Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\) (*)
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
ĐKXĐ:...
\(A=\left(\frac{\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+2\right)}-\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a}+1}\right).\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}.\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}=\frac{1}{a}\)
\(C=\left(\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(=\left(\frac{\left(\sqrt{x}+1\right)}{-\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)}.\frac{\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)}\)
\(=\left(-1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\right).\sqrt{x}=\left(\frac{-x-\sqrt{x}-1+x+\sqrt{x}}{x+\sqrt{x}+1}\right)\sqrt{x}=\frac{-\sqrt{x}}{x+\sqrt{x}+1}\)
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 3 : Áp dụng BĐT Bu - nhi - a cốp xki ta có :
\(A=\sqrt{x-2}+\sqrt{4-x}\le\sqrt{\left(1^2+1^2\right)\left(x-2+4-x\right)}=\sqrt{2.2}=2\)
Vậy GTLN của A là 2 . Dấu \("="\) xảy ra khi \(x=3\)
\(B=\sqrt{6-x}+\sqrt{x+2}\le\sqrt{\left(1^2+1^2\right)\left(6-x+x+2\right)}=\sqrt{2.8}=4\)
Vậy GTLN của B là 4 . Dấu \("="\) xảy ra khi \(x=2\)
\(C=\sqrt{x}+\sqrt{2-x}\le\sqrt{\left(1^2+1^2\right)\left(x+2-x\right)}=\sqrt{2.2}=2\)
Vậy GTLN của C là 2 . Dấu \("="\) xảy ra khi \(x=1\)
Bài 2:
a .\(\dfrac{a+b}{2}\ge\sqrt{ab}\Leftrightarrow a+b-2\sqrt{ab}\ge0\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)
\("="\Leftrightarrow a=b\)
b. \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\Leftrightarrow a+b< \left(\sqrt{a}+\sqrt{b}\right)^2\Leftrightarrow a+b< a+b+2\sqrt{ab}\left(a,b>0\right)\)
\(c.a+b+\dfrac{1}{2}\ge\sqrt{a}+\sqrt{b}\) ( t nghĩ là > thôi )
d. \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
\(\Leftrightarrow2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\Leftrightarrow\left(a-2\sqrt{ab}+b\right)+\left(b-2\sqrt{bc}+c\right)+\left(c-2\sqrt{ca}+a\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\)
\("="\Leftrightarrow a=b=c\)
e. \(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\)
\(\Leftrightarrow\dfrac{a+b}{2}-\dfrac{a+b+2\sqrt{ab}}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a+2b-a-b-2\sqrt{ab}}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{4}\ge0\) ( đúng)
\("="\Leftrightarrow a=b\)
Minh bi nham dau bai, chi co 1 thua so \(\dfrac{2}{x}\) thoi nhe!
=\(\frac{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a-b}\)-\(\frac{2b}{a-b}\)
=\(\frac{\sqrt{a}^2+\sqrt{ab}-\sqrt{ab}+\sqrt{b}^2}{a-b}\)-\(\frac{2b}{a-b}\)
=\(\frac{a+b}{a-b}\)-\(\frac{2b}{a-b}\)
=\(\frac{a+b-2b}{a-b}\)
=\(\frac{a-b}{a-b}\)
=1
mong là đúng đừng trách mình nếu sai nhé ^^