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=1,4 x\(\dfrac{15}{49}-\) \(\left(\dfrac{4}{5}+\dfrac{2}{3}\right)\) : 2\(\dfrac{1}{5}\)
= \(\dfrac{3}{7}\) - \(\dfrac{22}{15}\) : \(\dfrac{11}{5}\)
= \(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)
= \(-\dfrac{5}{21}\)
( 2\(\dfrac{1}{5}\) + \(\dfrac{3}{5}\) \(\times\) \(x\)) = \(\dfrac{3}{4}\)
\(\dfrac{11}{5}\) + \(\dfrac{3}{5}\)\(x\) = \(\dfrac{3}{4}\)
\(\dfrac{3}{5}\)\(x\) = \(\dfrac{3}{4}\) - \(\dfrac{11}{5}\)
\(\dfrac{3}{5}\)\(x\) = - \(\dfrac{29}{20}\)
\(x\) = -\(\dfrac{29}{12}\)
a) 2/3 x + 1/2 = 1/10
2/3 x = 1/10 - 1/2
2/3 x = -2/5
x = -2/5 : 2/3
x = -3/5
b) 2/3 x + 1/5 = 7/10
2/3 x = 7/10 - 1/5
2/3 x = 1/2
x = 1/2 : 2/3
x = 3/4
c) (3 4/5 - 2x) . 1 1/3 = 5
19/5 - 2x = 5 : 4/3
19/5 - 2x = 15/4
2x = 19/5 - 15/4
2x = 1/20
x = 1/20 : 2
x = 1/40
d) x/7 = 6/(-21)
x = 6.7/(-21)
x = -2
1) \(x+y=10\) mà \(x=y\) nên: \(x=y=\dfrac{10}{2}=5\)
2) \(2x+3y=180\) mà \(x=y\)
Ta có: \(2y+3y=180\Rightarrow5y=180\Rightarrow y=180:5=36\)
Vậy \(x=y=36\)
3) \(x+y=180\) mà \(x=y\) nên: \(x=y=\dfrac{180}{2}=90\)
4) \(3x+5y=13\) mà \(y=2x\) ta có:
\(3x+5\cdot2x=13\Rightarrow13x=13\Rightarrow x=1\)
\(y=2x=2\cdot1=2\)
Các câu còn lại bạn làm tương tự
a)\(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\text{⇔}\dfrac{7}{4}x-5=\dfrac{10}{3}\text{⇔}\dfrac{7}{4}x=\dfrac{25}{3}\text{⇔}x=\dfrac{100}{21}\)
b)\(\dfrac{2}{3}x+\dfrac{1}{4}=\dfrac{7}{12}\text{⇔}\dfrac{2}{3}x=\dfrac{1}{3}\text{⇔}x=\dfrac{1}{2}\)
c)\(\dfrac{1}{3}+\dfrac{2}{5}\left(x+1\right)=1\text{⇔}\dfrac{2}{5}\left(x+1\right)=\dfrac{2}{3}\text{⇔}x+1=\dfrac{5}{3}\text{⇔}x=\dfrac{2}{3}\)
d)\(\dfrac{1}{4}+\dfrac{1}{3}:3x=-5\text{⇔}\dfrac{1}{3}:3x=-\dfrac{21}{4}\text{⇔}\dfrac{1}{9x}=-\dfrac{21}{4}\text{⇔}9x=-\dfrac{4}{21}\text{⇔}x=-\dfrac{4}{189}\)
a, \(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\)
\(\Rightarrow\dfrac{7}{4}x=5+\dfrac{10}{3}=\dfrac{25}{3}\)
\(\Rightarrow x=\dfrac{25}{3}:\dfrac{7}{4}=\dfrac{100}{21}\)
Vậy ...
b, \(PT\Leftrightarrow\dfrac{2}{3}x=\dfrac{7}{12}-\dfrac{1}{4}=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{1}{3}:\dfrac{2}{3}=\dfrac{1}{2}\)
Vậy ....
c, \(PT\Leftrightarrow\dfrac{2}{5}\left(x+1\right)=1-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\Rightarrow x+1=\dfrac{2}{3}:\dfrac{2}{5}=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{3}-1=\dfrac{2}{3}\)
Vậy ...
d, \(PT\Leftrightarrow\dfrac{1}{3}:3x=-5-\dfrac{1}{4}=\dfrac{1}{9}x=-\dfrac{21}{4}\)
\(\Rightarrow x=-\dfrac{189}{4}\)
Vậy ...
(x+5)⋮(x+1)⇒x+5x+1∈Z(x+5)⋮(x+1)⇒x+5x+1∈Z
Ta co: x+5x+1=x+1+4x+1=x+1x+1+4x+1=1+4x+1x+5x+1=x+1+4x+1=x+1x+1+4x+1=1+4x+1
(1+4x+1)∈Z⇒(x+1)∈U(4)(1+4x+1)∈Z⇒(x+1)∈U(4)
=> x+1={±1;±2;±4}x+1={±1;±2;±4}
⇒x={−5;−3;−2;0;1;3}
tìm x nhé