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Lời giải:
\(x=\sqrt{4+\sqrt{8}}.\sqrt{(2+\sqrt{2+\sqrt{2}})(2-\sqrt{2+\sqrt{2}})}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-(2+\sqrt{2})}=\sqrt{2(2+\sqrt{2})}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}.\sqrt{(2+\sqrt{2})(2-\sqrt{2})}=\sqrt{2}.\sqrt{2^2-2}=2\)
\(y=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\frac{\frac{2}{3}(9\sqrt{2}-6\sqrt{3}+3\sqrt{5})}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\frac{2}{3}\)
Do đó:
\(E=\frac{1+xy}{x+y}-\frac{1-xy}{x-y}=\frac{1+\frac{4}{3}}{2+\frac{2}{3}}-\frac{1-\frac{4}{3}}{2-\frac{2}{3}}=\frac{9}{8}\)
Bài 1: Tính
a) Ta có: \(\left(\sqrt{3}+2\right)^2\)
\(=\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot2+2^2\)
\(=3+4\sqrt{3}+4\)
\(=7+4\sqrt{3}\)
b) Ta có: \(-\left(\sqrt{2}-1\right)^2\)
\(=-\left[\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2\right]\)
\(=-\left(2-2\sqrt{2}+1\right)\)
\(=-\left(3-2\sqrt{2}\right)\)
\(=2\sqrt{2}-3\)
Bài 2: Tính
a) Ta có: \(0.5\cdot\sqrt{100}-\sqrt{\frac{25}{4}}\)
\(=\frac{1}{2}\cdot10-\frac{5}{2}\)
\(=5-\frac{5}{2}\)
\(=\frac{5}{2}\)
b) Ta có: \(\left(\sqrt{1\frac{9}{16}}-\sqrt{\frac{9}{16}}\right):5\)
\(=\left(\sqrt{\frac{25}{16}}-\frac{3}{4}\right)\cdot\frac{1}{5}\)
\(=\left(\frac{5}{4}-\frac{3}{4}\right)\cdot\frac{1}{5}\)
\(=\frac{2}{4}\cdot\frac{1}{5}\)
\(=\frac{1}{10}\)
Bài 3: So sánh
a) Ta có: \(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{12}\)
mà \(\sqrt{18}>\sqrt{12}\)(Vì 18>12)
nên \(3\sqrt{2}>2\sqrt{3}\)
\(\Leftrightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
b) Ta có: \(\left(15-2\sqrt{10}\right)^2\)
\(=225-2\cdot15\cdot2\sqrt{10}+\left(2\sqrt{10}\right)^2\)
\(=225-60\sqrt{10}+40\)
\(=265-60\sqrt{10}\)
\(=135+130-60\sqrt{10}\)
Ta có: \(\left(3\sqrt{15}\right)^2=3^2\cdot\left(\sqrt{15}\right)^2=9\cdot15=135\)
Ta có: \(130-60\sqrt{10}\)
\(=\sqrt{16900}-\sqrt{36000}< 0\)(Vì 16900<36000)
\(\Leftrightarrow130-60\sqrt{10}+135< 135\)(cộng hai vế của BĐT cho 135)
\(\Leftrightarrow\left(15-2\sqrt{10}\right)^2< \left(3\sqrt{15}\right)^2\)
\(\Leftrightarrow15-2\sqrt{10}< 3\sqrt{15}\)
\(\Leftrightarrow\frac{15-2\sqrt{10}}{3}< \frac{3\sqrt{15}}{3}=\sqrt{15}\)
hay \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)
\(A=\sqrt{\dfrac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}}:\dfrac{2}{\sqrt{3}-1}\)
\(=\sqrt{\dfrac{\sqrt{10}\left(2+\sqrt{3}\right)-\sqrt{2}\left(2+\sqrt{3}\right)}{2\sqrt{2}\left(\sqrt{5}-1\right)}}.\dfrac{\sqrt{3}-1}{2}\)
\(=\sqrt{\dfrac{\left(2+\sqrt{3}\right)\sqrt{2}\left(\sqrt{5}-1\right)}{2\sqrt{2}\left(\sqrt{5}-1\right)}}.\dfrac{\sqrt{3}-1}{2}\)
\(=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{4}\)
\(=\dfrac{2\sqrt{3}-2+3-\sqrt{3}}{4}\)
\(=\dfrac{1+\sqrt{3}}{4}\)
b1. a)
Gỉa sử căn bậc 2 + căn bậc 3 lớn hơn hoặc bằng căn bậc 10
=> ( căn bậc 2 + căn bậc 3 )2 lớn hơn hoặc bằng căn bậc 102
2+ 2 * căn bậc 3 + 3 lớn hơn hoặc bằng 10
5 + 2 căn 6 lớn hơn hoặc bằng 10
2 căn 6 lớn hơn hoặc bằng 5
( 2 căn 6 )2 lớn hơn hoặc bằng 52
4 * 6 lớn hơn 25
24 lớn hơn hoặc bằng 25 (sai)
Vậy căn bậc 2 + căn bậc 3 nhỏ hơn căn bậc 10
(bài 1) a) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\) = \(\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{25-24}\)
= \(\dfrac{-4\sqrt{6}}{1}\) = \(-4\sqrt{6}\)
b) \(\sqrt{6+2\sqrt{5}}-\dfrac{\sqrt{15}-\sqrt{3}}{\sqrt{3}}\) = \(\sqrt{\left(\sqrt{5}+1\right)^2}-\dfrac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{3}}\)
= \(\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\) = \(\sqrt{5}+1-\sqrt{5}+1\) = \(2\)
c) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\) = \(\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\)
= \(\sqrt{6}.\sqrt{16}\) = \(4\sqrt{6}\)
d) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
= \(\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
= \(\sqrt{3}+2+\sqrt{2}-\dfrac{1}{2-\sqrt{3}}\) = \(\dfrac{\left(\sqrt{3}+2+\sqrt{2}\right)\left(2-\sqrt{3}\right)-1}{2-\sqrt{3}}\)
= \(\dfrac{2\sqrt{3}-3+4-2\sqrt{3}+2\sqrt{2}-\sqrt{6}-1}{2-\sqrt{3}}\)
= \(\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3}}\) = \(\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{2}}\) = \(\sqrt{2}\)
e) \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\) = \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}\)
= \(\dfrac{4}{1+\sqrt{3}}-\sqrt{3}\) = \(\dfrac{4-\sqrt{3}-3}{1+\sqrt{3}}\) = \(\dfrac{1-\sqrt{3}}{1+\sqrt{3}}\)
= \(\dfrac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{1-3}\) = \(\dfrac{1-2\sqrt{3}+3}{-2}\) = \(\dfrac{4-2\sqrt{3}}{-2}\)
= \(\dfrac{-2\left(-2+\sqrt{3}\right)}{-2}\) = \(\sqrt{3}-2\)
bài 2)
a)\(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(a+b-2\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}-2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{a\sqrt{a}+-a\sqrt{b}+b\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\) = \(\dfrac{a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\) = \(a-b\)
b) \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2\left(a-1\right)}{4\sqrt{a}}.\dfrac{-4a}{a-1}\) = \(-2\)
a) \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{2-\sqrt{3}}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+2\sqrt{12}+2\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+4\sqrt{3}+2\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(8+4\sqrt{3}\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}\)
\(=\sqrt{\left(4-3\right)\cdot4}\)
\(=\sqrt{1\cdot4}\)
\(=\sqrt{4}\)
\(=2\)
b) \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(5\sqrt{2}-7\right)\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-5\sqrt{2}+7\)
\(=0+14\)
\(=14\)
c) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
dài quá ==' cả d, e, f nữa ==' có j rảnh lm cho nhé :D
a, Điều kiện x ∉ {\(\frac{5}{3};\frac{1}{7}\)}
\(\sqrt{3x-5}=\sqrt{7x-1}\)
\(\left(\sqrt{3x-5}\right)^2=\left(\sqrt{7x-1}\right)^2\)
\(\left|3x-5\right|=\left|7x-1\right|\)
\(3x-5=7x-1\)
\(-4x=4\) => x = -1
\(\sqrt[3]{200+126\sqrt[3]{2}+\dfrac{54}{1+\sqrt[3]{2}}}+\sqrt[3]{\dfrac{18}{1+\sqrt[3]{2}}-6\sqrt[3]{2}}\)
\(=\sqrt[3]{200+126\sqrt[3]{2}+\dfrac{54\left(1-\sqrt[3]{2}+\sqrt[3]{4}\right)}{3}}+\sqrt[3]{\dfrac{18\left(1-\sqrt[3]{2}+\sqrt[3]{4}\right)}{3}-6\sqrt[3]{2}}\)
\(=\sqrt[3]{218+108\sqrt[3]{2}+18\sqrt[3]{4}}+\sqrt[3]{6-12\sqrt[3]{2}+6\sqrt[2]{4}}\)
\(=\sqrt[3]{216+3.6^2\sqrt[3]{2}+3.6\sqrt[3]{4}+2}+\sqrt[3]{8-3.2^2\sqrt[3]{2}+3.2\sqrt[2]{4}-2}\)
\(=\sqrt[3]{\left(6+\sqrt[3]{2}\right)^3}+\sqrt[3]{\left(2-\sqrt[3]{2}\right)^3}\)
\(=6+\sqrt[3]{2}+2-\sqrt[3]{2}=8\)
Xin chào bạn. Rất vui đc làm quen với bạn. Chúc bạn chăm chỉ học tập như hiện tại nhé!!
\(A=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}\)
\(=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)+\sqrt{2}\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)}{2\sqrt{3}}\)
\(=\dfrac{\sqrt{2}\left(2\sqrt{3}-2+3-\sqrt{3}+2\sqrt{3}+2-3-\sqrt{3}\right)}{2\sqrt{3}}\)
\(=\dfrac{4\sqrt{3}-2\sqrt{3}}{2\sqrt{3}}\cdot\sqrt{2}=\sqrt{2}\)