\(\sqrt{2017}\)- \(\sqrt{2015}\); B=
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a: \(=\left|x-4\right|-\left|x-2\right|\)

\(=\left|3\sqrt{2}-1-4\right|-\left|3\sqrt{2}-1-2\right|\)

\(=5-3\sqrt{2}-\left(3\sqrt{2}-3\right)=-6\sqrt{2}+8\)

b: \(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)

\(=\left|\sqrt{7}-1+1\right|+\left|\sqrt{7}-1-1\right|\)

\(=\sqrt{7}+4-\sqrt{7}=4\)

1 tháng 8 2018

a) Ta có: \(\left(\sqrt{2017}+\sqrt{2019}\right)^2=2017+2019+2\sqrt{2017.2019}\)

                                                              \(=4036+2\sqrt{\left(2018-1\right).\left(2018+1\right)}\)

                                                                \(=4036+2\sqrt{2018^2-1}< 4036+2\sqrt{2018^2}=2018.4=\left(2\sqrt{2018}\right)^2\)

Vậy x < y

25 tháng 7 2016

binh phuong len

25 tháng 7 2016

\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40}\)

\(\left(\sqrt{7}+\sqrt{6}\right)^2=13+2\sqrt{42}\)

vi

\(2\sqrt{40}< 2\sqrt{42}\)

nen \(\sqrt{8}+\sqrt{5}< \sqrt{7}+\sqrt{6}\)

\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

11 tháng 8 2017

\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\) \(=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

11 tháng 8 2017

a/ \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\) \(=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}=-2\sqrt{3}\).

b/ \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\Rightarrow A^2=8+2\sqrt{4^2-\left(\sqrt{10+2\sqrt{5}}\right)^2}=8+2\sqrt{6-2\sqrt{5}}\) \(=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)

\(\Rightarrow A=\sqrt{5}+1\)

c/ \(B=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\Rightarrow\sqrt{2}B=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\sqrt{5}+2=2\Rightarrow B=\sqrt{2}\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6\right)^2=\left(1-x\right)^2\\-3< =x< =1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6+x-1\right)\left(2x+6+1-x\right)=0\\-3< =x< =1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x+5\right)\left(x+7\right)=0\\-3< =x< =1\end{matrix}\right.\Leftrightarrow x=-\dfrac{5}{3}\)

b: \(\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=2x\)

\(\Leftrightarrow4\sqrt{x-3}=2x\)

\(\Leftrightarrow2\sqrt{x-3}=x\)

\(\Leftrightarrow\sqrt{4x-12}=x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\x^2=4x-12\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

26 tháng 3 2017

Dùng BĐT Bunhiacopski:

Ta có: \(ac+bd\le\sqrt{a^2+b^2}.\sqrt{c^2+d^2}\)

\(\left(a+c\right)^2+\left(b+d\right)^2\)

\(=a^2+b^2+2\left(ac+bd\right)+c^2+d^2\)

\(\le\left(a^2+b^2\right)+2\sqrt{a^2+b^2}.\sqrt{c^2+d^2}+c^2+d^2\)

\(\Rightarrow\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\le\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\) (Đpcm)

26 tháng 3 2017

Câu hỏi của Hoàng Khánh Linh - Toán lớp 8 - Học toán với OnlineMath copy nhớ ghi nguồn

3 tháng 7 2017

\(P=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(P=\dfrac{\sqrt{2}-1}{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{99}+\sqrt{100}\right)\left(\sqrt{100}-\sqrt{99}\right)}\)

\(P=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}\)

\(P=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)

\(P=-1+\sqrt{100}=-1+10=9\)

3 tháng 7 2017

Áp dụng:\(\dfrac{1}{\sqrt{a}+\sqrt{a+1}}=\dfrac{\sqrt{a+1}-\sqrt{a}}{\left(\sqrt{a}+\sqrt{a+1}\right)\left(\sqrt{a+1}-\sqrt{a}\right)}=\dfrac{\sqrt{a+1}-\sqrt{a}}{a+1-a}=\sqrt{a+1}-\sqrt{a}\)