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Câu 2:
a: \(\Leftrightarrow12x-60=7x-5\)
=>5x=55
=>x=11
b: \(\Leftrightarrow\left(2x-3\right)^{2010}\left[\left(2x-3\right)^2-1\right]=0\)
=>(2x-3)(2x-2)(2x-4)=0
hay \(x\in\left\{\dfrac{3}{2};1;2\right\}\)
\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)
\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)
=0
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
Ta có:B=1\(\dfrac{6}{41}\)( \(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\) )
B=\(\dfrac{47}{41}\) [\(\dfrac{12\left(1+\dfrac{1}{19}-\dfrac{1}{37}-\dfrac{1}{53}\right)}{3\left(1+\dfrac{1}{3}-\dfrac{1}{37}-\dfrac{1}{53}\right)}:\dfrac{4\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}\) B = \(\dfrac{47}{41}\) [ \(\dfrac{12}{3}:\dfrac{4}{5}\)]
B = \(\dfrac{47}{41}\)[ 4 . \(\dfrac{5}{4}\)]
B = \(\dfrac{47}{41}.5\)
B = \(\dfrac{235}{41}\)
Chúc bn hc tốt!!!
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
a. 7/9 - 16/9 = -9/9 = -1
b. 2/-15 + 7/10 = 17/30
c. (4 2/3 - 4 3/4) : -5/12 - 4/5
= (14/3 - 19/4) : (-5/12) - 4/5
= -1/12 : (-5/12) - 4/5
= 1/5 - 4/5
= -3/5
Bài 1:
a: \(\Leftrightarrow\dfrac{2}{3}\cdot\dfrac{6+9-4}{12}< =\dfrac{x}{18}< =\dfrac{7}{13}\cdot\dfrac{3-1}{6}\)
\(\Leftrightarrow\dfrac{22}{36}< =\dfrac{x}{18}< =\dfrac{14}{78}=\dfrac{7}{39}\)
\(\Leftrightarrow\dfrac{11}{9}< =\dfrac{x}{9}< =\dfrac{7}{13}\)
=>143<=x<=63
hay \(x\in\varnothing\)
b: \(\Leftrightarrow\dfrac{31\cdot9-26\cdot4}{180}\cdot\dfrac{-36}{35}< x< \dfrac{153+64+56}{168}\cdot\dfrac{8}{13}\)
\(\Leftrightarrow-1< x< 1\)
=>x=0
a: \(=\dfrac{-12}{7}\left(\dfrac{4}{35}+\dfrac{31}{35}\right)-\dfrac{2}{7}=\dfrac{-12}{7}-\dfrac{2}{7}=-2\)
b: =(-4)+(-4)+...+(-4)
=-4*25=-100
c: \(=157\cdot\left(-37\right)-41\cdot53+37\cdot157+51\cdot53\)
=10*53
=530