x/3=7/y

=>xy=3.7=21

=> x,y E Ư(21)

 Vì x,y E Z nên (x,y)=(7;3)=(3;7)=(1;21)=(21;1)=(-3;-7)=(-7;-3)=(-21;-1)=(-1;-21)

song ngư hợp với cự giải đó          

bạn boi thử mà xem

tìm bội chung sau đó cộng 28 là tìm được a

\(6y^2-5y-38=0\\ \Leftrightarrow y^2-\dfrac{5}{6}y-\dfrac{38}{6}=0\\ \Leftrightarrow y^2-\dfrac{5}{6}y+\dfrac{25}{144}-\dfrac{937}{144}=0\\ \Leftrightarrow\left(y^2-\dfrac{5}{6}y+\dfrac{25}{144}\right)-\dfrac{937}{144}=0\\ \Leftrightarrow\left(y-\dfrac{5}{12}\right)^2-\dfrac{937}{144}=0\\ \Leftrightarrow\left(y-\dfrac{5}{12}+\dfrac{\sqrt{937}}{12}\right)\left(y-\dfrac{5}{12}-\dfrac{\sqrt{937}}{12}\right)=0\\ \Leftrightarrow\left(y-\dfrac{5-\sqrt{937}}{12}\right)\left(y-\dfrac{5+\sqrt{937}}{12}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y-\dfrac{5-\sqrt{937}}{12}=0\\y-\dfrac{5+\sqrt{937}}{12}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{5-\sqrt{937}}{12}\\y=\dfrac{5+\sqrt{937}}{12}\end{matrix}\right.\)

Vậy tập nghiệm phương trình là \(S=\left\{\dfrac{5-\sqrt{937}}{12};\dfrac{5+\sqrt{937}}{12}\right\}\)

\(2x^4-x^3-6x^2-x+2=0\\ \Leftrightarrow x^2\left(2x^2-x-6-\dfrac{1}{x}+\dfrac{2}{x^2}\right)=0\\ \Leftrightarrow x^2\left[\left(2x^2+4+\dfrac{2}{x^2}\right)-\left(x+\dfrac{1}{x}\right)-10\right]=0\\ \Leftrightarrow x^2\left[2\left(x^2+2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)-10\right]=0\\ \Leftrightarrow x^2\left[2\left(x+\dfrac{1}{x}\right)^2-\left(x+\dfrac{1}{x}\right)-10\right]=0\)

Đặt \(x+\dfrac{1}{x}=y\)

\(\Leftrightarrow x^2\left(2y^2-y-10\right)=0\\ \Leftrightarrow x^2\left(2y^2-5y+4y-10\right)=0\\ \Leftrightarrow x^2\left[\left(2y^2-5y\right)+\left(4y-10\right)\right]=0\\ \Leftrightarrow x^2\left[y\left(2y-5\right)+2\left(2y-5\right)\right]\\ \Leftrightarrow x^2\left(y+2\right)\left(2y-5\right)=0\\ x^2\left(x+\dfrac{1}{x}+2\right)\left(2x+\dfrac{2}{x}-5\right)=0\\ \Leftrightarrow\left(x^2+1+2x\right)\left(2x^2+2-5x\right)=0\\ \Leftrightarrow\left(x^2+2x+1\right)\left(2x^2-x-4x+2\right)=0\\ \Leftrightarrow\left(x+1\right)^2\left[\left(2x^2-x\right)-\left(4x-2\right)\right]=0\\ \Leftrightarrow\left(x+1\right)^2\left[x\left(2x-1\right)-2\left(2x-1\right)\right]=0\\ \Leftrightarrow\left(x+1\right)^2\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x-2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x=2\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy tập nghiệm phương trình là \(S=\left\{-1;\dfrac{1}{2};2\right\}\)

\(P=a^3+b^3+3ab\\ =\left(a+b\right)^3-3ab\left(a+b\right)+3ab\\ =\left(a+b\right)^3-\left[3ab\left(a+b\right)-3ab\right]\\ =\left(a+b\right)^3-3ab\left(a+b-1\right)\\ Thay\text{ }a+b=1,ta\text{ }được:\\P =1^3-3ab\left(1-1\right)=1\)

\(A=a^3+a^2c-abc+b^2c+b^3\\ =\left(a^3+b^3\right)+\left(a^2c-abc+b^2c\right)\\ =\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a^2-ab+b^2\right)c\\ =\left(a+b+c\right)\left(a^2-ab+b^2\right)\\ Thay\text{ }a+b+c=0,\text{ }ta\text{ }được:\text{ }\\ A=\left(a+b+c\right)\left(a^2-ab+b^2\right)\\ =0\cdot\left(a^2-ab+b^2\right)\\ =0\)

Vậy \(A=0\) tại \(a+b+c=0\)

Thương của 630 và 5 là:
    630:5=126
=>số đó nhân với 3 có kết quả bằng 126
Vậy số đó là:
     126:3=42
Đáp số:42