Từ pt thứ 2, ta thấy \(y^2⋮9\Leftrightarrow y⋮3\) \(\Leftrightarrow y=3z\left(z\inℤ\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+3xz=2019\\9z^2-9xz=99\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+3xz=2019\\z^2-xz=11\end{matrix}\right.\) (*)

Từ pt đầu tiên của (*), ta thấy \(x⋮3\Leftrightarrow x=3t\left(t\inℤ\right)\)

Khi đó \(9t^2+9tz=2019\)  \(\Rightarrow2019⋮9\), vô lí. 

Do đó, pt đã cho không có nghiệm nguyên.

Bạn xem lại đề

\(...\Rightarrow x+x+\dfrac{x}{43}+\dfrac{x}{8}=14+148+\dfrac{10}{30}+\dfrac{5}{95}\)

\(\Rightarrow\left(1+1+\dfrac{1}{43}+\dfrac{1}{8}\right)x=162+\dfrac{1}{3}+\dfrac{1}{19}\)

\(\Rightarrow\left(\dfrac{2.43.8}{43.8}+\dfrac{1.8}{43.8}+\dfrac{1.43}{43.8}\right)x=\dfrac{162.3.19}{3.19}+\dfrac{1.19}{3.19}+\dfrac{1.3}{19.3}\)

\(\Rightarrow\left(\dfrac{688}{344}+\dfrac{8}{344}+\dfrac{43}{344}\right)x=\dfrac{9234}{57}+\dfrac{19}{57}+\dfrac{3}{57}\)

\(\Rightarrow\dfrac{739}{344}x=\dfrac{9256}{57}\)

\(\Rightarrow x=\dfrac{9256}{57}:\dfrac{739}{344}=\dfrac{9256}{57}.\dfrac{344}{739}=\dfrac{\text{3184064}}{\text{42123}}\)

ta có :

\(12\times x=42:\frac{1}{10}\)

ha y \(12\times x=42\times10=420\) nên : \(x=420:12=35\)

Vậy x = 35

x+4=8x-10

<=> x+4-8x+10=0

<=> -7x+14=0

<=> -7x=-14

<=> x=2

Vậy x=2

x + 4 = 8x - 10

4 + 10 = 8x - x

14        = 7x

x           =  14 : 7

x           = 2

Vậy x = 2

# HOK TỐT #

MNG GIẢI GIÚP MIK VS Ạ,MƠN MNG NHIỀU LMK,

Đề sai rồi bạn nhé

2 + 3 - 5 = 0 (ở dưới mẫu) thì vô lí nên đề sai  

Sửa đề: x+y+z=10

Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)

mà x+y+z=10

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y+z}{2+3+5}=\dfrac{10}{10}=1\)

Do đó: x=2; y=3; z=5

\(\Leftrightarrow9x\left(x+2\right)+9y\left(y-\dfrac{2}{3}\right)=10\\ \Leftrightarrow9x^2+18x+9y^2-6y-10=0\\ \Leftrightarrow\left(9x^2+18x+9\right)+\left(9y^2-6y+1\right)=0\\ \Leftrightarrow9\left(x+1\right)^2+\left(3y-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)

thấy sai sai bạn ạ

\(x^2+2x-10=0\)

\(\Leftrightarrow x^2+2x+1-9=0\)

\(\Leftrightarrow\left(x+1\right)^2-9=0\\\)

\(\Leftrightarrow\left(x+1\right)^2=9\)

\(\Leftrightarrow\left(x+1\right)^2=\pm\sqrt{9}\)

\(\Leftrightarrow\left(x+1\right)^2=\left(\pm3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3-1\\x=-3-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy S={2;-4}

\(x^2\) + 2\(x\) - 10 = 0

\(x^2\) + 2\(x\) + 1 - 11 = 0

(\(x^2\) + 2\(x\) + 1) - 11 = 0

(\(x\) + 1)² - 11 = 0

(\(x\) + 1)² = 11

\(\left[{}\begin{matrix}x+1=\sqrt{11}\\x+1=-\sqrt{11}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\sqrt{11}-1\\x=-\sqrt{11}-1\end{matrix}\right.\)

\(\frac{x}{27}=\frac{2}{3}\)

Ta có:

\(\frac{2}{3}=\frac{18}{27}\)

\(\Leftrightarrow\frac{x}{27}=\frac{18}{27}\)

\(\Leftrightarrow x=18\)

\(\frac{35}{49}=\frac{10}{y}\)

Ta có:

\(\frac{35}{49}=\frac{5}{7}=\frac{10}{14}\)

\(\Leftrightarrow\frac{10}{14}=\frac{10}{y}\)

\(\Leftrightarrow y=14\)

y =14 nha