5^2x-3 - 2.5²=5²

\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)

\(=\sqrt{12}+1=2\sqrt{3}+1\)

\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\sqrt{3}-1\)

\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)

\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)

2) biến đổi khúc sau như câu 1:

\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)

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

4) Ta có: \(\sqrt{30-2\sqrt{16+6\sqrt{11+4\sqrt{4-2\sqrt{3}}}}}\)

\(=\sqrt{30-2\sqrt{16+6\sqrt{11+4\left(\sqrt{3}-1\right)}}}\)

\(=\sqrt{30-2\sqrt{16+6\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{30-2\sqrt{16+6\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{30-2\sqrt{28+6\sqrt{3}}}\)

\(=\sqrt{30-2\left(3\sqrt{3}+1\right)}\)

\(=\sqrt{28-6\sqrt{3}}=3\sqrt{3}-1\)

5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)

\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{75}-5\sqrt{2}}\)

\(=\dfrac{5\left(\sqrt{3}-\sqrt{2}\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=1\)

5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)

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

=1

Lần sau bạn chú ý viết đầy đủ đề.

1.

\(\sqrt{9+4\sqrt{5}-\sqrt{9-4\sqrt{5}}}=\sqrt{9+4\sqrt{5}-\sqrt{5-2\sqrt{4.5}+4}}\)

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

\(=\sqrt{9+4\sqrt{5}-\sqrt{5}+2}=\sqrt{11+3\sqrt{5}}\)

2.

\(\sqrt{8-2\sqrt{7}-\sqrt{8+2\sqrt{7}}}=\sqrt{8-2\sqrt{7}-\sqrt{7+2\sqrt{7}+1}}\)

\(=\sqrt{8-2\sqrt{7}-\sqrt{(\sqrt{7}+1)^2}}\)

\(=\sqrt{8-2\sqrt{7}-\sqrt{7}-1}=\sqrt{7-3\sqrt{7}}\)

Phạm Mạnh Kiên: sửa lại theo ý bạn thì làm như sau:

1.

\(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{5+2\sqrt{5}.\sqrt{4}+4}-\sqrt{5-2\sqrt{5}.\sqrt{4}+4}\)

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

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

2.

\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)

\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}=|\sqrt{7}-1|-|\sqrt{7}+1|\)

\(=-2\)

`5/4=15/12`

`3/2=18/12`

`5/2=30/12`

`7/3=28/12`

`-> 15/12 < 18/12 < 28/12 < 30/12`

`->` Thứ tự sắp xếp theo thứ tự từ bé đến lớn là: `5/4, 3/2, 7/3, 5/2`

\(\dfrac{3}{2}\),\(\dfrac{5}{2}\), \(\dfrac{7}{3}\) , \(\dfrac{5}{4}\)

\(\Rightarrow\) Sắp xếp theo thứ tự từ bé đến lớn:\(\dfrac{5}{4},\dfrac{3}{2},\dfrac{7}{3},\dfrac{5}{2}\)

Lời giải:

1.

$3^{x+2}+4.3^{x+1}=7.3^6$

$3^{x+1}.3+4.3^{x+1}=7.3^6$

$3^{x+1}(3+4)=7.3^6$

$3^{x+1}.7=7.3^6$

$\Rightarrow 3^{x+1}=3^6$

$\Rightarrow x+1=6$

$\Rightarrow x=5$

2.

$5^{x+4}-3.5^{x+3}=2.5^{11}$

$5^{x+3}.5-3.5^{x+3}=2.5^{11}$

$5^{x+3}(5-3)=2.5^{11}$

$2.5^{x+3}=2.5^{11}$

$\Rightarrow 5^{x+3}=5^{11}$

$\Rightarrow x+3=11$

$\Rightarrow x=8$

3.

$4^{x+3}-3.4^{x+1}=13.4^{11}$

$4^{x+1}.4^2-3.4^{x+1}=13.4^{11}$

$4^{x+1}.16-3.4^{x+1}=13.4^{11}$

$13.4^{x+1}=13.4^{11}$

$\Rightarrow 4^{x+1}=4^{11}$

$\Rightarrow x+1=11$
$\Rightarrow x=10$

k minh minh giai

* (-6 - 2) . (-6 + 2)

=    -8   .     -4

=           32

*(7 . 3 - 3) : (-6)

=      18    : (-6)

=            -3

k cho mik nhe

(-6-2).(-6+2)

=-8.(-4)

=32

(7.3-3):(-6)

=(21-3):(-6)

=18:(-6)

=-3

(-5+9).(-4)

=4.(-4)

=-16

72:(-6.2+4)

=72.(-12+4)

=72.8

-3.7-4.(-5)+1

=-21+20+1

=-1+1

=0

18-10:(+2)-7

=18-5-7

=13-7

=6

15:(-5).(-3)-8

=-3.(-3)-8

=9-8

=1

(6.8-10:5)+3.(-7)

=(48-2)+3.(-7)

=46+(-21)

=25