1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)

3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)

\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)

\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)

\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

b.

ĐKXĐ: \(x\ge2\)

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

\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)

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

\(\Rightarrow x=2\)

Bạn xem lại câu 5 xem có sai đề không chứ mình tính mãi không ra

Đề câu 5 k sai nhé. Dùng Mode 5 3 vẫn ra.

\(\sqrt{29-12\sqrt{5}}=\sqrt{20-2.2\sqrt{5}.3+9}=\sqrt{\left(\sqrt{20}-3\right)^2}=\sqrt{20}-3=2\sqrt{5}-3\)

Đúng cho mình đi dẫ

\(\sqrt{3\cdot27}-\sqrt{\dfrac{144}{36}}\)=\(\sqrt{81}-\sqrt{4}\)=9-2=7

\(\dfrac{2\cdot3+3\cdot6}{4}\)=6

\(\sqrt{7}-\sqrt{7-2\cdot\sqrt{7}+1}\)=\(\sqrt{7}-\left(\sqrt{7}-1\right)\)=1

\(\dfrac{\sqrt{3-2\cdot\sqrt{3}+1}}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}\)=\(\dfrac{\sqrt{3}-1}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}\)=\(\dfrac{1}{\sqrt{2}}\)

\(\dfrac{\sqrt{5}\cdot\left(\sqrt{5}+3\right)}{\sqrt{5}}\)+\(\dfrac{\sqrt{3}\cdot\left(1+\sqrt{3}\right)}{\sqrt{3}+1}\)-(\(\sqrt{5}+3\))

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

\(\sqrt{3}\cdot\sqrt{9}+5\cdot\sqrt{4}\cdot3-2\sqrt{3}\)

=\(\sqrt{3}\cdot\left(3+10-2\right)\)

=\(11\sqrt{3}\)

ta có ;\(36-16\sqrt{5}=16-2\cdot4\cdot2\sqrt{5}+20=\left(2\sqrt{5}-4\right)^2\)

         \(12+2\sqrt{35}=7+2\sqrt{7}\cdot\sqrt{5}+5=\left(\sqrt{7}+\sqrt{5}\right)^2\)

        \(81-36\sqrt{5}=36-2\cdot6\cdot3\sqrt{5}+45=\left(3\sqrt{5}-6\right)^2\)

        \(11+4\sqrt{7}=\sqrt{7}+2\cdot2\cdot\sqrt{7}+4=\left(\sqrt{7}+2\right)^2\)

TỪ ĐÓ TÍNH RA

Bạn bình phươg A lên nha

s2 Lắc Lư s2 lười làm