\(\left(B\right)\dfrac{70}{3}\)

Gọi số bị trừ là a, số trừ là b, hiệu là c

Ta có: a+b+c=1700     (1)

c-b=243 (2)

Từ (1) và (2) ta có:

b+c+b+c=1700

=> b+b+243+b+b+243=1700

=> bx4=1700-243-243

=> bx4=1214

=> b=1214:4=303,5

Vậy a= 303,5+(303,5+243)=850

a) \(\sqrt{2}.x-\sqrt{50}=0\Leftrightarrow\sqrt{2}\left(x-\sqrt{25}\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)

b) \(\sqrt{3}.x+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3}\left(x+1\right)=\sqrt{3}\left(\sqrt{4}+\sqrt{9}\right)\Leftrightarrow x+1=2+3\Leftrightarrow x=4\)

c) \(\sqrt{3}.x^2-\sqrt{12}=0\Leftrightarrow\sqrt{3}\left(x^2-\sqrt{4}\right)=0\Leftrightarrow x^2-2=0\Leftrightarrow x=\pm\sqrt{2}\)

d) \(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\Leftrightarrow\dfrac{x^2}{\sqrt{5}}=\sqrt{20}\Leftrightarrow x^2=10\Leftrightarrow x=\pm\sqrt{10}\)

a) \(\sqrt{\dfrac{289}{225}}=\dfrac{\sqrt{289}}{\sqrt{225}}=\dfrac{17}{15}\)

b) \(\sqrt{2\dfrac{14}{25}}=\dfrac{\sqrt{64}}{\sqrt{25}}=\dfrac{8}{5}\)

c) \(\sqrt{\dfrac{0,25}{9}}=\dfrac{\sqrt{0,25}}{\sqrt{9}}=\dfrac{0,5}{3}=\dfrac{1}{6}\)

d) \(\sqrt{\dfrac{8,1}{1,6}}=\dfrac{\sqrt{81}}{\sqrt{16}}=\dfrac{9}{4}\)

a) \(x+8+x-22\)

Tại X= -98

\(-98+8-\left(-98\right)-22=-98+8-98-22=8-22=-14\)

b) \(-x-a+12+a\)

\(\Leftrightarrow-\left(-98\right)-61+12+61=12+98=110\)

c) \(a-m+7-8+m\)

\(\Leftrightarrow61-\left(-25\right)+7-8+\left(-25\right)=61+25+7-8-25=61-1=60\)d) \(m-24-x+24+x\)

\(\Leftrightarrow-25-24-\left(-98\right)+24+\left(-98\right)=-25+98-98=-25\)

a) \(\sqrt{45.80}=\sqrt{9.400}=\sqrt{9}.\sqrt{400}=3.20=60\)

b)\(\sqrt{75.48}=\sqrt{25.3.3.16}=5.3.4=60\)

c)\(\sqrt{90.6,4}=\sqrt{9.64}=3.8=24\)

d)\(\sqrt{2,5.14,4}=\sqrt{\dfrac{25}{10}.\dfrac{144}{10}}=\sqrt{\dfrac{25.144}{100}=\dfrac{5.12}{10}=\dfrac{60}{10}=6}\)

a) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}=\dfrac{1}{\sqrt{2}}\)

b) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

=\(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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