Lời giải:

a)

\(\frac{4}{\sqrt{10}}(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}})=\frac{4}{\sqrt{20}}(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}})\)

\(=\frac{4}{2\sqrt{5}}(\sqrt{5+1+2\sqrt{5}}+\sqrt{5+1-2\sqrt{5}})=\frac{2}{\sqrt{5}}[\sqrt{(\sqrt{5}+1)^2}+\sqrt{(\sqrt{5}-1)^2}]\)

\(=\frac{2}{\sqrt{5}}(\sqrt{5}+1+\sqrt{5}-1)=\frac{2}{\sqrt{5}}.2\sqrt{5}=4\)

b)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})\)

\(=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)

c)

\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+8\sqrt{3}+18}=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(3+1+2\sqrt{3})+2}\)

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

\(=\sqrt{(2\sqrt{3}+2)^2+(\sqrt{2})^2+2.(2\sqrt{3}+2).\sqrt{2}}\)

\(=\sqrt{(2\sqrt{3}+2+\sqrt{2})^2}=2\sqrt{3}+2+\sqrt{2}\)

9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)

10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)

12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)

\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)

=9-3=6

13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)

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

bạn tách từng câu ra mik suy nghĩ từng câu

bạn trả lời từng câu cũng được mà :) làm được câu nào thì giúp mình nhé. Tks!

Câu 3: đề là \(\sqrt{x+5}-\sqrt{x-2}\) hay \(\sqrt{x+5}-\sqrt{x+2}\)?

Câu 4:

ĐKXĐ: \(x\le9\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x-4}=a\\\sqrt{9-x}=b\end{matrix}\right.\) ta có hệ:

\(\left\{{}\begin{matrix}a-b=-1\\a^3+b^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=a+1\\a^3+b^2=5\end{matrix}\right.\)

\(\Rightarrow a^3+\left(a+1\right)^2=5\)

\(\Leftrightarrow a^3+a^2+2a-4=0\) \(\Rightarrow a=1\)

\(\Rightarrow\sqrt[3]{x-4}=1\Rightarrow x-4=1\Rightarrow x=5\)

5.

ĐKXĐ: \(x\ge-\frac{17}{16}\)

\(\Leftrightarrow8x^2-15x-23-\left(x+1\right)\sqrt{16x+17}=0\)

\(\Leftrightarrow\left(x+1\right)\left(8x-23\right)-\left(x+1\right)\sqrt{16x+17}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\8x-23=\sqrt{16x+17}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow16x+17-2\sqrt{16x+17}-63=0\)

Đặt \(\sqrt{16x+17}=t\ge0\)

\(\Rightarrow t^2-2t-63=0\Rightarrow\left[{}\begin{matrix}t=9\\t=-7\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{16x+17}=9\Leftrightarrow x=\frac{32}{3}\)

mình cần phần 3 4 5 nữa thui ạ

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) Ta có: \(\sqrt{96}\cdot\sqrt{125}\)

\(=\sqrt{16}\cdot\sqrt{6}\cdot\sqrt{25}\cdot\sqrt{5}\)

\(=20\cdot\sqrt{30}\)

b) Ta có: \(\sqrt{a^4\cdot6^5}\)

\(=a^2\cdot36\cdot\sqrt{6}\)

c) Ta có: \(\sqrt{a^6\cdot b^{11}}\)

\(=\sqrt{a^6}\cdot\sqrt{b^{11}}\)

\(=\left|a^3\right|\cdot\left|b^5\right|\cdot\sqrt{b}\)

\(=a^3b^5\cdot\sqrt{b}\)

d) Ta có: \(\sqrt{a^3\left(1-a\right)^4}\)

\(=\sqrt{a^3}\cdot\sqrt{\left(1-a\right)^4}\)

\(=a\sqrt{a}\cdot\left(1-a\right)^2\)

a, \(=7\sqrt{2}-6\sqrt{2}+\frac{1}{2}.2\sqrt{2}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

b, \(=4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}=4\sqrt{a}-5\sqrt{10a}\)

c, \(=6+\sqrt{15}-\sqrt{60}=6+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)

Rút gọn

a) Ta có: \(\sqrt{98}-\sqrt{72}+\frac{1}{2}\sqrt{8}\)

\(=\sqrt{2}\left(\sqrt{49}-\sqrt{36}+\frac{1}{2}\sqrt{4}\right)\)

\(=\sqrt{2}\left(7-6+\frac{1}{2}\cdot2\right)\)

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

b) Ta có: \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\)

\(=\sqrt{a}\left(\sqrt{16}+2\sqrt{40}-3\sqrt{90}\right)\)

\(=\sqrt{a}\left(4+4\sqrt{10}-9\sqrt{10}\right)\)

\(=\sqrt{a}\left(4-5\sqrt{10}\right)\)

\(=4\sqrt{a}-5\sqrt{10a}\)

c) Ta có: \(\left(2\sqrt{3}+\sqrt{5}\right)\cdot\sqrt{3}-\sqrt{60}\)

\(=6+\sqrt{15}-\sqrt{60}\)

\(=6-\sqrt{15}\)

\(\sqrt[3]{\left(1-\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\sqrt[3]{\left(1-\sqrt{3}\right)\left(\sqrt{3}-1\right)^2}\)=\(\sqrt[3]{\left(1-\sqrt{3}\right)^3}\)=1-\(\sqrt{3}\)

\(\sqrt[3]{\left(1-\sqrt{5}\right)\left(6-2\sqrt{5}\right)}=\sqrt[3]{\left(1-\sqrt{5}\right)\left(\sqrt{5}-1\right)^2}\)=\(\sqrt[3]{\left(1-\sqrt{5}\right)^3}\)=1-\(\sqrt{5}\)

Ta thấy \(\sqrt{5}>\sqrt{3}\)nên 1-\(\sqrt{3}\)>\(1-\sqrt{5}\)

Vậy \(\sqrt[3]{\left(1-\sqrt{3}\right)\left(4-2\sqrt{3}\right)}\)>\(\sqrt[3]{\left(1-\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)