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.

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

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

\(=-2\)

Đặt \(a=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\Rightarrow a^2=8+2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}=8+6=14\Rightarrow a=\sqrt{14}\)(Dễ thấy a > 0) 

vừa chị làm rồi em ko hiểu à?

Ta có: \(\dfrac{-3}{7}-\dfrac{4}{7}:x=-2\)

\(\Leftrightarrow\dfrac{4}{7}:x=\dfrac{-3}{7}+2=\dfrac{11}{7}\)

hay \(x=\dfrac{11}{7}:\dfrac{4}{7}=\dfrac{11}{7}\cdot\dfrac{7}{4}=\dfrac{11}{4}\)

Vậy: \(x=\dfrac{11}{4}\)

`-2/5 : x=1/2`

`=> x= -2/5 : 1/2`

`=> x= -2/5 xx 2`

`=>x= -4/5`

__

`7/6 : x = 7/4`

`=>x= 7/6 : 7/4`

`=>x=7/6 xx 4/7`

`=>x= 28/42`

`=>x=2/3`

tìm x hả

\(x=\dfrac{-2}{5}:\dfrac{1}{2}\)

\(x=\dfrac{-4}{5}\)

b. \(x=\dfrac{7}{6}:\dfrac{7}{4}\)

\(x=\dfrac{2}{3}\)

Lời giải:
$S=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+...+\frac{69}{7^{70}}$

$7S=\frac{1}{7}+\frac{2}{7^2}+\frac{3}{7^3}+...+\frac{69}{7^{69}}$

$6S=7S-S=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{69}}-\frac{69}{7^{70}}$

$42S=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{68}}-\frac{69}{7^{69}}$

$\Rightarrow 42S-6S=(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{68}}-\frac{69}{7^{69}})-(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{69}}-\frac{69}{7^{70}})$

$\Rightarrow 36S=1-\frac{69}{7^{69}}-\frac{1}{7^{69}}+\frac{69}{7^{70}}$

Hay $36S=1-\frac{69.7-7-69}{7^{70}}=1-\frac{407}{7^{70}}$

$\Rightarrow S=\frac{1}{36}(1-\frac{407}{7^{70}})$

1) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=\left(\sqrt{19}\right)^2-3^2=19-9=10\)

2) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{\dfrac{8+2\sqrt{7}}{2}}-\sqrt{\dfrac{8-2\sqrt{7}}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}\)

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

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

3) \(\sqrt{8+\sqrt{60}}+\sqrt{45}-\sqrt{12}=\sqrt{8+\sqrt{4.15}}+\sqrt{9.5}-\sqrt{4.3}\)

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

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

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

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

4) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}\right)^2-2.2.\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2+2.2.\sqrt{5}+2^2}\)

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

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

1) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=19-9=10\)

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

\(C=\dfrac{-5}{7}+\dfrac{-2}{7}+\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{-1}{5}=-1+1-\dfrac{1}{5}=\dfrac{-1}{5}\)

x : 3/4 = 1 1/4 : 2/7

x : 3/4 = 35/8

x        = 105/32

hok tốt

x = 1 1/4 : 2/7 : 3/7 = 5/4 x 7/2 x 7/3 = 245/24 = 10 5/24

x : 3/7 = 1 1/4 : 2/7

x : 3/7 = 35/8

x         = 15/8

hok tốt

\(\dfrac{4}{7}-\dfrac{4}{9}\)

\(=\dfrac{36}{63}-\dfrac{28}{63}\)

\(=\dfrac{8}{63}\)