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\(A=\left|x+1\right|+24\)
\(\Leftrightarrow\left|x+1\right|\ge0\)
\(\Leftrightarrow\left|x+1\right|+24\ge24\)
\(\Leftrightarrow\) MAXA = 24 => x = - 1
a)
ta có:
/x+1/>hoặc = 0 nên A=/x+1/+24 < hoặc = 24
=>A>hoặc = 24
=> A đạt giá trị lớn nhất là 24 thì x+1=0<=>x=1
\(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\\ A=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\\ A=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
\(A=\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}=\dfrac{3\left(0,125-0,1+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(0,125-0,1+\dfrac{5}{11}+\dfrac{5}{12}\right)}+\dfrac{\dfrac{3}{5}\left(2,5+\dfrac{5}{3}-1,25\right)}{2,5+\dfrac{5}{3}-1,25}=-\dfrac{3}{5}+\dfrac{3}{5}=0\)
\(a)196:4-12.\left(-5\right)=49+60=109.\\ b)2^2.5+\left(49-17^2\right)=4.5+49-289=20-240=-220.\\ c)29.\left(15-34\right)-15.\left(29-34\right)=29.15-29.34-15.29+15.34=-29.34+15.34=-476.\)
Bài 2:
Gọi số sách là x
Theo đề, ta có: \(x\in BC\left(10;12;15\right)\)
mà 100<=x<=150
nên x=120
c) 0,1+0,2-0,3-0,4+0,5+0,6-0,7-0,8+0,9+0,10
=(0,1+0,2-0,3-0,4)+(0,5+0,6-0,7-0,8)+(0,9+0,1)
=-0,4+(-0,4)+1
=-0,8+1
=-0,2
ok rồi nha
6,25 . 1,25 . 2,4 . 0,8
= (6,25 . 2,4) . (1,25 . 0,8)
= 15 . 1
= 15
Các câu sau tương tự
Chúc bạn học tốt
a: \(=\dfrac{15}{4}:\dfrac{2-7}{16}-\dfrac{5}{9}\cdot\left(\dfrac{3}{5}+\dfrac{4}{5}\right)\)
\(=\dfrac{15}{4}\cdot\dfrac{16}{-5}-\dfrac{5}{9}\cdot\dfrac{7}{5}\)
\(=\dfrac{-240}{20}-\dfrac{7}{9}=-12-\dfrac{7}{9}=\dfrac{-115}{9}\)
c: \(=\dfrac{1}{5}-\dfrac{7}{2}-\dfrac{3}{2}+\dfrac{5}{4}\)
\(=\dfrac{4+25}{20}-5=\dfrac{29}{20}-\dfrac{100}{20}=\dfrac{-71}{20}\)
d: \(=\dfrac{12}{17}\left(1-\dfrac{1}{15}-\dfrac{4}{5}+1\right)=\dfrac{12}{17}\cdot\dfrac{17}{15}=\dfrac{12}{15}=\dfrac{4}{5}\)
e: \(=\dfrac{2}{15}\cdot\dfrac{9}{8}-\dfrac{7}{4}\cdot\dfrac{9}{8}+\dfrac{25}{36}\)
\(=\dfrac{9}{8}\left(\dfrac{2}{15}-\dfrac{7}{4}\right)+\dfrac{25}{36}\)
\(=\dfrac{9}{8}\cdot\dfrac{8-105}{60}+\dfrac{25}{36}\)
\(=\dfrac{9}{8}\cdot\dfrac{-97}{60}+\dfrac{25}{36}=\dfrac{-1619}{1440}\)
\(a.\)\(\frac{2}{7}+x=1,25\)
\(\Leftrightarrow x=1,25-\frac{2}{7}\)
\(\Leftrightarrow x=\frac{5}{4}-\frac{2}{7}\)
\(\Leftrightarrow x=\frac{35}{28}-\frac{8}{28}\)
\(\Leftrightarrow x=\frac{27}{28}\)
\(b.\)\(x:8,5=2,2\)
\(\Leftrightarrow x=2,2.8,5\)
\(\Leftrightarrow x=18,7\)
\(c.\)\(-4\frac{1}{5}-x=2,25\)
\(\Leftrightarrow-\frac{21}{5}-x=\frac{9}{4}\)
\(\Leftrightarrow x=-\frac{21}{5}-\frac{9}{4}\)
\(\Leftrightarrow x=-\frac{84}{20}-\frac{45}{20}\)
\(\Leftrightarrow x=-\frac{129}{45}\)
\(d.\)\(x=11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(\Leftrightarrow x=\frac{45}{4}-\left(\frac{19}{7}+\frac{21}{4}\right)\)
\(\Leftrightarrow x=\frac{45}{4}-\frac{19}{7}-\frac{21}{4}\)
\(\Leftrightarrow x=\left(\frac{45}{4}-\frac{21}{4}\right)-\frac{19}{7}\)
\(\Leftrightarrow x=6-\frac{19}{7}\)
\(\Leftrightarrow x=\frac{42}{7}-\frac{19}{7}\)
\(\Leftrightarrow x=\frac{23}{7}\)
a) x = \(\frac{5}{4}-\frac{2}{7}\)
x = \(\frac{35}{28}-\frac{8}{28}\)
x = \(\frac{27}{28}\)
b) x = 2,2 x 8,5 = 18,7
c) -0,8 - x = 2,25
x = -3,05
d) x = dư nhiều bn ơi
Hội con 🐄 chúc bạn học tốt!!!
Ta có : \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
\(=\left(\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}+\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}\right).\frac{2005}{1890}+115\)
\(=\left(\frac{3}{5}-\frac{3}{5}\right).\frac{2005}{1890}+115=0+115=115\)
= ( \(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)+ \(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)) x \(\frac{2005}{1890}\)+ 115
= ( \(\frac{3(\frac{1}{2}+\frac{1}{3}-\frac{1}{4})}{5(\frac{1}{1}+\frac{1}{3}-\frac{1}{4})}\)+ \(\frac{3(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12})}{-5(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12})}\)) x \(\frac{2005}{1890}\)+ 115
=( \(\frac{3}{5}\)+\(\frac{3}{-5}\)) x \(\frac{2005}{1890}\)+115 = 0 +115 = 115
a) \(\left(\frac{2x}{5}-1\right):\left(-5\right)=\frac{1}{7}\)
\(\frac{2x}{5}-1=\frac{1}{7}.\left(-5\right)\)
\(\frac{2x}{5}-1=\frac{-5}{7}\)
\(\frac{2x}{5}=\frac{-5}{7}+\frac{7}{7}\)
\(\frac{2x}{5}=\frac{2}{7}\)
\(=>2x.7=2.5\)
\(=>14x=10\)
\(=>x=\frac{5}{7}\)
c) \(\left|3,5+2,5x\right|-2,5=3,5\)
\(\left|3,5+2,5x\right|=3,5+2,5\)
\(\left|3,5+2,5x\right|=6\)
\(TH1\) \(3,5+2,5x=6\) \(TH2\) \(3,5+2,5x=-6\)
\(2,5x=6-3,5\) \(2,5x=-6-3,5\)
\(2,5x=2,5\) \(2,5x=-9.5\)
\(x=1\) \(x=-3,8\)
vậy \(x=1\) hoặc \(x=-3,8\)
câu d) làm tương tự như câu c)
a: \(3\cdot2,25-0,75=6,75-0,75=6\)
b: \(\left(-1,25\right)+3,5+1,25+36,5\)
\(=\left(-1,25+1,25\right)+\left(3,5+36,5\right)\)
=0+40
=40
mik camon^^