b)

\(\sqrt{\left(4-2\sqrt{3}\right)\left(13+4\sqrt{3}\right)}\)

\(=\sqrt{\left(3-2\sqrt{3}+1\right)\left(12+4\sqrt{3}+1\right)}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2\cdot\left(\sqrt{12}+1\right)^2}\)

\(=\left|\sqrt{3}-1\right|\cdot\left|\sqrt{12}+1\right|\)

\(=\left(\sqrt{3}-1\right)\left(2\sqrt{3}+1\right)\) ( vì \(\sqrt{3}-1>0;2\sqrt{3}+1>0\) )

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

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

a)

\(\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{6-2\sqrt{5}}}\)

\(=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{5-2\sqrt{5}+1}}\)

\(=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\left|\sqrt{5}-1\right|}\)

\(=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{5}-1}\) (vì \(\sqrt{5}-1>0\) )

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

\(=\left(\sqrt{10}-\sqrt{2}\right)\cdot\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\left(\sqrt{10}-\sqrt{2}\right)\cdot\dfrac{\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}\)

\(=\left(\sqrt{10}-\sqrt{2}\right)\cdot\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\)

\(=\left(\sqrt{10}-\sqrt{2}\right)\cdot\dfrac{\left|\sqrt{5}+1\right|}{\sqrt{2}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}{\sqrt{2}}\) (vì \(\sqrt{5}+1>0\) )

`=5-1`

`=4`

Khi viết thêm số 3 vào bên phải 215 ta được 2153

Số đã tăng số đơn vị là

`2153-215=1938` (đơn vị)

điều kiện thiếu kìa bạn

mik quên mất=)

\(\dfrac{\sqrt{5x+7}}{\sqrt{x+3}}=4\left(x\ge-\dfrac{7}{5};x\ne-3\right)\)

\(< =>\sqrt{5x+7}=4\sqrt{x+3}\)

\(< =>5x+7=16\left(x+3\right)\)

`<=>5x+7=16x+48`

`<=>5x-16x=48-7`

`<=>-11x=41`

`<=>x=-41/11(ktm)`

Vậy pt vô nghiệm