Bài 11: Số vô tỉ. Khái niệm về căn bậc hai

để B nguyên thì \(5⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

ta có bảng sau

vậy x \(\in\left\{0;4;36\right\}\)

\(\sqrt{50}\)+ \(\sqrt{26+1}\) > \(\sqrt{165}\)

\(\sqrt{\dfrac{169}{64}}=\sqrt{\dfrac{13^2}{8^2}}=\dfrac{13}{8}\)

\(\dfrac{\sqrt{169}}{\sqrt{64}}=\dfrac{\sqrt{13^2}}{\sqrt{8^2}}=\dfrac{13}{8}\)

Vậy \(\sqrt{\dfrac{169}{64}}=\dfrac{\sqrt{169}}{\sqrt{64}}\)

Tương tự

Bị đao không hai căn bậc bằng nhau hết mà tính làm gì nhìn vô là biết bằng roy :V

\(a)\sqrt{\dfrac{169}{64}}\) và \(\dfrac{\sqrt{169}}{\sqrt{64}}\)

\(\sqrt{\dfrac{169}{45}}=\sqrt{\dfrac{13^2}{8^2}}=\dfrac{13}{8}\)

\(\dfrac{\sqrt{169}}{\sqrt{64}}=\dfrac{\sqrt{13^2}}{\sqrt{8^2}}=\dfrac{13}{8}\)

Vì \(\dfrac{13}{8}=\dfrac{13}{8}\)

\(\Rightarrow\sqrt{\dfrac{169}{64}}=\dfrac{\sqrt{169}}{\sqrt{64}}\)

\(b)\sqrt{\dfrac{2,25}{2,56}}\) và \(\dfrac{\sqrt{2,25}}{\sqrt{2,56}}\)

\(\sqrt{\dfrac{2,25}{2,56}}=\dfrac{15}{16}\)

\(\dfrac{\sqrt{2,25}}{\sqrt{2,56}}=\dfrac{15}{16}\)

Vì \(\dfrac{15}{16}=\dfrac{15}{16}\)

\(\Rightarrow\sqrt{\dfrac{2,25}{2,56}}=\dfrac{\sqrt{2,25}}{\sqrt{2,56}}\)

Chúc bạn học tốt!

a/ \(x^2=5\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)

vậy .....

b/ \(x^2-9=0\)

\(\Leftrightarrow x^2=9\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=3^2\\x^2=\left(-3\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy .......( nhầm cái ngoặc)

c/ \(x^2+1=0\)

\(\Leftrightarrow x^2=-1\)

Mà \(x^2\ge0\Leftrightarrow x\in\varnothing\)

Vậy ....

d/ \(\left(x-1\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=3^2\\\left(x-1\right)^2=\left(-3\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

Vậy ...

e/ \(\left(2x+3\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x+3\right)^2=5^2\\\left(2x+3\right)^2=\left(-5\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

Vậy .....

f/ Ta có :

\(x^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=1^2\\x^2=\left(-1\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy ...

\(x^2=5\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)

\(\left(x-1\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

\(x^2-9=0\Leftrightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(\left(2x+3\right)^2=25\)

\(\Rightarrow\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varnothing\)

\(x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

\(\sqrt{\left(3-\sqrt{5}\right)^2}=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)

\(\sqrt{\left(\sqrt{3-\sqrt{4}}\right)^2}=\left|\sqrt{3-\sqrt{4}}\right|=\sqrt{3-\sqrt{4}}\)

\(\sqrt{\left(7-\sqrt{34}\right)^2}=\left|7-\sqrt{34}\right|=7-\sqrt{34}\)

\(\sqrt{4}+\sqrt{9}+\sqrt{16}+\sqrt{25}=2+3+4+5=14\)

\(\sqrt{4}\) + \(\sqrt{9}\) + \(\sqrt{16}\) + \(\sqrt{25}\)

= 2 + 3 + 4 + 5

=     5   + 4 + 5

= 9 + 5 = 14 

a)\(\sqrt{1}\)+\(\sqrt{9}\)+\(\sqrt{25}\)+\(\sqrt{49}\)+\(\sqrt{81}\)

=1+3+5+7+9

=25

b)=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{4}\)

=\(\dfrac{6}{12}\)+\(\dfrac{4}{12}\)+\(\dfrac{2}{12}\)+\(\dfrac{3}{12}\)

=\(\dfrac{15}{12}\)

c) =0,2+0.3+0,4

= 0.9

d) =9-8+7

=8

j) =1,2-1,3+1.4

= (-0,1)+1,4

=1,4

g) \(\dfrac{2}{5}\)+\(\dfrac{5}{2}\)+\(\dfrac{9}{10}\)+\(\dfrac{3}{4}\)

= (\(\dfrac{4}{10}\)+\(\dfrac{15}{10}\)+\(\dfrac{9}{10}\))+\(\dfrac{3}{4}\)

= \(\dfrac{14}{5}\)+\(\dfrac{3}{4}\)

=\(\dfrac{56}{20}\)+\(\dfrac{15}{20}\)

= \(\dfrac{71}{20}\)

Nhớ tick cho mk nha~

5,2 \(\in\)Q

4,6351 [ ] I

-7,0903... [ ] Q

1,333 [ ] I

\(\left(2x^2-3\right)\left(3x^2-\dfrac{1}{0,12}\right)\left(x^2+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x^2-3=0\\3x^2-\dfrac{1}{0,12}=0\\x^2+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x^2=3\\3x^2=\dfrac{1}{0,12}\\x^2=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x^2=3\Rightarrow x^2=1,5\\3x^2=\dfrac{1}{0,12}\Rightarrow x^2=\dfrac{25}{9}\\x^2=-1\Rightarrow x\in\varnothing\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\pm\sqrt{1,5}\\x=\pm\dfrac{5}{3}\end{matrix}\right.\)

\(linh=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}\)

\(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}\\\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}\\.............\\\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}\end{matrix}\right.\)

Suy ra:

\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+....+\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{99}}>\dfrac{99}{\sqrt{100}}\)

\(linh=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.....+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}>\dfrac{99}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}\)

\(\)\(linh>10\left(đpcm\right)\)

Bài này ko phải 100 nhé

bạn nào giải giúp mình với