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Bài 1 : Ta có :
\(A=\sqrt{3x+\sqrt{6x-1}}+\sqrt{3x-\sqrt{6x-1}}\)
\(A\sqrt{2}=\sqrt{6x+2\sqrt{6x-1}}+\sqrt{6x-2\sqrt{6x-1}}\)
\(=\sqrt{6x-1+2\sqrt{6x-1}+1}+\sqrt{6x-1-2\sqrt{6x-1}+1}\)
\(=\sqrt{\left(\sqrt{6x-1}+1\right)^2}+\sqrt{\left(\sqrt{6x-1}-1\right)^2}\)
\(=\left|\sqrt{6x-1}+1\right|+\left|\sqrt{6x-1}-1\right|\)
\(=\sqrt{6x-1}+1+\sqrt{6x-1}-1\)
\(=2\sqrt{6x-1}\)
\(\Rightarrow A=\sqrt{2}\left(\sqrt{6x-1}\right)\)
Thay \(x=4+\sqrt{10}\) vào A ta được :
\(A=\sqrt{2}.\sqrt{6\left(4+\sqrt{10}\right)-1}=\sqrt{2}.\sqrt{24+6\sqrt{10}-1}\)
\(=\sqrt{2}.\sqrt{23+6\sqrt{10}}=\sqrt{46+12\sqrt{10}}\)
\(=\sqrt{36+12\sqrt{10}+10}=\sqrt{\left(6+\sqrt{10}\right)^2}=6+\sqrt{10}\)
Vậy \(A=6+\sqrt{10}\) tại \(x=4+\sqrt{10}\)
1/ \(a+1=\sqrt[4]{\frac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}-\sqrt[4]{\frac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}=\sqrt{\frac{\sqrt{3}+1}{\sqrt{3}-1}}-\sqrt{\frac{\sqrt{3}-1}{\sqrt{3}+1}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}}=\frac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
2/ \(a+b=5\Leftrightarrow\left(a+b\right)^3=125\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=125\)
\(\Rightarrow a^3+b^3=125-3ab\left(a+b\right)=125-3.1.5=110\)
3/ \(mn\left(mn+1\right)^2-\left(m+n\right)^2.mn\)
\(=mn\left(\left(mn+1\right)^2-\left(m+n\right)^2\right)\)
\(=mn\left(mn+1-m-n\right)\left(mn+1+m+n\right)\)
\(=mn\left(m-1\right)\left(n-1\right)\left(m+1\right)\left(n+1\right)\)
\(=\left(m-1\right)m\left(m+1\right)\left(n-1\right)n\left(n+1\right)\)
Do \(\left(m-1\right)m\left(m+1\right)\) và \(\left(n-1\right)n\left(n+1\right)\) đều là tích của 3 số nguyên liên tiếp nên chúng đều chia hết cho 3 \(\Rightarrow\) tích của chúng chia hết cho 36
4/
Do \(0\le x\le1\Rightarrow\left\{{}\begin{matrix}x\ge0\\x-1\le0\end{matrix}\right.\) \(\Rightarrow x\left(x-1\right)\le0\)
\(\Leftrightarrow x^2-x\le0\Leftrightarrow x^2\le x\)
Dấu "=" xảy ra khi \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
5/ Đặt \(\left\{{}\begin{matrix}\sqrt{5a+4}=x\\\sqrt{5b+4}=y\\\sqrt{5c+4}=z\end{matrix}\right.\)
Do \(a+b+c=1\Rightarrow0\le a;b;c\le1\)
\(\Rightarrow2\le x;y;z\le3\) và \(x^2+y^2+z^2=5\left(a+b+c\right)+12=17\)
Khi đó ta có:
Do \(2\le x\le3\Rightarrow\left(x-2\right)\left(x-3\right)\le0\)
\(\Leftrightarrow x^2-5x+6\le0\Leftrightarrow x\ge\frac{x^2+6}{5}\)
Tương tự: \(y\ge\frac{y^2+6}{5}\) ; \(z\ge\frac{z^2+6}{5}\)
Cộng vế với vế:
\(A=x+y+z\ge\frac{x^2+y^2+z^2+18}{5}=\frac{17+18}{5}=7\)
\(\Rightarrow A_{min}=7\) khi \(\left(x;y;z\right)=\left(2;2;3\right)\) và các hoán vị hay \(\left(a;b;c\right)=\left(0;0;1\right)\) và các hoán vị
\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2-2xy\ge0\)
\(\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow x^2+y^2+2xy\ge4xy\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\Rightarrow1\ge4xy\Leftrightarrow xy\le\frac{1}{4}\)(1)
\(\left(x-y\right)^2\ge0\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow\left(x+y\right)^2\ge2\Leftrightarrow x+y\ge\sqrt{2}\)
Từ phần a ta có \(x+y\le\sqrt{2}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{2x+1}+\sqrt{2y+1}\right)^2\)
\(\le\left(1+1\right)\left(2\left(x+y\right)+2\right)\)
\(=2\cdot\left(2\left(x+y\right)+2\right)\le2\cdot\left(2\sqrt{2}+2\right)\)
\(=4\sqrt{2}+4=VP^2\)
Suy ra \(VT\ge VP\) (ĐPCM)
\(x\ne1\)
\(VT=\frac{2\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}=\frac{2\sqrt{x}}{1+\sqrt{x}}=\frac{2\sqrt{x}+2-2}{1+\sqrt{x}}=2-\frac{2}{1+\sqrt{x}}< 2\)
Mặt khác \(\left(\sqrt{3}+\sqrt{2}\right)^2=5+2\sqrt{6}>5+2\sqrt{4}=9\)
\(\Rightarrow\sqrt{3}+\sqrt{2}>3\Rightarrow\sqrt{3}+\sqrt{2}-1>2\Rightarrow VP>2\)
\(\Rightarrow VP>VT\)
BĐT sai do dấu "=" ko xảy ra
Lời giải:
Áp dụng BĐT Bunhiacopxky:
$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq (6x+3y+2z)(\frac{1}{6}+\frac{1}{3}+\frac{1}{2})$
Mà: $6x+3y+2z=3x+(x+y)+2(x+y+z)\leq 3.1+5+2.14=36$
Do đó: $(\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq 36.(\frac{1}{6}+\frac{1}{3}+\frac{1}{2})=36$
$\Rightarrow \sqrt{x}+\sqrt{y}+\sqrt{z}\leq 6$ (đpcm)
Dấu "=" xảy ra khi $x=1; y=2; z=3$
Lời giải:
Áp dụng BĐT Bunhiacopxky:
$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq (6x+3y+2z)(\frac{1}{6}+\frac{1}{3}+\frac{1}{2})$
Mà: $6x+3y+2z=3x+(x+y)+2(x+y+z)\leq 3.1+5+2.14=36$
Do đó: $(\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq 36.(\frac{1}{6}+\frac{1}{3}+\frac{1}{2})=36$
$\Rightarrow \sqrt{x}+\sqrt{y}+\sqrt{z}\leq 6$ (đpcm)
Dấu "=" xảy ra khi $x=1; y=2; z=3$
Ta có \(B=\sqrt{x+3}+\sqrt{5-x}\Leftrightarrow B^2=x+3+5-x+2\sqrt{\left(x+3\right)\left(5-x\right)}=8+2\sqrt{\left(x+3\right)\left(5-x\right)}\) Ta có \(\sqrt{\left(x+3\right)\left(5-x\right)}\ge0\Leftrightarrow2\sqrt{\left(x+3\right)\left(5-x\right)}\ge0\Leftrightarrow8+2\sqrt{\left(x+3\right)\left(5-x\right)}\ge8\Leftrightarrow B^2\ge8\Leftrightarrow B\ge2\sqrt{2}\)Vậy \(2\sqrt{2}\le B\)(1)
Áp dụng bđt Bunhia copski ta có
\(B^2=\left(\sqrt{x+3}+\sqrt{5-x}\right)^2=\left(\sqrt{x+3}.1+\sqrt{5-x}.1\right)^2\le\left[\left(\sqrt{x+3}\right)^2+\left(\sqrt{5-x}\right)^2\right]\left(1^2+1^2\right)=8.2=16\Leftrightarrow B^2\le16\Leftrightarrow B\le4\)(2)
Từ (1),(2)\(\Rightarrow2\sqrt{2}\le B\le4\)