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A) \(\frac{2}{3}+\frac{1}{5}-\frac{10}{7}=\frac{13}{15}-\frac{10}{7}=\frac{-59}{105}\)
B) \(\frac{7}{12}-\frac{27}{18}.\frac{2}{18}=\frac{7}{12}-\frac{3}{2}.\frac{2}{18}=\frac{7}{12}-\frac{1}{6}=\frac{5}{12}\)
C) \(\left(\frac{23}{41}-\frac{15}{82}\right).\frac{41}{25}=\frac{31}{82}.\frac{41}{25}=\frac{31}{50}\)
D) \(\left(\frac{4}{5}+\frac{1}{2}\right).\left(\frac{3}{13}-\frac{8}{13}\right)=\frac{13}{10}.\frac{-5}{13}=\frac{-1}{2}\)
a, = \(\frac{-59}{105}\)
b = \(\frac{7}{144}\)
c= \(\frac{31}{50}\)
d=\(\frac{-1}{2}\)
a,ĐK : x \(\ne\)3/7
\(\frac{24}{7x-3}=-\frac{4}{25}\Leftrightarrow600=-28x+12\Leftrightarrow-28x=588\Leftrightarrow x=-21\)
b, ĐK : x;y \(\ne\)6
Xét : \(\frac{4}{x-6}=-\frac{12}{18}\Leftrightarrow72=-12x+72\Leftrightarrow x=0\)
Xét : \(\frac{y}{24}=-\frac{12}{18}\Leftrightarrow18y=-288\Leftrightarrow y=-16\)
\(\frac{24}{7.x-3}=-\frac{4}{25}\)
24.25=7.x-3.-4
600=7.x-3.-4
7.x-3.-4=600
7.x-3=600:-4
7.x-3=-150
7.x=-150+3
7.x=-147
x=-147:7
x=-21
vậy x=-21
Bài 1:
\(g.\frac{5}{11}+\frac{6}{11}=\frac{5+6}{11}=\frac{11}{11}=1\)\(\)
\(e.\frac{-17}{25}.\frac{20}{33}+\frac{-17}{25}.\frac{13}{33}+\frac{-3}{25}=\frac{-17}{25}.\left(\frac{20}{33}+\frac{13}{33}\right)+\frac{-3}{25}\)
\(=\frac{-17}{25}.1+\frac{-3}{25}=\frac{-17}{25}+\frac{-3}{25}=\frac{-17-3}{25}=\frac{-20}{25}=\frac{-4}{5}\)
\(d.\frac{5}{7}.\frac{19}{23}+\frac{5}{7}.\frac{5}{23}-\frac{5}{7}.\frac{1}{23}=\frac{5}{7}\left(\frac{19}{23}+\frac{5}{23}-\frac{1}{23}\right)\)
\(=\frac{5}{7}\left(\frac{19+5-1}{23}\right)=\frac{5}{7}.1=\frac{5}{7}\)
\(c.\left(-11\right).\frac{9}{22}=\frac{\left(-11\right).9}{22}=\frac{-99}{22}=\frac{-9}{2}\)
\(b.\frac{5}{6}-\frac{1}{8}=\frac{5.4}{6.4}-\frac{1.3}{8.3}=\frac{20}{24}-\frac{3}{24}=\frac{17}{24}\)
\(a.\frac{2}{3}+\frac{1}{5}-\frac{1}{6}=\frac{2.10}{3.10}+\frac{1.6}{5.6}-\frac{1.5}{6.5}=\frac{20}{30}+\frac{6}{30}-\frac{5}{30}\)
\(=\frac{20+6-5}{30}=\frac{21}{30}=\frac{7}{10}\)
Bài 2:
\(a.\frac{3}{4}+x=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{11}{12}-\frac{3}{4}\)
\(\Leftrightarrow x=\frac{11}{12}-\frac{9}{12}\)
\(\Leftrightarrow x=\frac{2}{12}=\frac{1}{6}\)
\(b.x-\frac{11}{12}=0,5\)
\(\Leftrightarrow x=\frac{1}{2}-\frac{11}{12}\)
\(\Leftrightarrow x=\frac{6}{12}+\frac{11}{12}\)
\(\Leftrightarrow x=\frac{17}{12}\)
a/ \(\frac{x+2}{27}=\frac{x}{9}\)
=> 9(x + 2) = 27x
=> 9x + 18 = 27x
=> 9x + 18 - 27x = 0
=> 9x - 27x + 18 = 0
=> -18x = -18
=> x = 1
b/ \(\frac{-7}{x}=\frac{21}{34-x}\)
=> -7(34 - x) = 21x
=> -238 + 7x = 21x
=> 21x - 7x = -238
=> -14x = 238
=> x = -17
c) \(\frac{-8}{15}< \frac{x}{40}< \frac{-7}{15}\)
Ta có BCNN(15,40,15) = 120
=> \(\frac{-64}{120}< \frac{3x}{120}< \frac{-56}{120}\)
=> -64 < 3x < -56
=> x \(\in\){ -19;-20;-21}
Câu d tương tự
Bài1
a) 25/42 - 20/63 =5/18
b) 9/50 - 13/75 - 1/6 = -4/25
c) 2/15 - 2/65 - 4/39 = 0
Bài2
a) x + 7/12 =17/18-1/9 b) 29/30 - (18/23 + x)=7/69
x + 7/12 = 5/6 18/23 + x =29/30 - 7/69
x =5/6 - 7/12 18/23 +x = 199/230
x = 1/4 x = 199/230 - 18/23
x= 19/230
a) \(\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)\)
\(=\frac{31}{23}-\frac{15}{23}\)
\(=\frac{16}{23}\)
b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{-67}{67}+\frac{41}{41}\)
\(=\frac{1}{3}-1+1\)
\(=\frac{1}{3}\)
c) \(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{52}-\frac{3}{11}\right)\)
\(=\frac{38}{45}-\frac{8}{45}+\frac{17}{52}+\frac{3}{11}\)
\(=\frac{30}{45}+\frac{17}{52}+\frac{3}{11}\)
\(=\frac{2}{3}+\frac{17}{52}+\frac{3}{11}\)
\(=\frac{104+51}{156}+\frac{3}{11}\)
\(=\frac{155}{156}+\frac{3}{11}\)
\(=\frac{156}{156}-\frac{1}{156}+\frac{3}{11}\)
\(=1-\frac{1}{156}+\frac{3}{11}\)
\(=1-\left(\frac{11-468}{1716}\right)\)
\(=1-\frac{-457}{1716}\)
\(=1+\frac{457}{1716}\)
\(=\frac{2173}{1716}\)
a)31/23-(7/32+8/23)=31/23-7/32-8/23=(31/23-8/23)-7/32=1-7/32=25/32
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
a) \(\left(x+\frac{1}{4}\right)^2+\frac{11}{25}=\frac{18}{25}\)
\(\Rightarrow\left(x+\frac{1}{4}\right)^2=\frac{7}{25}\)
\(\Rightarrow\) Không có x