Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x+y}{4+5}=\dfrac{18}{9}=2\)

Do đó: x=8; y=10

tìm các số nguyên x để mỗi phân số sau đây là số nguyên

Ngô Hải Nam ơi bn trả lời giúp mik ik

bài đó là bài 4^* tìm các số nguyên x để mỗi phân số sau đây là số nguyên

a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh

a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)

\(x-\dfrac{23}{14}=1\)

\(x=1+\dfrac{23}{14}\)

\(x=\dfrac{37}{14}\)

b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)

\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)

\(x+\dfrac{9}{5}=\dfrac{11}{3}\)

\(x=\dfrac{11}{3}-\dfrac{9}{5}\)

\(x=\dfrac{28}{15}\)

a) \(x=\dfrac{-2}{7}+\dfrac{9}{7}=1\) 

b) \(\dfrac{x}{3}=\dfrac{2}{5}+\dfrac{-4}{3}\) 

     \(\dfrac{x}{3}=\dfrac{-14}{15}\) 

\(\Rightarrow x=\dfrac{3.-14}{15}=\dfrac{-14}{5}\)

\(x=\dfrac{-2}{7}+\dfrac{9}{7}\) 

\(x=1\)

x=1

a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

⇔\(7\left(x-3\right)=5\left(x+5\right)\)

⇔\(7x-21=5x+25\)

⇔\(7x-21-5x-25=0\)

⇔\(2x-46=0\)

⇔\(2x=46\)

⇔\(x=23\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

⇔\(\left(x+1\right)\left(x-1\right)=7.9\)

⇔\(x^2-1=63\)

⇔\(x^2=64=8^2\)

⇔\(\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

\(a.\dfrac{x-3}{x+5}=\dfrac{5}{7}\)                          

    \(\left(x-3\right).7=\left(x+5\right).5\) 

    \(\left(x.7\right)+\left[\left(-3\right).7\right]=\left(x.5\right)+\left(5.5\right)\)

    \(7x-21=5x+25\)

    \(7x-5x=25+21\)

             \(2x=46\)

               \(x=46:2\)

               \(x=23\)

Câu b/ cứ làm theo câu a/ là được

AcCl₃ và NiCl₂

khó quá ??????????????????????????????????

a) (-2) . ( x+7 ) + (-5) = 7 

<=>(-2).(x+7)=7+5

<=>x+7=12:(-2)

<=>x+7=-6

<=>x=(-6)-7

<=>x=-13

Vậy x=-13

b)(x+4) : (-7) = 14

<=>x+4=14 x (-7)

<=>x+4=-98

<=>x=-98-4

<=>x=-102

Vậy x= -102

c) 72 : ( x+5) - 4 = -12 

<=>72:(x+5)=(-12)+4

<=>x+5=72:(-8)

<=>x+5=-9

<=>x=-9-5

<=>x=-14

Vậy x= -14

d) (x+3) : (-6 ) + 12 = 8 

<=>(x+3) :(-6)=8-12

<=>x+3=(-4)x(-6)

<=>x+3=24

<=>x=24-3

<=>x=21

Vậy x= 21 

a)

ta có \(\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{1}{13}.\dfrac{1}{14}=\dfrac{3}{7}.9.\dfrac{1}{2}.\dfrac{1}{13}-\dfrac{1}{13}.\dfrac{1}{14}\)\(=\dfrac{1}{13}.\left(\dfrac{3}{7}.\dfrac{9}{2}-\dfrac{1}{14}\right)=\dfrac{1}{13}.\dfrac{26}{14}=\dfrac{1.26}{13.14}\)\(=\dfrac{1.13.2}{13.7.2}=\dfrac{1}{7}\)

b)\(x-\left(\dfrac{5}{2}+2x\right)=x-\dfrac{5}{2}-2x=-x-\dfrac{5}{2}=\dfrac{7}{4}\)

\(\Rightarrow-x=\dfrac{7}{4}+\dfrac{5}{2}=\dfrac{17}{4}\)

\(\Rightarrow x=-\dfrac{17}{4}\)(vì -x là số đối của x)

`a, 2/3 +3/4 = (8+9)/12=17/12.`

`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.

`=> x=2.`

`b, 5/6-1/4=(20-6)/24=7/12`.

`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.

`=> x=1`.

a) \(\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{8+9}{12}=\dfrac{17}{12}\)

-> 1 1/3 + 4/5 = 4/3 + 4/5 =  20+12/15 = 32/15

vậy x có thể = 14/14 = 1 (x thuộc N)

7:

a: ĐKXĐ: x>=0; x<>1

\(D=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)

\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-1}{\sqrt{x}+1}\)

b: Khi x=4/9 thì \(D=\dfrac{-1}{\dfrac{2}{3}+1}=-1:\dfrac{5}{3}=-\dfrac{3}{5}\)

c: |D|=1/3

=>D=-1/3 hoặc D=1/3

=>\(\left[{}\begin{matrix}\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{3}\\\dfrac{-1}{\sqrt{x}+1}=\dfrac{1}{3}\left(loại\right)\end{matrix}\right.\)

=>\(\sqrt{x}+1=3\)

=>\(\sqrt{x}=2\)

=>x=4

6:

a: \(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\)

\(=\dfrac{3\left(\sqrt{x}+3\right)}{3+\sqrt{x}}\cdot\dfrac{-\sqrt{x}}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)

b: C<-1

=>C+1<0

=>\(\dfrac{-3\sqrt{x}+2\sqrt{x}+4}{2\sqrt{x}+4}< 0\)

=>\(-\sqrt{x}+4< 0\)

=>\(-\sqrt{x}< -4\)

=>\(\sqrt{x}>4\)

=>x>16

\(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\\ =\dfrac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{3\left(\sqrt{x}+3\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)

Để `C < -1` Ta có :

 \(\dfrac{-3}{2\sqrt{x}+4}< -1\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+1< 0\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+\dfrac{2\sqrt{x}+4}{2\sqrt{x}+4}< 0\\ \Leftrightarrow-3+2\sqrt{x}+4< 0\\ \Leftrightarrow2\sqrt{x}+1< 0\\ \Leftrightarrow2\sqrt{x}< -1\\ \Leftrightarrow\sqrt{x}< -\dfrac{1}{2}\\ \Leftrightarrow x< \dfrac{1}{4}\)