Ta có: \(\frac{x}{5}=\frac{y}{2}=\frac{z}{7}\) => \(\frac{2x}{10}=\frac{y}{2}=\frac{z}{7}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

    \(\frac{2x}{10}=\frac{y}{2}=\frac{z}{7}=\frac{2x+y-z}{10+2-7}=\frac{53}{5}\)

=> \(\hept{\begin{cases}\frac{x}{5}=\frac{53}{5}\\\frac{y}{2}=\frac{53}{5}\\\frac{z}{7}=\frac{53}{5}\end{cases}}\) => \(\hept{\begin{cases}x=\frac{53}{5}.5=53\\y=\frac{53}{5}.2=\frac{106}{5}\\z=\frac{53}{5}.7=\frac{371}{5}\end{cases}}\)

Vậy ...

Ta có : x/5=y/2=z/7

=> 2x/10=y/2=z/7

= 2x +y -z / 10 + 2 - 7

=53/5

=> x= 53/5 . 5 = 53

     y=53/5 . 2 = 106/5

     z=53/5 . 7 = 371/5

a) \(x^2+7x+12\)

\(=x^2+3x+4x+12\)

\(=x\left(x+3\right)+4\left(x+3\right)\)

\(=\left(x+3\right)\left(x+4\right)\)

b) \(3x^2-5x+2\)

\(=3x^2-3x-2x+2\)

\(=3x\left(x-1\right)-2\left(x-1\right)\)

\(=\left(x-1\right)\left(3x-2\right)\)

Xét ∆ABC có : 

MN//BC ( I \(\in\)MN )

M là trung điểm AB )

=> N là trung điểm AC 

=> AN = NC

a,

Ta có: \(a\left(b+1\right)b\left(a+1\right)=\left(a+1\right)\left(b+1\right)\)

\(\Rightarrow ab=\left(a+1\right)\left(b+1\right):\left(a+1\right)\left(b+1\right)=1\)

=>đpcm

b,

Ta có: \(2\left(a+1\right)\left(a+b\right)=\left(a+b\right)\left(a+b+2\right)\)

\(\Rightarrow2a+2=a+b+2\)

\(\Rightarrow a-b=0\)

\(\Rightarrow a^2+b^2=2ab\)

\(\Rightarrow a^2+b^2=2\) (đpcm)

\(\left|x^2-9\right|=\left|-7\right|\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-9=7\\x^2-9=-7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=16\\x^2=2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\pm4\\x=\pm\sqrt{2}\end{cases}}\)

x² + 6x - 16 > 2x - 7

<=> x² + 6x - 2x > -7 + 16

<=> x² + 4x > 9

<=> x² + 4x + 4 > 9 + 4

<=> ( x + 2 )² > 13

<=> ( x + 2 )² > \(\left(\pm\sqrt{13}\right)^2\)

<=> \(\orbr{\begin{cases}x+2>\sqrt{13}\\x+2>-\sqrt{13}\end{cases}\Rightarrow}\orbr{\begin{cases}x>\sqrt{13}-2\\x>-2-\sqrt{13}\end{cases}}\)

=x(x+1)

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

\(=x+1-x^2+1\)

\(=x-x^2\)

\(x^3+6x^2+12x+8=\left(x-2\right)^3\)

cái thứ 2 số xấu quá

Mik nghĩ đề câu sau là thek này:

\(x^3+6x^2+3x+1\)

\(=\left(x+1\right)\left(x^2+x+1\right)+6x\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+7x+1\right)\)

\(x^3-x+y^3-y=x^3+y^3-x-y=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)

\(x^2-x-y^2-y=x^2-y^2-x-y=\left(x+y\right)\left(x-y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)

\(5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(5+a\right)\)