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TT
1
19 tháng 5 2017
\(\sqrt{53-20\sqrt{4-\sqrt{9+4\sqrt{2}}}}=\sqrt{53-20\sqrt{4-\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}.1+1}}}=\sqrt{53-20\sqrt{4-\sqrt{\left(2\sqrt{2}+1\right)^2}}}=\sqrt{53-20\sqrt{4-2\sqrt{2}-1}}=\sqrt{53-20\sqrt{2-2\sqrt{2}+1}}=\sqrt{53-20\sqrt{\left(\sqrt{2}-1\right)^2}}=\sqrt{53-20\left(\sqrt{2}-1\right)}=\sqrt{53-20\sqrt{2}-20}=\sqrt{25-2.5.2\sqrt{2}+8}=\sqrt{\left(5-2\sqrt{2}\right)^2}=5-2\sqrt{2}\)
S
13 tháng 6 2019
\(A=2\sqrt{5}-\sqrt{45}+2\sqrt{20}=2\sqrt{5}-\sqrt{3^2.5}+2\sqrt{2^2.5}=2\sqrt{5}-3\sqrt{5}+4\sqrt{5}=3\sqrt{5}\)
\(B=\left(\sqrt{18}-\frac{1}{2}\cdot\sqrt{32}+12\sqrt{2}\right):\sqrt{2}=\left(3\sqrt{2}-\frac{1}{2}\cdot4\sqrt{2}+12\sqrt{2}\right):\sqrt{2}\)
\(=13\sqrt{2}:\sqrt{2}=13\)
\(C=\left(\sqrt{12}+2\sqrt{27}-3\sqrt{3}\right)\cdot\sqrt{3}=\left(2\sqrt{3}+6\sqrt{3}-3\sqrt{3}\right)\cdot\sqrt{3}=5\sqrt{3}\cdot\sqrt{3}=15\)
\(D=\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=-\sqrt{5}+15\sqrt{2}\)
18 tháng 12 2022
a: \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Leftrightarrow A^3=9+4\sqrt{5}+9-4\sqrt{5}+3\cdot A\)
=>A^3-3A-18=0
=>A=3
b: \(B=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
=>\(B^3=5\sqrt{2}+7-5\sqrt{2}+7+3B\)
=>B^3-3B-14=0
=>B=2,82
c: \(C^3=20+14\sqrt{2}-14\sqrt{2}+20-6C\)
=>C^3+6C-40=0
=>C=2,84
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 4 2019
\(A=\sqrt[3]{2^3+3.2^2.\sqrt{2}+3.2.\sqrt{2}^2+\sqrt{2}^3}+\sqrt[3]{\sqrt{2}^3-3.\sqrt{2}^2.2+3.\sqrt{2}.2^2-2^3}\)
\(A=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(\sqrt{2}-2\right)^3}\)
\(A=2+\sqrt{2}+\sqrt{2}-2=2\sqrt{2}\)
\(X=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(\Rightarrow X^3=\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)^3\)
\(\Rightarrow X^3=2+3\sqrt[3]{1-\frac{84}{81}}\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)\)
\(\Rightarrow X^3=2-3\sqrt[3]{\frac{1}{27}}.X\)
\(\Rightarrow X^3=2-X\)
\(\Rightarrow X^3+X-2=0\)
\(\Rightarrow\left(X-1\right)\left(X^2+2X+2\right)=0\)
\(\Rightarrow X=1\) (do \(X^2+2X+2=\left(X+1\right)^2+1>0\) \(\forall X\))
AH
Akai Haruma
Giáo viên
30 tháng 9 2020
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:
\(E=\frac{1+xy}{x+y}-\frac{1-xy}{x-y}=\frac{1+\frac{4}{3}}{2+\frac{2}{3}}-\frac{1-\frac{4}{3}}{2-\frac{2}{3}}=\frac{9}{8}\)
27 tháng 10 2019
a, c.Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
CW
25 tháng 6 2018
a) \(2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\) (*)
đk: x >/ 0
(*) \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)
\(\Leftrightarrow13\sqrt{2x}=28\) \(\Leftrightarrow\sqrt{2x}=\dfrac{28}{13}\Leftrightarrow2x=\left(\dfrac{28}{13}\right)^2\Leftrightarrow x=\dfrac{392}{169}\left(N\right)\)
Kl: \(x=\dfrac{392}{169}\)
b) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\) (*)
đk: x >/ 5
(*) \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\left(N\right)\)
Kl: x=9
c) \(\sqrt{\dfrac{3x-2}{x+1}}=2\) (*)
Đk: \(\left[{}\begin{matrix}x< -1\\x\ge\dfrac{2}{3}\end{matrix}\right.\)
(*) \(\Leftrightarrow\dfrac{3x-2}{x+1}=4\Leftrightarrow3x-2=4x+4\Leftrightarrow x=-6\left(N\right)\)
Kl: x=-6
d) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (*)
Đk: \(x\ge\dfrac{4}{5}\)
(*) \(\Leftrightarrow\sqrt{5x-4}=2\sqrt{x+2}\Leftrightarrow5x-4=4x+8\Leftrightarrow x=12\left(N\right)\)
Kl: x=12
9 tháng 9 2020
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 \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)
\(\sqrt{9+2\sqrt{20}}-\sqrt{9-2\sqrt{20}}=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|=\sqrt{5}+2-\sqrt{5}+2=4\)
= \(\sqrt{\left(2+\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=2+\sqrt{5}+2-\sqrt{5}=4\)