a) Ta có: \(2=\sqrt{4}\)

Vì \(4>3\Rightarrow\sqrt{4}>\sqrt{3}\Rightarrow2>\sqrt{3}\Rightarrow1>\sqrt{3}-1\)

b) \(\left\{{}\begin{matrix}2\sqrt{31}=\sqrt{4.31}=\sqrt{124}\\10=\sqrt{100}\end{matrix}\right.\)

Vì \(124>100\Rightarrow\sqrt{124}>\sqrt{100}\Rightarrow2\sqrt{31}>10\)

c) Vì \(15< 16\Rightarrow\sqrt{15}< \sqrt{16}\Rightarrow\sqrt{15}-1< \sqrt{16}-1\)

\(\Rightarrow\sqrt{15}-1< 4-1\Rightarrow\sqrt{15}-1< 3\)

Lại có: \(10>9\Rightarrow\sqrt{10}>\sqrt{9}\Rightarrow\sqrt{10}>3\)

\(\Rightarrow\sqrt{10}>\sqrt{15}-1\)

a) \(\sqrt{0,16}+\sqrt{0,04}-\sqrt{0,25}\)

= 0,4 + 0,2 - 0,5 

= 0,1

b) \(\sqrt{85^2-84^2}-\sqrt{26^2-24^2}\)

= \(\sqrt{\left(85-84\right)\left(85+84\right)}\) - \(\sqrt{\left(26-24\right)\left(26+24\right)}\)

= \(\sqrt{169}\) - \(\sqrt{2.50}\)

= 13 - 10

= 3 

 Chúc bạn học tốt

a) Ta có: \(\sqrt{0.16}+\sqrt{0.04}-\sqrt{0.25}\)

\(=0,4+0,2-0,5\)

=0,1

b) Ta có: \(\sqrt{85^2-84^2}-\sqrt{26^2-24^2}\)

=13-10

=3

Bài 1: 

Để M có nghĩa thì \(\left\{{}\begin{matrix}x+4\ge0\\2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\le2\end{matrix}\right.\Leftrightarrow-4\le x\le2\)

Số giá trị nguyên thỏa mãn điều kiện là:

\(\left(2+4\right)+1=7\)

các bạn giúp mk để mk ăn tết cho zui

luong thuy anh giúp mk vs

Đăng ra nhiều câu nhé bạn,ví dụ mỗi lần bạn đăng 2 bài để mọi người biết mà giúp chứ đăng thế này ai cũng ngán dù họ biết làm=((

a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)

\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)

\(=33\sqrt{3}\cdot\sqrt{3}\)

=99

b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)

\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)

\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)

c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=36-36\sqrt{2}+18\sqrt{3}\)

d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)

\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)

a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)

   \(=28.3+9.3-4.3=99\)

b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)

  \(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)

d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

        \(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

        \(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)

        \(=19\sqrt{3\sqrt{2}}\)

a: \(B=\dfrac{1}{\sqrt{x}+1}\)

\(B-1=\dfrac{\sqrt{x}+1-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}>=0\)

=>B>=1

b: \(P=\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

\(P\cdot\sqrt{x}+2x-\sqrt{x}=3x-2\sqrt{x-4}+3\)

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

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

=>x-4=1

=>x=5

a: Ta có: \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{6\sqrt{a}-6+10-2\sqrt{a}}{\left(\sqrt{a}-1\right)^2\cdot\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{4\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\cdot\dfrac{1}{4\sqrt{a}}\)

\(=\dfrac{1}{\sqrt{a}}\)

a) \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{6\left(\sqrt{a}-1\right)+10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{4\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{1}{\sqrt{a}}\)

b) \(C=B.\left(a-\sqrt{a}+1\right)=\dfrac{a-\sqrt{a}+1}{\sqrt{a}}=\sqrt{a}-1+\dfrac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\dfrac{1}{\sqrt{a}}}-1=1\)(bất đẳng thức Cauchy cho 2 số dương)