= -2.094740617

b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)

\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)

\(=4\left(7+3\sqrt{5}\right)\)

\(=28+12\sqrt{5}\)

Lời giải:

a. 

$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$

$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$

$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$

$=2\sqrt{5}-5\sqrt{10}$

$\Rightarrow A=\sqrt{10}-5\sqrt{5}$

b.

$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$

$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$

$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$

$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$

$\Rightarrow B=28+12\sqrt{5}$

c.

$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$

$=(7-5)(6-\sqrt{35})$

$=2(6-\sqrt{35})=12-2\sqrt{35}$

a) \(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}+2\frac{14}{90}\)

= \(\left(74\frac{19}{35}+15\frac{16}{35}\right).\frac{7}{90}+2\frac{14}{90}\)

=  90 . 7/90 + 194/90

= 630/90 + 194/90

= 824/90 = 412/45

b) (-2/5 + 3/7) - (4/9 + 12/20 - 13/35) + 7/35

= -2/5 + 3/7 - 4/9 -  3/5 + 13/35 + 7/35

= (-2/5 - 3/5) + 3/7 - 4/9 + (13/35 + 7/35)

= -1 + 3/7 - 4/9 + 4/7

= -1 + (3/7 + 4/7) - 4/9

= -1 + 1 - 4/9 = -4/9

Các câu sai: a, c, d, f

Các câu đúng: b, e

Các câu đúng: b,e
Các câu sai: a, c, d; f.
a) \(\left(-5\right)^2.\left(-5\right)^3=\left(-5\right)^5\);
c) \(\left(0,2\right)^{10}:\left(0,2\right)^5=\left(0,2\right)^{10-5}=0,2^5\);
d) \(\left[\left(-\dfrac{1}{7}\right)^2\right]^4=\left(-\dfrac{1}{7}\right)^{2.4}=\left(-\dfrac{1}{7}\right)^8\)
f \(\dfrac{8^{10}}{4^8}=\dfrac{\left(2^3\right)^5}{\left(2^2\right)^8}=\dfrac{2^{15}}{2^{16}}=\dfrac{1}{2}\)

a) \(\left(-5\right)^2\).\(\left(-5\right)^3\) = -3125

5 x [(85 - 35 : 7) : 8 + 90] - 50

= 5 x [(85 - 5) : 8 + 90] - 50

= 5 x [80 : 8 + 90] - 50

= 5 x [10 + 90] - 50

= 5 x 100 - 50

= 500 - 50

= 450

\(5\times\left[\left(85-35:7\right):8+90\right]-50\\ =5\times\left[\left(85-5\right):8+90\right]-50\\ =5\times\left[80:8+90\right]-50\\ =5\times\left[10+90\right]-50\\ =5\times100-50\\ =500-50=450\)

5[(85 - 35 : 7) : 8 + 90] - 50

= 5[(85 - 5) : 8 + 90] -  50

= 5.[ 80 : 8 + 90] - 50

= 5.[ 10 + 90 ] - 50

= 5. 100 - 50

= 500 - 50

= 450

5[(85 - 35 : 7) : 8 + 90] - 50

= 5[(85 - 5) : 8 + 90] -  50

= 5.[ 80 : 8 + 90] - 50

= 5.[ 10 + 90 ] - 50

= 5. 100 - 50

= 500 - 50

= 450

Làm 1 câu các câu còn lại tương tự!

a, \(\left(x+2\right)\left(x-4\right)\left(x+6\right)\left(x-12\right)+36x^2\)

\(=\left[\left(x+2\right)\left(x-12\right)\right]\left[\left(x-4\right)\left(x+6\right)\right]+36x^2\)

\(=\left(x^2-12x+2x-24\right)\left(x^2+6x-4x-24\right)+36x^2\)

\(=\left(x^2-10x-24\right)\left(x^2+2x-24\right)+36x^2\)(1)

Đặt \(a=x^2-10x-24\Rightarrow a+12x=x^2+2x-24\)

\(\Rightarrow\left(1\right)=a\left(a+12x\right)+36x^2=a^2+12ax+36x^2\)

\(=a^2+6ax+6ax+36x^2=a\left(a+6x\right)+6x\left(a+6x\right)\)

\(=\left(a+6x\right)^2\)(*)

Vì \(a=x^2-10x-24\) nên

(*)\(=\left(x^2-10x-24+6x\right)^2=\left(x^2-4x-24\right)^2\)

Vậy...........

làm phần b đi