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\(\sqrt{\left(x+1\right)^2}-2\sqrt{x+1}=0\)\(\Leftrightarrow\left|x+1\right|-2\sqrt{x+1}=0\)
\(\Leftrightarrow\left|x+1\right|=2\sqrt{x+1}\)\(\Leftrightarrow\left|x+1\right|^2=\left(2\sqrt{x+1}\right)^2\)
\(\Leftrightarrow x^2+2x+1=4x+4\)\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow\left(x-1\right)^2-4=0\)\(\Leftrightarrow\left(x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=-2\\x-1=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy ..............
\(\sqrt{3000}.\sqrt{9000}+\sqrt{x}=30000\)
\(5196,15242271+\sqrt{x}=30000\)
\(\sqrt{x}=30000-5196,15242271\)
\(\sqrt{x}=24803,8475773\)
\(x=155,18971479225033\)
\(Vậy\)\(x=155,18971479225033\)
Bài1:
Ta có:
a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)
c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)
Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
Bài 2:
Không có đề bài à bạn?
Bài 3:
a)\(\sqrt{x}-1=4\)
\(\Rightarrow\sqrt{x}=5\)
\(\Rightarrow x=\sqrt{25}\)
\(\Rightarrow x=5\)
b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)
Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=4^2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
\(\sqrt{3333333}+\sqrt{33333333}+\sqrt{x}=2007\)
\(\Rightarrow\sqrt{33333333}+\sqrt{x}=181,2582329365414\)
\(\Rightarrow\sqrt{x}=-5592,24443009\)
\(\Rightarrow x=\sqrt{-5592,24443009}\)
\(\sqrt{3333333}+\sqrt{33333333}+\sqrt{x}=2007\)
\(\Rightarrow\sqrt{33333333}+\sqrt{x}=181,2582329365414\)
\(\Rightarrow\sqrt{x}=-5592,24443009\)
\(\Rightarrow x=\sqrt{-5592,24443009}\)
1. a)\(2\&\sqrt{5}\)
\(2=\sqrt{4}\)
=> \(2< \sqrt{5}\)
b)\(5\&\sqrt{23}\)
\(5=\sqrt{25}\)
=> \(5>\sqrt{23}\)
c) \(\sqrt{23}+\sqrt{13}\&\sqrt{83}\)
\(\left(\sqrt{23}+\sqrt{13}\right)^2=36+2\sqrt{229}\)
\(\left(\sqrt{83}\right)^2=83\)
\(\Rightarrow36+2\sqrt{299}< 83\)
=> \(\sqrt{23}+\sqrt{13}< \sqrt{83}\)
2. a) \(\sqrt{x}=5;x\ge0\)
=> x = 25
b) \(3\sqrt{x}=6;x\ge0\)
=> x = 4
c) trùng
d) \(3-\sqrt{3+1}=1\)
\(3-\sqrt{3+1}=3-2=1\)
1)
a)\(2=\sqrt{4}< \sqrt{5}\)
b) \(5=\sqrt{25}>\sqrt{23}\)
c) \(\sqrt{83}>\sqrt{81}=9\)
\(\left\{{}\begin{matrix}\sqrt{23}< \sqrt{25}=5\\\sqrt{13}< \sqrt{16}=4\end{matrix}\right.\)
\(\sqrt{23}+\sqrt{13}< 4+5=9\)
Vậy \(\sqrt{23}+\sqrt{13}< \sqrt{83}\)
2) Ta có:
\(\sqrt{x}=5\Rightarrow x=25\)
\(3\sqrt{x}=6\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(3-\sqrt{3+1}=1\)
Nên:
\(3-2=1\)(luôn đúng)
Bài 1:
a) Ta có: \(6=\sqrt{36}< \sqrt{37}\)
Vậy \(6< \sqrt{37}\)
b) Ta có: \(2\sqrt{3}=\sqrt{4}.\sqrt{3}=\sqrt{12}< \sqrt{18}=\sqrt{9}.\sqrt{2}=3\sqrt{2}\)
Vậy \(2\sqrt{3}< 3\sqrt{2}\)
p/s: Bạn có thể lấy số gần mà tính cũng được do mình nghĩ lớp 7 chưa học mà học rồi thì làm cách trên cho nhanh nhé.
c) Ta có: \(\sqrt{63}\approx7,4;\sqrt{35}\approx6\)
Mà \(7,4+6=13,4< 14\Rightarrow\sqrt{63}+\sqrt{35}< 14\)
Câu 2: a) \(\sqrt{x-1}=\frac{1}{2}\Rightarrow\left(\sqrt{x-1}\right)^2=\left(\frac{1}{2}\right)^2\Rightarrow x-1=\frac{1}{4}\Rightarrow x=\frac{5}{4}\)
b) \(\sqrt{\left(x-1\right)^2}=9=\sqrt{81}\Rightarrow\left(x-1\right)^2=81\Rightarrow x-1\in\left\{\pm9\right\}\Rightarrow x\in\left\{10;-8\right\}\)
c) \(2\sqrt{3x-2}=3\Rightarrow\sqrt{3x-2}=\frac{3}{2}=\sqrt{\frac{9}{4}}\Rightarrow3x-2=\frac{9}{4}\Rightarrow x=\frac{17}{12}\)
\(\sqrt{300000}+\sqrt{123456789}+\sqrt{x}=1234567897\)
\(\Leftrightarrow\sqrt{123456789}+\sqrt{x}=123456241,277\)
\(\Leftrightarrow\sqrt{x}=123467352,389\)
\(\Leftrightarrow x=17,5244187271828182846\)
\(\sqrt{300000}+\sqrt{123456789}+\sqrt{x}=1234567897\)
\(\Leftrightarrow\sqrt{123456789}+\sqrt{x}=123456241,277\)
\(\Leftrightarrow\sqrt{x}=123467352,389\)
\(\Leftrightarrow x=123467352,389^2\)