Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
3.a.Ta có :\(\sqrt{40^2-24^2}=\sqrt{\left(40-24\right)\left(40+24\right)}=\sqrt{16.64}=4.8=32\)
b.Ta có :\(\sqrt{52^2-48^2}=\sqrt{\left(52-48\right)\left(52+48\right)}=\sqrt{4.100}=2.10=20\)
4.a)Ta có :
\(\sqrt{4x}=8\Leftrightarrow4x=8^2\Leftrightarrow4x=64\Leftrightarrow x=16\left(tm\right)\)
Vậy x=16
b)Ta có :
\(\sqrt{0,7x}=6\Leftrightarrow0,7x=36\Leftrightarrow x=\dfrac{36}{0.7}\left(tm\right)\)
Vậy x=\(\dfrac{36}{0.7}\)
c)Ta có:
\(9-4\sqrt{x}=1\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
Vậy x=4
d)Ta có :
\(\sqrt{5x}< 6\Leftrightarrow5x< 36\Leftrightarrow x< \dfrac{36}{5}\)
vậy 0≤x<\(\dfrac{36}{5}\)
Bài 3
a) \(\sqrt{40^2-24^2}\)
\(=\sqrt{\left(40+24\right)\left(40-24\right)}\)
=\(\sqrt{64.16}=\sqrt{64}.\sqrt{16}\)
\(=8.4=24\)
b)\(\sqrt{52^2-48^2}\)
\(=\sqrt{\left(52+48\right)\left(52-48\right)}\)
\(=\sqrt{100.4}=\sqrt{100}.\sqrt{4}\)
=10.2=20
Bài 4
a)\(\sqrt{4x}=8\)
\(\Leftrightarrow2\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)(TM)
b)\(\sqrt{0,7x}=6\)
\(\Leftrightarrow\sqrt{\left(0,7x\right)^2}=6^2\)
\(\Leftrightarrow\left|0,7x\right|=36\)
\(\Leftrightarrow\left[{}\begin{matrix}0,7x=36\\0,7x=-36\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{360}{7}\left(TM\right)\\x=-\dfrac{360}{7}\left(KTM\right)\end{matrix}\right.\)
c)\(9-4\sqrt{x}=1\)
\(\Leftrightarrow4\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(TM)
d)\(\sqrt{5x}< 6\)
Cau 1:
a: \(A=\dfrac{\left(\sqrt{a}-2\right)\left(a+2\sqrt{a}+4\right)+2\sqrt{a}\left(\sqrt{a}-2\right)}{a-4}\)
\(=\dfrac{\left(\sqrt{a}-2\right)\left(a+4\sqrt{a}+4\right)}{a-4}=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}=\sqrt{a}+2\)
c: \(=\dfrac{\left|c+1\right|}{\left|c\right|-1}\)
TH1: c>0
\(C=\dfrac{c+1}{c-1}\)
TH2: c<0
\(C=\dfrac{\left|c+1\right|}{-\left(c+1\right)}=\pm1\)
giai phuong trinh
\(\sqrt{x}-5+\dfrac{1}{3}\sqrt{9x}-45=\dfrac{1}{5}\sqrt{25x}-125=6\)
giup minh voi
Sửa đề: \(\sqrt{x-5}+\dfrac{1}{3}\sqrt{9x-45}=\dfrac{1}{5}\sqrt{25x-125}+6\)
\(\Leftrightarrow\sqrt{x-5}+\dfrac{1}{3}\cdot3\cdot\sqrt{x-5}-\dfrac{1}{5}\cdot5\sqrt{x-5}=6\)
\(\Leftrightarrow\sqrt{x-5}=6\)
=>x-5=36
hay x=41
dk \(1\le x\le3\)
\(P^2=x-1+3-x+2\sqrt{\left(x-1\right)\left(3-x\right)}\) =\(2+2\sqrt{\left(x-1\right)\left(3-x\right)}\)
ta co \(p^2\ge2\Rightarrow p\ge\sqrt{2}\) dau = xay ra khi \(\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
\(P^2=2+2\sqrt{\left(x-1\right)\left(3-x\right)}\le2+x-1+3-x=4\) (ap dung bdt amgm)\(\Rightarrow p\le2\)
dau = xay ra khi \(x-1=3-x\Leftrightarrow x=2\)
kl min p= \(\sqrt{2}khi\orbr{\begin{cases}x=1\\x=3\end{cases}}\) maxp= 2 khix=2
\(B=\sqrt{8-2\sqrt{5}-\sqrt{8+2\sqrt{5}}}\)
\(B=3,662841501-4,728708045\)
\(B=-1,653330227\)
\(\sqrt{9+3+4+5}=\sqrt{21}\)
\(\sqrt{9+3+4+5}\)
=\(\sqrt{21}\)
=4,582575695
ai k mình mình k lại