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b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{2}+2}-\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{2}\right|-\left|\sqrt{5}+\sqrt{2}\right|\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\) (vì \(\sqrt{5}\ge\sqrt{2}\)
=0
c) \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}-1+\sqrt{3+1}\) (vì \(\sqrt{3}\ge1\))
\(=2\sqrt{3}\)
a)\(\sqrt{5+2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}-\sqrt{3-2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\) (vì \(\sqrt{3}\ge\sqrt{2}\))
=0
rút gọn :
A=\(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\)
B=\((a^5+2a^4-13a^3-a^2+18a-17)^{2017}\)
\(A=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\)
\(\Rightarrow A^3=14+3\sqrt[3]{\left(7+5\sqrt{2}\right)\left(7-5\sqrt{2}\right)}\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\)
<=>A3=14-3A
<=>A3+3A-14=0
<=>A3-4A+7A-14=0
<=>A.(A-2)(A+2)+7.(A-2)=0
<=>(A-2)(A2+9A-14)=0
<=>A=2(nhận)
Vậy A=2
a) \(\sqrt{3-2\sqrt{2}}\)+\(\sqrt{3+2\sqrt{2}}\)
= \(\sqrt{2-2\sqrt{2}+1}\)+ \(\sqrt{2+2\sqrt{2}+1}\)
= \(\sqrt{\left(\sqrt{2}-1\right)^2}\)+ \(\sqrt{\left(\sqrt{2}+1\right)^2}\)
= \(\sqrt{2}\)-1+\(\sqrt{2}\)+1
=2\(\sqrt{2}\)
b) \(\sqrt{9-4\sqrt{5}}\)+ \(\sqrt{6+2\sqrt{5}}\)
= \(\sqrt{4-4\sqrt{5}+5}\)+\(\sqrt{4+2\sqrt{5}+2}\)
= \(\sqrt{\left(2-\sqrt{5}\right)^2}\)+\(\sqrt{\left(2+\sqrt{2}\right)^2}\)
= 2-\(\sqrt{5}\)+2+\(\sqrt{2}\)
= 4-\(\sqrt{5}\)+\(\sqrt{2}\)
\(A=\sqrt[3]{2-\sqrt{5}}\left(\sqrt[6]{\left(2+\sqrt{5}\right)^2}+\sqrt[3]{2+\sqrt{5}}\right)\)
\(=\sqrt[3]{2-\sqrt{5}}.2\sqrt[3]{2+\sqrt{5}}=2\sqrt[3]{4-5}=-2\)
\(B=\sqrt[4]{\left(3-2\sqrt{2}\right)^2}-\sqrt{2}=\sqrt{3-2\sqrt{2}}-\sqrt{2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{2}=\sqrt{2}-1-\sqrt{2}=-1\)
\(C=\sqrt[4]{\left(6-2\sqrt{5}\right)^2}=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
\(D=1+\sqrt[4]{\left(4-2\sqrt{3}\right)^2}=1+\sqrt{4-2\sqrt{3}}\)
\(=1+\sqrt{\left(\sqrt{3}-1\right)^2}=1+\sqrt{3}-1=\sqrt{3}\)
Câu e lấy nguyên văn từ sách thầy Vũ Hữu Bình:
Đặt \(x=\sqrt[4]{5}\Rightarrow x^4=5\Rightarrow5-x^4=0\)
\(E=\frac{2}{\sqrt{4-3x+2x^2-x^3}}=\frac{2\left(x+1\right)}{\sqrt{\left(x+1\right)^2\left(4-3x+2x^2-x^3\right)}}=\frac{2\left(x+1\right)}{\sqrt{-x^5+5x+4}}\)
\(E=\frac{2\left(x+1\right)}{\sqrt{x\left(5-x^4\right)+4}}=\frac{2\left(x+1\right)}{\sqrt{4}}=x+1=\sqrt[4]{5}+1\)
Không hiểu ý tưởng nhân cả tử và mẫu với \(x+1\) từ đâu ra luôn
a, ĐK: \(x\le-1,x\ge3\)
\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x^2-2x-3}+3\right).\left(\sqrt{x^2-2x-3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x-3}=-\dfrac{3}{2}\left(l\right)\\\sqrt{x^2-2x-3}=1\end{matrix}\right.\)
\(\Leftrightarrow x^2-2x-3=1\)
\(\Leftrightarrow x^2-2x-4=0\)
\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)
b, ĐK: \(-2\le x\le2\)
Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)
Khi đó phương trình tương đương:
\(3t-t^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2+x}-2\sqrt{2-x}=0\\\sqrt{2+x}-2\sqrt{2-x}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)
Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm
Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)
1) ĐK:x\(\ge\frac{1}{2}\)
PT\(\Leftrightarrow\sqrt{2x-1}=x\)
\(\Leftrightarrow\begin{cases}x\ge0\\2x-1=x^2\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge0\\x=1\end{cases}\)
\(\Leftrightarrow x=1\) (thỏa mãn)
\(A=\frac{\left(3+\sqrt{5}\right)^2+\left(3-\sqrt{5}\right)^2}{\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(A=\frac{18+10}{4}\)
\(A=7\)
