Bài 1: Căn bậc hai

a) \(A=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\)

\(=\sqrt{3}-1+-\sqrt{3}-1=-2\)

b) \(B=\sqrt{11-6\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

\(=\sqrt{3^2-2.3.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)

\(=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\left|3-\sqrt{2}\right|-\left|\sqrt{2}-1\right|\)

\(=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)

c) \(C=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\)

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

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

\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{2}\right)=5-\sqrt{10}+\sqrt{15}-\sqrt{6}\)

a)Pt \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\dfrac{1}{3}+\dfrac{1}{2}\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{5}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{5}{6}\\2x-1=-\dfrac{5}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=\dfrac{1}{12}\end{matrix}\right.\)

Vậy...

b)Đk:\(x\ge3\)

Pt \(\Leftrightarrow\sqrt{x-3}\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\x-4=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=4\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)

Vậy...

c)Đk:\(x\ge1\)

\(x+\sqrt{x-1}=13\)

\(\Leftrightarrow\sqrt{x-1}=13-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}13-x\ge0\\x-1=x^2-26x+169\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-27x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-17x-10x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left(x-17\right)\left(x-10\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left[{}\begin{matrix}x=17\\x=10\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=10\) (tm)

Vậy...

Đề có vấn đề gì không bạn???

Đk: \(x^2-3x+7\ge0\)

\(\Leftrightarrow x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}+\dfrac{19}{4}\ge0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge0\) (lđ với mọi x)

Vậy biểu thức luôn xác định với mọi x

\(\sqrt{x^2-3x+7}\)

Có \(x^2-3x+7=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{19}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{19}{4}>0\forall x\)

ĐKXĐ: \(x\in R\)

14) Ta có: \(\sqrt{\left(x-2\right)^2}+\dfrac{x-2}{\sqrt{\left(x-2\right)^2}}\)

\(=x-2+\dfrac{x-2}{x-2}\)

\(=x-2+1=x-1\)

16) Ta có: \(\sqrt{m}-\sqrt{m-2\sqrt{m}+1}\)

\(=\sqrt{m}-\left(\sqrt{m}-1\right)\)

\(=\sqrt{m}-\sqrt{m}+1\)

=1

18) Ta có: \(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

\(=\sqrt{x-4+2\cdot\sqrt{x-4}\cdot2+4}+\sqrt{x-4-2\cdot\sqrt{x-4}\cdot2+4}\)

\(=\sqrt{x-4}+2+\sqrt{x-4}-2\)

\(=2\sqrt{x-4}\)

2) Ta có: \(\sqrt{\left(x-3\right)^2}-x\)

\(=x-3-x\)

=-3

4) Ta có: \(\sqrt{m^2-6m+9}-2m\)

\(=m-3-2m\)

\(=-m-3\)

6) Ta có: \(2x-\sqrt{4x^2+4x+1}\)

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

=-1

8) Ta có: \(x+y-\sqrt{x^2-2xy+y^2}\)

\(=x+y-x+y\)

\(=2y\)

10) Ta có: \(m+2n-\sqrt{m^2-4mn+4n^2}\)

\(=m+2n-m+2n\)

=4n

12) Ta có: \(1+\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}\)

\(=1+\dfrac{x-1}{x-1}\)

=1+1=2

a, Ta có : \(\left\{{}\begin{matrix}\left(\sqrt{26}+3\right)^2=35+6\sqrt{26}=35+\sqrt{936}\\\left(\sqrt{63}\right)^2=63=35+\sqrt{784}\end{matrix}\right.\)

Thấy : \(936>784\)

\(\Rightarrow\sqrt{26}+3>\sqrt{63}\)

b, Ta có : \(3< 4\)

\(\Rightarrow\dfrac{\sqrt{3}}{2}< \dfrac{\sqrt{4}}{2}=1\)

\(\Rightarrow\dfrac{\sqrt{3}-1}{2}< 1-\dfrac{1}{2}=\dfrac{1}{2}\)

Vậy ...

a) Ta có: \(\left(\sqrt{26}+3\right)^2=35+6\sqrt{26}\)

\(\left(\sqrt{63}\right)^2=63=35+28\)

mà \(6\sqrt{26}=\sqrt{936}>\sqrt{784}=28\)

nên \(\sqrt{26}+3>\sqrt{63}\)

b) Ta có: \(1-\sqrt{3}+1=2-\sqrt{3}>0\)

nên \(1>\sqrt{3}-1\)

\(\Leftrightarrow\dfrac{1}{2}>\dfrac{\sqrt{3}-1}{2}\)

`D=(sqrtx+1+x)/sqrtx(x>0)`

`D=1+1/sqrtx+sqrtx`

Áp dụng bđt cosi ta có:`sqrtx+1/sqrtx>=2`

`=>D>=2+1=3`

Dấu "=" xảy ra khi `sqrtx=1/sqrtx<=>x=1`

\(B=\dfrac{2}{a^2+a+1}=\dfrac{2}{\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{8}{3}\)

Dấu = xảy ra khi a = -1/2