Bài 1: (cái này là khai căn nên làm tắt xíu nha)

\(a.\sqrt{3}-\frac{1}{3}\sqrt{27}+2\sqrt{507}\\ =\sqrt{3}-\frac{1}{3}\sqrt{9\cdot3}+2\sqrt{169\cdot3}\\ =\sqrt{3}-\frac{1}{3}\cdot3\sqrt{3}+2\cdot13\sqrt{3}\\ =\sqrt{3}-\sqrt{3}+26\sqrt{3}=26\sqrt{3}\)

\(b.\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =\left(\sqrt{4\cdot7}-\sqrt{4\cdot3}-\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =\left(\sqrt{7}\right)^2-2\sqrt{21}+2\sqrt{21}=7\)

\(c.2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\\ =2\sqrt{40\sqrt{4\cdot3}}-2\sqrt{\sqrt{25\cdot3}}-3\sqrt{5\sqrt{16\cdot3}}\\ =2\sqrt{16\cdot5\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\cdot4\sqrt{3}}\\ =8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)

Bài 2:

a. ĐKXĐ: \(x\ge0\)

\(5\sqrt{12x}-4\sqrt{3x}+2\sqrt{48x}=14\\ \Leftrightarrow5\sqrt{4\cdot3x}-4\sqrt{3x}+2\sqrt{16\cdot3x}=14\\ \Leftrightarrow10\sqrt{3x}-4\sqrt{3x}+8\sqrt{3x}=14\\ \Leftrightarrow14\sqrt{3x}=14\\ \Leftrightarrow\sqrt{3x}=1\\ \Leftrightarrow3x=1\Leftrightarrow x=\frac{1}{3}\left(tm\right)\)

b. ĐKXĐ: \(x\ge5\)

\(\sqrt{4x-20}+\sqrt{x-5}-\frac{1}{3}\sqrt{9x-45}=4\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

a)\(\sqrt{45}:\sqrt{80}\)

= \(\sqrt{45:80}\)

=\(\sqrt{9:16}\)

= \(\sqrt{9}:\sqrt{16}\)

= \(\frac{3}{4}\)

b)\(\sqrt{\frac{3}{15}}:\sqrt{\frac{36}{45}}\)

= \(\sqrt{\frac{1}{5}}:\sqrt{\frac{4}{5}}\)

= \(\sqrt{\frac{1}{5}.\frac{5}{4}}\)

= \(\sqrt{\frac{1}{4}}\)

=\(\frac{1}{2}\)

c)\(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}\)

= \(\left(7\sqrt{4^2.3}+3\sqrt{3^2.3}-2\sqrt{2^2.3}\right):\sqrt{3}\)

=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}\)

=28+9-4

=33

d) \(\sqrt{\frac{125}{245}}\)

= \(\sqrt{\frac{25}{49}}\)

= \(\frac{\sqrt{25}}{\sqrt{49}}\)

= \(\frac{5}{7}\)

a,

\(\frac{5\sqrt{60}\cdot3\sqrt{15}}{15\sqrt{50}\cdot2\sqrt{18}}\\ =\frac{5\cdot\sqrt{2^2\cdot15}\cdot3\sqrt{15}}{15\sqrt{2\cdot5^2}\cdot2\sqrt{2\cdot3^2}}\\ =\frac{5\cdot2\cdot3\cdot15}{15\cdot5\cdot2\cdot3\cdot3}=\frac{1}{3}\)

b,

\(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}\\ =\frac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}\\ =\frac{6}{3^2-2}=\frac{6}{7}\)

c,

\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\\ =\frac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\\ =\frac{5-2\sqrt{15}+3+5+2\sqrt{15}+3}{5-3}\\ =\frac{16}{2}=8\)

d, Với \(x,y\ge0;x\ne y\), ta được:

\(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\\ =\frac{\sqrt{x\cdot x^2}-\sqrt{y\cdot y^2}}{\sqrt{x}-\sqrt{y}}\\ =\frac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\\ =\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}^3\right)}{\sqrt{x}-\sqrt{y}}\\ =\frac{\left(\sqrt{x}-\sqrt{y}\right)\left[\left(\sqrt{x}\right)^2+\sqrt{x\cdot y}+\left(\sqrt{y}\right)^2\right]}{\sqrt{x}-\sqrt{y}}\\ =x+y+\sqrt{xy}\)

Chúc bạn học tốt nha.

câu a đoạn \(\frac{5.2.3.15}{15.5.2.3.3}\) bạn làm cách nào vậy

Em không chắc câu c, d đâu nha

a) ĐK: \(1\ne\sqrt{x-1}\text{và }x\ge1\Leftrightarrow x\ne2;x\ge1\)

b) \(ĐK:\left\{{}\begin{matrix}x\ge\frac{5}{2}\\\sqrt{x+3}\ne\sqrt{2x-5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{5}{2}\\x\ne8\end{matrix}\right.\)

c) ĐK: \(\left\{{}\begin{matrix}x^2\ge36\left(1\right)\\x\ge-6\left(2\right)\end{matrix}\right..\text{Giải (1) }\Leftrightarrow\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)

Kết hợp (2) suy ra \(x=-6\text{ hoặc }x\ge6\)

d) ĐK: \(\frac{3x-2}{x+4}\ge0\). tức là 3x - 2 và x + 4 đồng dấu.

TH1: \(\left\{{}\begin{matrix}3x-2\ge0\\x+4>0\left(\text{do x phải khác -4}\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{2}{3}\\x>-4\end{matrix}\right.\Leftrightarrow x\ge\frac{2}{3}\)

TH2: \(\left\{{}\begin{matrix}3x-2< 0\\x+4< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \frac{2}{3}\\x< -4\end{matrix}\right.\Leftrightarrow x< -4\)

Do vậy ĐKXĐ: x >= 2/3 hoặc x<-4

e) ĐK: \(x\in\mathbb{R}\)

~~woa ~~ đây là cách tìm điều kiện ( gần giống trên youtube :))

\(\sqrt[3]{\left(a+1\right)^3}\)+\(\sqrt[3]{\left(a-1\right)^3}\)

=(a+1)+(a-1)

=a+1+a-1

=a+a

=2a

2/ 7\(\sqrt[3]{8}\) và 8\(\sqrt[3]{7}\)

\(\left(7\sqrt[3]{8}\right)^3\) và

343.8 và 512.7

343.8 và (343+169).7

343.8 và 343.7+169.7

So sánh 343.8 và 343.7

343.8 lớn hơn 343.7 là 343

169.7>343

\(\Rightarrow\) 8\(\sqrt[3]{7}\)>7 ​\(\sqrt[3]{8}\)

​3/ \(\sqrt[3]{3x-1}\) =2

3x-1=\(2^3\)=8

3x=8+1=9

x=9:3=3

Lời giải :

a) \(\sqrt{\left(0,1-\sqrt{0,1}\right)^2}\)

\(=0,1-\sqrt{0,1}\)

b) \(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

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

d) \(\sqrt{9-4\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)

e) \(\sqrt{16-6\sqrt{7}}=\sqrt{9-2\cdot3\cdot\sqrt{7}+7}=\sqrt{\left(3-\sqrt{7}\right)^2}=3-\sqrt{7}\)