\(\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{2}.\sqrt{4-\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{2}}=\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

ngu thế

gọi số lớn là a 

số bé là b 

ta có  : trùng bình cộng của hai số = 435 nên a +b / 2 = 435 suy ra a+b = 432*2 = 870

mà số bé là số lớn nhất có hai chữ số nên số bé bằng 99 số lơn bằng 771

Tổng 2 số là : 435 x 2 = 870

Vì số bé là số lớn nhất có 2 chữ số nên số bé là : 99

Số lớn là : 870 - 99 = 771

\(\sqrt{21-12\sqrt{3}}\)

\(=\sqrt{12-2.2\sqrt{3}.3+9}\)

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

\(=2\sqrt{3}-3\)

\(\sqrt{21-12\sqrt{3}}=\sqrt{\left(2\sqrt{3}-3\right)^2}=\left|2\sqrt{3}-3\right|=2\sqrt{3}-3\)

7x-5=16

7x=16+5

7x=21

x=21:7

x=3

7x_16+5

7x_21

x_21/7
x_3

Đề bài ko rõ nên làm 2 trường hợp :

+) \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\) ( Điều kiện : x > 0 )

Trừ A đi 1 ta có :

\(\dfrac{\sqrt{x}-1}{\sqrt{x}}-1=\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}}=\dfrac{-1}{\sqrt{x}}< 0\left(\sqrt{x}>0\right)\)

\(\Leftrightarrow A-1< 0\Leftrightarrow A< 1\)

+) \(A=\dfrac{\sqrt{x-1}}{\sqrt{x}}\) ( Điều kiện : x > 1 )

Trừ A đi 1 , ta được : 

\(\dfrac{\sqrt{x-1}}{\sqrt{x}}-1=\dfrac{\sqrt{x-1}-\sqrt{x}}{\sqrt{x}}\) 

Nhận xét : Dễ thấy rằng , với x > 1 thì \(\sqrt{x-1}< \sqrt{x}\)

\(\Rightarrow\sqrt{x-1}-\sqrt{x}< 0\)

Mặt khác \(\sqrt{x}>0\left(x>1\right)\)

\(\Rightarrow A< 1\)

Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64

⇒ 2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32

⇒ 2A - A = (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32) - (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64) 

⇒ A = 1 - 1/64

⇒ A = 64/64 - 1/64

⇒ A = 63/64

0.984375

A = \(-\dfrac{3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)

A = \(-\dfrac{3}{17}+\dfrac{43}{51}\)

A = \(\dfrac{2}{3}\)

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

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

B = \(\dfrac{1}{5}\)

C = \(\left(-\dfrac{9}{25}\right)\) x \(\dfrac{53}{3}-\left(-\dfrac{3}{5}\right)2\) x \(\dfrac{2}{3}2\)

C = \(-\dfrac{9}{25}\) x \(\dfrac{53}{3}-\left(-\dfrac{3}{5}\right)^2\left(\dfrac{2}{3}\right)^2\)

C = \(-\dfrac{159}{25}-\left(-\dfrac{3}{5}\right)^2\left(\dfrac{2}{3}\right)^2\)

C = \(-\dfrac{159}{25}-\left(\dfrac{3}{5}\right)^2\left(\dfrac{2}{3}\right)^2\)

C = \(\dfrac{159}{25}-\dfrac{3^2}{5^2}\left(\dfrac{2}{3}\right)^2\)

C = \(-\dfrac{159}{25}-\dfrac{9}{25}\left(\dfrac{2}{3}\right)^2\)

C = \(-\dfrac{159}{25}-\dfrac{9}{25}\) x \(\dfrac{2^2}{3^2}\)

C = \(-\dfrac{159}{25}-\dfrac{9}{25}\) x \(\dfrac{4}{9}\)

C = \(-\dfrac{159}{25}-\dfrac{4}{25}\)

C = \(-\dfrac{163}{25}\)

\(\dfrac{x+140}{x}.260=71+65.4\) ( ĐK : x khác 0 )

\(\Leftrightarrow\dfrac{x+140}{x}.260=331\)

\(\Leftrightarrow\dfrac{x+140}{x}=\dfrac{331}{260}\)

\(\Rightarrow260\left(x+140\right)=331x\)

\(\Leftrightarrow x\approx512,7\)

$\genfrac{}{}{0ex}{}{x+140}{x}.260=71+65.4$ ( ĐK : x khác 0 )

$\iff \genfrac{}{}{0ex}{}{x+140}{x}.260=331$

$\iff \genfrac{}{}{0ex}{}{x+140}{x}=\genfrac{}{}{0ex}{}{331}{260}$

$\Rightarrow 260\left(x+140\right)=331x$

$\iff x\; sấp\; sỉ\; =512,7$

Ta có : \(\left\{{}\begin{matrix}\dfrac{1006}{1007}< 1\\\dfrac{1007}{1008}< 1\\\dfrac{1008}{1009}< 1\end{matrix}\right.\)

Cộng vế theo vế của 3 bất đẳng thức trên , ta có : 

\(\dfrac{1006}{1007}+\dfrac{1007}{1008}+\dfrac{1008}{1009}< 1+1+1=3\)

Ta có : \(\dfrac{1006}{1007}< 1,\dfrac{1007}{1008}< 1,\dfrac{1008}{1009}< 1\)

\(\Rightarrow\dfrac{1006}{1007}+\dfrac{1007}{1008}+\dfrac{1008}{1009}< 1+1+1=3\)

Do đó \(\dfrac{1006}{1007}+\dfrac{1007}{1008}+\dfrac{1008}{1009}< 3\)

\(\dfrac{6^9.2^{10}+12^{10}}{2^{19}.27^3+15.4^9.9^4}\)

\(=\dfrac{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}\)

\(=\dfrac{2^{19}.3^9+2^{20}.3^{10}}{2^{19}.3^9+5.2^{18}.3^9}\)

\(=\dfrac{2^{19}.3^9\left(1+2.3\right)}{2^{19}.3^9\left(1+5.2\right)}\)

\(=\dfrac{1+6}{1+10}=\dfrac{7}{11}\)

\(\dfrac{6^9.2^{10}+12^{10}}{2^{19}.27^3+15.4^9.9^4}\\ =\dfrac{2^9.3^9.2^{10}+\left(2^2.3\right)^{10}}{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9\left(3^2\right)^4}\\ =\dfrac{2^{19}.3^9+2^{20}.3^{10}}{2^{19}.3^9+5.2^{18}.3^9}\\ =\dfrac{2^{18}.3^9.\left(2+2^2.3\right)}{2^{18}.3^9.\left(2+5\right)}\\ =\dfrac{14}{7}\\ =2\)