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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
Bài 1 :
\(6xy\cdot\sqrt{\frac{9x^2}{16y^2}}=6xy\cdot\frac{3x}{4y}=\frac{18x^2y}{4y}=\frac{9}{2}x^2\)
\(\sqrt{\frac{4+20a+25a^2}{b^4}}=\sqrt{\frac{\left(2+5a\right)^2}{\left(b^2\right)^2}}=\frac{2+5a}{b^2}\)
\(\left(m-n\right).\sqrt{\frac{m-n}{\left(m-n\right)^2}}=\sqrt{\left(m-n\right)^2}\cdot\sqrt{\frac{1}{m-n}}=\sqrt{\frac{\left(m-n\right)^2}{m-n}}=\sqrt{m-n}\)
Bài 2 :
1. \(\left(2\sqrt{3}-\sqrt{12}\right):5\sqrt{3}=\left(2\sqrt{3}-2\sqrt{3}\right):5\sqrt{3}=0:5\sqrt{3}=0\)
2. \(\sqrt{\frac{317^2-302^2}{1013^2-1012^2}}=\frac{\sqrt{\left(317+302\right)\left(317-302\right)}}{\sqrt{\left(1013+1012\right)\left(1013-1012\right)}}=\frac{\sqrt{619}\cdot\sqrt{15}}{\sqrt{2025}}=\sqrt{\frac{619}{135}}\)(check lại)
3. \(\sqrt{27\left(1-\sqrt{3}\right)^2}:3\sqrt{75}\)
\(=\sqrt{27}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=3\sqrt{3}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=\frac{1-\sqrt{3}}{5}\)
4.\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\frac{5}{\sqrt{5}}+\frac{\sqrt{20}}{2}-\frac{\frac{5}{4}\cdot2}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\frac{2\sqrt{5}}{2}-\frac{\frac{5}{2}}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\sqrt{5}+\frac{\sqrt{5}}{2}+\sqrt{5}\right):2\sqrt{5}\)
\(=\frac{7}{2}\sqrt{5}:2\sqrt{5}\)
\(=\frac{7}{4}\)
Nhân tử và mẫu của biểu thức với \(\sqrt{m}+\sqrt{n}-\sqrt{m+n}.\)
\(\Rightarrow\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{\left(\sqrt{m}+\sqrt{n}+\sqrt{m+n}\right)\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}\)
\(=\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{\left(\sqrt{m}+\sqrt{n}\right)^2-\left(\sqrt{m+n}\right)^2}\)
\(=\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{m+n+2\sqrt{mn}-m-n}=\sqrt{m}+\sqrt{n}-\sqrt{m+n}\)
Ta có: \(\frac{2\sqrt{mn}}{\sqrt{m}+\sqrt{n}+\sqrt{m+n}}=\frac{2\sqrt{mn}.\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{(\sqrt{m}+\sqrt{n}+\sqrt{m+n})\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}\)
\(=\frac{2\sqrt{mn}.\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{\left(\sqrt{m}+\sqrt{n}\right)^2-\left(\sqrt{m+n}\right)^2}=\frac{2\sqrt{mn}.\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{m+2\sqrt{mn}+n-m-n}\)
\(=\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{2\sqrt{mn}}=\sqrt{m}+\sqrt{n}-\sqrt{m+n}\)( đpcm )
Áp dụng: Với \(m=2\)và \(n=5\)và \(mn=10\); \(m+n=7\)ta có:
\(\frac{2\sqrt{10}}{\sqrt{2}+\sqrt{5}+\sqrt{7}}=\sqrt{2}+\sqrt{5}-\sqrt{2+5}=\sqrt{2}+\sqrt{5}-\sqrt{7}\)
thực hiện phép tính
\(\frac{\sqrt{m^3}+4\sqrt{mn^2}-4\sqrt{m^2n}}{\sqrt{m^2n}-2\sqrt{mn^2}},m>0,n>o\)
\(\frac{\sqrt{m^3}+4\sqrt{mn^2}-4\sqrt{m^2n}}{\sqrt{m^2n}-2\sqrt{mn^2}}=\frac{\sqrt{m}\left(m+4n-4\sqrt{m}\sqrt{n}\right)}{\sqrt{m}\left(\sqrt{mn}-2n\right)}=\frac{\left(\sqrt{m}-2\sqrt{n}\right)^2}{\sqrt{n}\left(\sqrt{m}-2\sqrt{n}\right)}=\frac{\sqrt{m}-2\sqrt{n}}{\sqrt{n}}\)
thực hiện phép tính
\(\frac{\sqrt{m^3}+4\sqrt{mn^2}-4\sqrt{m^2n}}{\sqrt{m^2n}-2\sqrt{mn^2}},m>0,n>0\)