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a, \(\frac{17}{y}=\frac{-7}{11}\)
\(\Rightarrow17\cdot11=-7\cdot y\)
\(\Rightarrow187=-7\cdot y\)
\(\Rightarrow\frac{187}{-7}=y\)
b, \(\frac{-8}{3x-1}=\frac{4}{7}\)
\(\Rightarrow\frac{-8}{3x-1}=\frac{-8}{-14}\)
\(\Rightarrow3x-1=-14\)
\(\Rightarrow3x=-14+1\)
\(\Rightarrow3x=-13\)
\(\Rightarrow x=\frac{-13}{3}\)
c, \(\frac{x}{-3}=\frac{-3}{x}\)
\(\Rightarrow x\cdot x=-3\cdot\left(-3\right)\)
\(\Rightarrow x^2=9\)
\(\Rightarrow x^2=\left(\pm3\right)^2\)
\(\Rightarrow x=\pm3\)
d, \(\frac{-4}{y}=\frac{x}{2}\)
\(\Rightarrow-4\cdot2=x\cdot y\)
\(\Rightarrow-8=x\cdot y\)
\(\Rightarrow x;y\inƯ\left(-8\right)=\left\{-1;1;-2;2;-4;4;-8;8\right\}\)
ta có bảng :
x | -1 | -8 | -2 | -4 |
y | 8 | 1 | 4 | 2 |
a)\(\frac{14}{y}\)\(=\) \(\frac{-7}{11}\)
\(\Rightarrow\)\(14\cdot11=y\cdot\left(-7\right)\)
\(y=\)\(\frac{14\cdot11}{-7}\)
\(y=22\)
c) \(\frac{x}{-3}\) = \(\frac{-3}{x}\)
\(\Rightarrow\) \(x\cdot x=\left(-3\right)\cdot\left(-3\right)\)
\(\Rightarrow\)\(x^2=9\)
\(\Rightarrow\)\(x^2=9\)hoặc \(x^2=-9\)
\(TH1:\) \(x^2=9\)
\(\Rightarrow\)\(x=3\)
\(TH2:\)\(x^2=-9\)
\(\Rightarrow\)\(x=-3\)
a) \(x+\frac{3}{4}=\frac{-3}{7}\Leftrightarrow x=\frac{-3}{7}-\frac{3}{4}=-\frac{33}{28}\)
b) \(\frac{5}{2}-\left(\frac{3}{2}+x\right)=\frac{7}{4}+\frac{3}{5}=\frac{47}{20}\Rightarrow\frac{3}{2}+x=\frac{5}{2}-\frac{47}{20}=\frac{3}{20}\Rightarrow x=\frac{3}{20}-\frac{3}{2}=-\frac{27}{20}\)
c) \(\frac{3}{7}-\left(\frac{7}{4}-x\right)=\frac{5}{14}\Leftrightarrow\frac{7}{4}-x=\frac{3}{7}-\frac{5}{14}=\frac{1}{14}\Leftrightarrow x=\frac{7}{4}-\frac{1}{14}=\frac{47}{28}\)
d) \(\left[\frac{8}{3}-\left(-x\right)\right]-\left(\frac{-1}{2}\right)=\frac{5}{3}+\frac{1}{6}=\frac{11}{6}\Rightarrow\frac{8}{3}+x+\frac{1}{2}=\frac{11}{6}\)
\(\Rightarrow\frac{8}{3}+x=\frac{11}{6}-\frac{1}{2}=\frac{4}{3}\Rightarrow x=\frac{4}{3}-\frac{8}{3}=-\frac{4}{3}\)
e) \(\frac{3-x}{9}=\frac{4}{3-x}\Leftrightarrow\left(3-x\right)^2=4.9=36\Rightarrow3-x=6\Leftrightarrow x=3-6=-3\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{6}{6}=1\)
\(\frac{13}{14}+\frac{14}{8}=\frac{13.4}{14.4}+\frac{14.7}{8.7}=\frac{52}{56}+\frac{98}{56}=\frac{150}{56}\simeq2,68\)
Như vậy: \(1\le x\le2,68\)
Mà x thuộc N => x=1 và x=2
Đáp số: x=1 và x=2
\(2\frac{3}{4}x=3\frac{1}{7}:0,01\)
=> \(\frac{11}{4}x=\frac{22}{7}:\frac{1}{100}\)
=> \(x=\frac{\frac{22}{7}\cdot100}{\frac{11}{4}}=\frac{\frac{2200}{7}}{\frac{11}{4}}=\frac{2200}{7}\cdot\frac{4}{11}=\frac{800}{7}\)
\(2\frac{1}{3}:\frac{1}{3}=\frac{7}{9}:x\)
=> \(\frac{7}{3}:\frac{1}{3}=\frac{7}{9}:x\)
=> \(\frac{7}{3}\cdot\frac{3}{1}=\frac{7}{9}:x\)
=> \(\frac{7}{9}:x=7\)
=> \(x=\frac{7}{9}:7=\frac{7}{9}\cdot\frac{1}{7}=\frac{1}{9}\)
\(3,2x+\left(-1,2\right)x+2,7=-4,9\)
=> \(\left[3,2+\left(-1,2\right)\right]x=-7,6\)
=> 2.x = -7,6
=> x = -3,8
\(x:\frac{9}{14}=\frac{7}{3}:x\)
=> \(x\cdot\frac{14}{9}=\frac{7}{3}\cdot\frac{1}{x}\)
=> \(\frac{14x}{9}=\frac{7}{3x}\)
nên sửa lại đề này
\(\frac{37-x}{x+13}=\frac{3}{7}\)
=> 7(37 - x) = 3(x + 13)
=> 259 - 7x = 3x + 39
=> 259 - 7x - 3x - 39 = 0
=> 220 - 10x = 0
=> 220 = 10x
=> x = 22
\(\frac{x}{15}=\frac{-60}{x}\)=> x2 = -900 => x không thỏa mãn
a) \(2\frac{3}{4}.x=3\frac{1}{7}:0,01\)
\(=>2\frac{3}{4}.x=2200=>\frac{11}{4}x=2200=>x=800\)
b) \(2\frac{1}{3}:\frac{1}{3}=\frac{7}{9}:x\)
\(=>7=\frac{7}{9}:x\)
hay \(\frac{7}{9}:x=7=>x=\frac{1}{9}\)
c) \(3,2x+\left(-1,2\right)x+2,7=-4,9\)
\(=>2x+2,7=-4,9\)
\(=>2x=-7,6=>x=-3,8\)
d) \(x:\frac{9}{14}=\frac{7}{3}:x\)
\(=>x.x=\frac{7}{3}.\frac{9}{14}\)
\(=>x^2=\frac{3}{2}\)
....................( ko hiểu)
e) \(\frac{37-x}{x+13}=\frac{3}{7}\)
\(=>7.\left(37-x\right)=3.\left(x+13\right)\)
\(=>259-7x=3x+39\)
\(-7x-3x=39-269\)
\(=>-10x=-220=>x=22\)
f) \(\frac{x}{15}=\frac{-60}{x}\)
\(=>x.x=-60.15=>x^2=-900=>x=-30\)
cậu có thể tham khảo bài alfm trên đây ạ, chúc học tốt:>
\(a/\frac{7}{9}-\frac{x}{3}=\frac{1}{9}\)
\(\Rightarrow\frac{x}{3}=\frac{7}{9}-\frac{1}{9}\)
\(\Rightarrow\frac{x}{3}=\frac{2}{3}\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(b/\frac{1}{x}-\frac{-2}{15}=\frac{7}{15}\)
\(\Rightarrow\frac{1}{x}=\frac{7}{15}+\frac{-2}{15}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{3}\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
\(c/\frac{-11}{14}-\frac{-4}{x}=\frac{-3}{14}\)
\(\Rightarrow\frac{-4}{x}=\frac{-11}{14}-\frac{-3}{14}\)
\(\Rightarrow\frac{-4}{x}=\frac{-4}{7}\)
\(\Rightarrow x=7\)
Vậy \(x=7\)
\(d/\frac{x}{21}-\frac{2}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{x}{21}=\frac{5}{21}+\frac{2}{3}\)
\(\Rightarrow\frac{x}{21}=\frac{19}{21}\)
\(\Rightarrow x=19\)
Vậy \(x=19\)
#Mạt Mạt#
Giải:
a) \(\left(4,5-2x\right).\left(-1\dfrac{4}{7}\right)=\dfrac{11}{14}\)
\(\Leftrightarrow\left(4,5-2x\right).\left(-\dfrac{3}{7}\right)=\dfrac{11}{14}\)
\(\Leftrightarrow4,5-2x=\dfrac{11}{14}:\left(-\dfrac{3}{7}\right)=-\dfrac{11}{6}\)
\(\Leftrightarrow2x=4,5-\left(-\dfrac{11}{6}\right)\)
\(\Leftrightarrow2x=\dfrac{19}{3}\)
\(\Leftrightarrow x=\dfrac{19}{3}:2=\dfrac{19}{6}\)
Vậy ...
b) \(\dfrac{4}{9}x=\dfrac{9}{8}-0,125\)
\(\Leftrightarrow\dfrac{4}{9}x=\dfrac{9}{8}-\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{9}x=1\)
\(\Leftrightarrow x=1:\dfrac{4}{9}=\dfrac{9}{4}\)
Vậy ...
Các câu còn lại làm tương tự.
=>\(\frac{2x-2}{14}=\frac{3}{14}\Rightarrow2x-2=3\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\)
\(\frac{x-1}{7}\)= \(\frac{3}{14}\)
=> 2( x-1 ) = 3
<=> 2x - 2 = 3
<=> 2x = 5
<=> x = \(\frac{5}{2}\)
Vậy x = \(\frac{5}{2}\)
#Rin - Học tốt nha !! ^^