chịu

1. A

2. B

3. A

4. B

1.    444 x 5 = 222 x 2 x 5 = 222 x 10 = 2220
2.    282 x 5 = 280 + 2 x 5 = 280 x 10 = 2800
3.    4 x 8 x 7 x 25 = (8 x 7) x (25 x4) = 56 x 100 = 5600
4.    25 x 8 x 9 = (25 x 4) x (4 x 9) = 100 x 36 = 3600
a.Đúng         b.Sai

1. sai

2. Sai

3. đúng

4.sai

P>1/3

=>P-1/3>0

=>\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{1}{3}>0\)

=>\(\dfrac{3\sqrt{x}-3-\sqrt{x}-2}{3\left(\sqrt{x}+2\right)}>0\)

=>2 căn x-5>0

=>x>25/4

\(25.5.4.125.20.8\)

\(=\left(25.4\right).\left(5.20\right).\left(125.8\right)\)

\(=100.100.1000\)

\(=10000000\)

25*(5+5)*4*20*8

=250*20+4+8

= 5000*12

=60000

đúng mk tích cho .

nó khó nhìn thiệt ha

rốt cục là hỏi j, hỏi hay trả lời đó

\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)

Đề có sai không bạn?

1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)

\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)

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

\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)

2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)

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

\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)

3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)

\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)

\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)

\(=\dfrac{17}{7}\)

4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)

\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)

\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)

\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)

đK: \(x\ge0;x\ne25;x\ne9\)

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right]:\left[\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right]\)

\(=\left[\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right]:\dfrac{25-x-\left(x-9\right)+\left(x-25\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{9-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{-\sqrt{x}-3}{\sqrt{x}+5}\)

\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{\sqrt{x}+5}{-\left(\sqrt{x}+3\right)}=\dfrac{5}{\sqrt{x}+3}\)

Chỗ căn bị thiếu là √x nha.

Ta có: \(\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}:\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{5}{\sqrt{x}+3}\)