrt{y}\right)\)

\(\Leftrightarrow x-4\sqrt{x}+4+y-6\sqrt{y}+9=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)^2=0\\\left(\sqrt{y}-3\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2=0}{\sqrt{y}-3=0}\)

\(\Leftrightarrow\dfrac{x=4}{y=9}\)

KL: x = 4, y = 9

Áp dụng bất đẳng thức Bunhiacopxki ta có:

\(D^2\le3\left[a+\dfrac{\left(b-c\right)^2}{12}+\dfrac{\left(\sqrt{b}+\sqrt{c}\right)^2}{4}+\dfrac{\left(\sqrt{b}+\sqrt{c}\right)^2}{4}\right]\)

\(\Rightarrow D^2\le3\left[a+\dfrac{\left(b-c\right)^2}{12}+\dfrac{6\left(\sqrt{b}+\sqrt{c}\right)^2}{12}\right]\)

\(\Rightarrow D^2\le3\left[a+\dfrac{\left(b-c\right)^2+6\left(\sqrt{b}+\sqrt{c}\right)^2}{12}\right]\)

Ta sẽ C/m \(\dfrac{\left(b-c\right)^2+6\left(\sqrt{b}+\sqrt{c}\right)^2}{12}\le b+c\) (1)

Thật vậy \(\dfrac{\left(b-c\right)^2+6\left(\sqrt{b}+\sqrt{c}\right)^2}{12}\le b+c\)

\(\Leftrightarrow\left(b-c\right)^2+6\left(\sqrt{b}+\sqrt{c}\right)^2\le12\left(b+c\right)\)

\(\Leftrightarrow\left(b-c\right)^2+6b+6c+12\sqrt{bc}\le12\left(b+c\right)\)

\(\Leftrightarrow\left(b-c\right)^2\le6\left(b+c\right)-12\sqrt{bc}\)

\(\Leftrightarrow\left(\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{b}-\sqrt{c}\right)^2\le6\left(\sqrt{b}-\sqrt{c}\right)^2\)

\(\Leftrightarrow\left(\sqrt{b}-\sqrt{c}\right)^2\left(6-\left(\sqrt{b}+\sqrt{c}\right)^2\right)\ge0\) (2)

Ta có: \(\left(\sqrt{b}+\sqrt{c}\right)^2\le2\left(b+c\right)< 2\left(a+b+c\right)=6\)

Do đó (2) đúng nên (1) đúng

\(\Rightarrow D^2\le3\left(a+b+c\right)=9\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=1\)

Vậy max \(D=3\)

Trước tiên ta chứng minh:

\(\left(b-c\right)^2\le6\left(\sqrt{b}-\sqrt{c}\right)^2\)

\(\Leftrightarrow\left(\sqrt{b}-\sqrt{c}\right)^2\left(6-\left(\sqrt{b}+\sqrt{c}\right)^2\right)\ge0\) (đúng)

Áp dụng vào bài toán ta được:

\(D\le\sqrt{a+\dfrac{6\left(\sqrt{b}-\sqrt{c}\right)^2}{12}+\sqrt{b}+\sqrt{c}}\)

\(=\sqrt{a+\dfrac{b+c-2\sqrt{bc}}{2}+\sqrt{b}+\sqrt{c}}\)

\(=\sqrt{a+b+c-\left(\left(\dfrac{\sqrt{b}+\sqrt{c}}{\sqrt{2}}\right)^2-2.\dfrac{\sqrt{b}+\sqrt{c}}{\sqrt{2}}.\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}\right)+\dfrac{1}{2}}\)

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

\(\le\sqrt{3+\dfrac{1}{2}}=\sqrt{\dfrac{7}{2}}\)

Dấu = xảy ra khi: \(\left\{{}\begin{matrix}b=c\\\sqrt{b}+\sqrt{c}=1\\a+b+c=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2,5\\b=c=0,25\end{matrix}\right.\)

PS: Bài giải đây nhé Ace Legona

Đề sai sửa lại là:

\(x=\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\dfrac{125}{27}}}\)

\(\Leftrightarrow x=\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}+\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\)

\(\Leftrightarrow x^3=3+\sqrt{9+\dfrac{125}{27}}+3-\sqrt{9+\dfrac{125}{27}}+3.\left(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}+\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\right)\left(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}.\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\right)\)

\(\Leftrightarrow x^3=6+3x.\left(\dfrac{-5}{3}\right)\)

\(\Leftrightarrow x^3+5x-6=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)

\(\Leftrightarrow x=1\)

Vậy x là số nguyên

đề sai à??

Sửa đề:

\(x=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)

\(=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+12\sqrt{5}-8}}{\sqrt{5}+\sqrt{9-6\sqrt{5}+5}}\)

\(=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}\)

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

Thế vô A ta được

\(A=\left(3.\dfrac{1}{3^3}+8.\dfrac{1}{3^2}+2\right)^{2018}=3^{2018}\)

Thích không lập phương thì không lập phương. T dễ tính lắm

\(A=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(=\dfrac{1}{2}.\left(\sqrt[3]{40+16\sqrt{13}}+\sqrt[3]{40-16\sqrt{13}}\right)\)

\(=\dfrac{1}{2}.\left(\sqrt[3]{1+3\sqrt{13}+39+13\sqrt{13}}+\sqrt[3]{1-3\sqrt{13}+39-16\sqrt{13}}\right)\)

\(=\dfrac{1}{2}.\left(\sqrt[3]{\left(1+\sqrt{13}\right)^3}+\sqrt[3]{\left(1-\sqrt{13}\right)^3}\right)\)

\(=\dfrac{1}{2}.\left(1+\sqrt{13}+1-\sqrt{13}\right)=\dfrac{2}{2}=1\)

\(A=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(\sqrt[3]{5+2\sqrt{13}}=a\)

\(\sqrt[3]{5-2\sqrt{13}}=b\)

\(a^3+b^3=5+2\sqrt{13}+5-2\sqrt{13}=10\)

\(ab=\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}=\sqrt[3]{25-52}=\sqrt[3]{-27}=-3\)

\(A^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
\(A^3=10-9A\)
\(A^3+9a-10=0\)
\(\left(A-1\right)\left(A^2+A+10\right)=0\)
\(A^2+A+10>0\) mọi A
\(A-1=0\Rightarrow A=1\) là nghiệm duy nhất

KL: A = 1

\(\left\{{}\begin{matrix}2x^2+3xy-2y^2-5\left(2x-y\right)=0\\x^2-2xy-3y^2+15=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(x+2y\right)-5\left(2x-y\right)=0\\x^2-2xy-3y^2+15=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(x+2y-5\right)=0\left(1\right)\\x^2-2xy-3y^2+15=0\left(2\right)\end{matrix}\right.\)

\(PT\left(1\right)\Leftrightarrow\left[{}\begin{matrix}2x-y=0\\x+2y-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{y}{2}\\x=5-2y\end{matrix}\right.\)

Với \(x=\dfrac{y}{2}\) : \(PT\left(2\right)\Leftrightarrow\dfrac{y^2}{4}-y^2-3y^2+15=0\)

\(\Leftrightarrow-15y^2+60=0\)

\(\Leftrightarrow y^2-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Với \(x=5-2y\) : \(PT\left(2\right)\Leftrightarrow\left(5-2y\right)^2-2y\left(5-2y\right)-3y^2+15=0\)

\(\Leftrightarrow4y^2-20y+25+4y^2-10y-3y^2+15=0\)

\(\Leftrightarrow5y^2-30y+40=0\)

\(\Leftrightarrow y^2-6y+8=0\)

\(\Leftrightarrow\left(y-2\right)\left(y-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=2\\y=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

Vậy phương trình có 3 cặp nghiệm : \(\left[{}\begin{matrix}\left(x;y\right)=\left(-1;-2\right)\\\left(x;y\right)=\left(1;2\right)\\\left(x;y\right)=\left(-3;4\right)\end{matrix}\right.\)

Câu 1/ Ta có:

\(\left\{{}\begin{matrix}\sqrt{x^2-4x+5}=\sqrt{\left(x-2\right)^2+1}\ge1\\\sqrt{x^2-4x+8}=\sqrt{\left(x-2\right)^2+4}\ge2\\\sqrt{x^2-4x+9}=\sqrt{\left(x-2\right)^2+5}\ge\sqrt{5}\end{matrix}\right.\)

\(\Rightarrow VT\ge1+2+\sqrt{5}=VP\)

Dấu = xảy ra khi x = 2

PS: Câu còn lại thì chỉ cần phân tích cái trong căn thành số chính phương là xong.

Câu 2/ Sửa đề

\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)

Điều kiện: \(x\ge1\)

\(\Leftrightarrow\sqrt{\left(x-1\right)-4\sqrt{x-1}+4}+\sqrt{\left(x-1\right)+6\sqrt{x-1}+9}=5\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}=5\)

\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\sqrt{x-1}+3=5\)

Tới đây thì đơn giản rồi

Góc B bao nhiêu độ

Ta có hình vẽ như sau:

Trong tam giác vuông ACH có:

AC²=AH²+HC²=AH²+(BC-BH)²=AH²+BC²+BH²-2BCBH

Trong tam giác vuông ABH có:

AH²+BH²=AB² và BH=AB. cosB hay BH=c.cosB=> ĐPCM