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Chứng minh đẳng thức"
\(\dfrac{A+\sqrt{A}}{1+\sqrt{A}}=\dfrac{\sqrt{A}-A}{1-\sqrt{A}}\) (với A không âm và A khác 1)
giúp mình với ạ
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhân 2 vế với √2 ta có:
=> √2A = √2√(3 + √5) - √2√(3 - √5)
√2A = √(6 + 2√5) - √(6 - 2√5)
= √(5 + 2√5 + 1) - √(5 - 2√5 + 1)
= √(√5 + 1)² - √(√5 - 1)²
= (√5 + 1) - (√5 - 1)
= √5 + 1 - √5 + 1
= 2
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ Đặt \(\hept{\begin{cases}\sqrt{3+\sqrt{5}}=a\\\sqrt{3-\sqrt{5}}=b\end{cases}}\)
Khi đó ta có a2 + b2 = 6; ab = 2; a + b = \(\sqrt{10}\) ; a - b = \(\sqrt{2}\); a2 - b2 = \(2\sqrt{5}\)
Ta có cái ban đầu
\(=\frac{a^2}{\sqrt{10}+a}-\frac{b^2}{\sqrt{10}+b}\)=
\(\frac{\sqrt{10}a^2+a^2b-\sqrt{10}b^2-ab^2}{10+\sqrt{10}a+\sqrt{10}b+ab}\)
\(=\frac{10\sqrt{2}+2\sqrt{2}}{10+10+2}=\frac{6\sqrt{2}}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\) ta có:
\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
\(A^3=2+\sqrt{5}+2-\sqrt{5}+3\sqrt[3]{4-5}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3A\)
Giải PT:
\(A^3+3A-4=0\Leftrightarrow A^3-1+3A-3=0\)\(\Leftrightarrow\left(A-1\right)\left(A^2+A+1\right)+3\left(A-1\right)=0\)\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}A-1=0\\A^2+A+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}A=1\\A^2+2.\frac{1}{2}A+\frac{1}{4}-\frac{1}{4}+4=0\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}A=1\\\left(A+\frac{1}{2}\right)^2+\frac{15}{4}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}A=1\\\left(A+\frac{1}{2}\right)^2=-\frac{15}{4}\left(L\right)\end{cases}}}\)
Vậy \(A=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
\(\Leftrightarrow A^3=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\cdot\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(\Leftrightarrow A^3=4+3\cdot\left(-1\right)\cdot A\)
\(\Leftrightarrow A^3=4-3A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A^2+A^2-A+4A-4=0\)
\(\Leftrightarrow A^2\left(A-1\right)+A\left(A-1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\) ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))
\(=\sqrt{\left(2+\sqrt{7}+2-\sqrt{7}\right)\left(2+\sqrt{7}-2+\sqrt{7}\right)}\)
\(=\sqrt{4\cdot\sqrt{7}}\)
\(2,\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}-\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}=\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2=\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2-\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(=\left(3\sqrt{5}-5\sqrt{2}+5\sqrt{2}+3\sqrt{5}\right)\left(3\sqrt{5}-5\sqrt{2}-5\sqrt{2}-3\sqrt{5}\right)\)
\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)
\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}\)
cuối lười tính nên thôi nhá :>
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}-\dfrac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\dfrac{6+2\sqrt{5}}{2\sqrt{2}+\sqrt{2}\cdot\left(\sqrt{5}+1\right)}-\dfrac{6-2\sqrt{5}}{2\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)}\)
\(=\dfrac{6+2\sqrt{5}}{2\sqrt{2}+\sqrt{10}+\sqrt{2}}-\dfrac{6-2\sqrt{5}}{2\sqrt{2}-\sqrt{10}+\sqrt{2}}\)
\(=\dfrac{6+2\sqrt{5}}{3\sqrt{2}+\sqrt{10}}-\dfrac{6-2\sqrt{5}}{3\sqrt{2}-\sqrt{10}}\)
\(=\dfrac{\left(6+2\sqrt{5}\right)\left(3\sqrt{2}-\sqrt{10}\right)-\left(6-2\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)}{8}\)
\(=\dfrac{18\sqrt{2}-6\sqrt{10}+6\sqrt{10}-10\sqrt{2}-18\sqrt{2}-6\sqrt{10}+6\sqrt{10}+10\sqrt{2}}{8}\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,ĐK:x\le\dfrac{5}{3}\\ PT\Leftrightarrow-3x+5=49\\ \Leftrightarrow x=-\dfrac{44}{3}\left(tm\right)\\ b,ĐK:x\ge-12\\ PT\Leftrightarrow\dfrac{1}{2}x+6=2\\ \Leftrightarrow\dfrac{1}{2}x=-4\\ \Leftrightarrow x=-8\left(tm\right)\\ c,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow2x+1=13+4\sqrt{3}\\ \Leftrightarrow x=\dfrac{12+4\sqrt{3}}{2}=6+2\sqrt{3}\left(tm\right)\\ d,PT\Leftrightarrow\left|3x-1\right|=8\Leftrightarrow\left[{}\begin{matrix}3x-1=8\\1-3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)
Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(\Rightarrow A^2=3+\sqrt{5}+3-\sqrt{5}+2\sqrt{9-5}\)
\(=6+2\sqrt{4}\)
\(=10\)
Mà \(A>0\Rightarrow A=\sqrt{10}\)