7x=3y=>x/3=y/7=k và x-y=16

=>x=3k;y=7k

có x-y=3k-7k=-4k=16

=>k=-4

x/3=-4=>x=-12

y/7=-4=>y=-28

dễ thui mà bạn Nguyễn Tiến Lực  trả lời rồi

\(f,\dfrac{x^2-6x+9}{x^2-8x+15}\\ =\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x-5\right)}\\ =\dfrac{x-3}{x-5}\\ l,\dfrac{5xy+5x+3+3y}{10xy-15x-9+6y}\\ =\dfrac{5x\left(y+1\right)+3\left(y+1\right)}{5x\left(2y-3\right)+3\left(2y-3\right)}\\ =\dfrac{\left(y+1\right)\left(5x+3\right)}{\left(2y-3\right)\left(5y+3\right)}\\ =\dfrac{y+1}{2y-3}\)

a)4x²+8x+3=0

<=>(4x²+2x)+(6x+3)=0

<=>2x(2x+1)+3(2x+1)=0

<=>(2x+1)(2x+3)=0

<=>2x+1=0 hoặc 2x+3=0

<=>x=-1/2 hoặc x=-3/2

b)(2x+3)²=(x-6)²

<=>(2x+3)²-(x-6)²=0

<=>(2x-3-x+6)(2x+3+x-6)=0

<=>(x+3)(3x-3)=0

<=>x+3=0 hoặc 3x-3=0

<=>x=-3 hoặc x=1

c)x³-7x²+15x-9=0

<=>(x³-6x²+9x)-(x²-6x+9)=0

<=>x(x-3)²-(x-3)²=0

<=>(x-3)²(x-1)=0

<=>(x-3)²=0 hoặc x-1=0

<=>x=3 hoặc x=1

2: Ta có: \(x^4-4x^3-9x^2+8x+4=0\)

\(\Leftrightarrow x^4-x^3-3x^3+3x^2-12x^2+12x-4x+4=0\)

\(\Leftrightarrow x^3\left(x-1\right)-3x^2\left(x-1\right)-12x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2-12x-4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-5x^2-10x-2x-4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)-5x\left(x+2\right)-2\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-5x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x^2-5x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{5-\sqrt{33}}{2}\\x=\dfrac{5+\sqrt{33}}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-2;\dfrac{5-\sqrt{33}}{2};\dfrac{5+\sqrt{33}}{2}\right\}\)

1: Ta có: \(x^4+5x^3+10x^2+15x+9=0\)

\(\Leftrightarrow x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0\)

\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)+6x\left(x+1\right)+9\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2+6x+9\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[x^3+3x^2+x^2+6x+9\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x+3\right)+\left(x+3\right)^2\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\left(x^2+x+3\right)=0\)

mà \(x^2+x+3>0\forall x\)

nên (x+1)(x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

Vậy: S={-1;-3}

a) Ta có: \(x^2-x-12=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

b) Ta có: \(x^2+3x-18=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=3\end{matrix}\right.\)

8x + 15x - 3x = -400

=> (8 + 15 - 3)x = -400

=> 20x = -400

=> x = -400 : 20

=> x = -20

    8x + 15x - 3x = - 400

=> (8 + 15 - 3)x = - 400

              => 20x = - 400

                 => x = - 400 : 20

=> x = - 20

\(8x+15x-3x=-400\)

\(\Rightarrow x\left(8+15-3\right)=-400\)

\(\Rightarrow20x=-400\)

\(\Rightarrow x=-400:20\)

\(\Rightarrow x=-20\)

Vậy x=-20

C(x)= 2x-3=0 hoac 5x+7=0

        2x=0+3        5x=0-7

        2x=3            5x=-7

         x=3:2            x=-7:5

          x=1.5            x=-1.4

a.

\(\left(2x-3\right)\times\left(5x+7\right)=0\)

TH1:

\(2x-3=0\)

\(2x=3\)

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

TH2:

\(5x+7=0\)

\(5x=-7\)

\(x=-\frac{7}{5}\)

Vậy \(C\left(x\right)\) có nghiệm là \(\frac{3}{2}\) hoặc \(-\frac{7}{5}\)

b.

\(\left(15x^5+4x^2-8\right)-\left(15x^5-x-8\right)=0\)

\(15x^5+4x^2-8-15x^5+x+8=0\)

\(\left(15x^5-15x^5\right)+4x^2+x+\left(8-8\right)=0\)

\(x\left(4x-1\right)=0\)

TH1:

\(x=0\)

TH2:

\(4x-1=0\)

\(4x=1\)

\(x=\frac{1}{4}\)

Vậy \(D\left(x\right)\) có nghiệm là \(0\) hoặc \(\frac{1}{4}\)

c.

\(\left(5x^7-8x^2\right)-\left(4x^7+4^2\right)-\left(x^7+4\right)=0\)

\(5x^7-8x^2-4x^7-16-x^7-4=0\)

\(\left(5x^7-4x^7-x^7\right)-8x^2-\left(16-4\right)=0\)

\(-8x^2-12=0\)

\(-8x^2=12\)

\(x^2=-\frac{12}{8}\)

mà \(x^2\ge0\) với mọi x

=> \(E\left(x\right)\) vô nghiệm

\(a,C\left(x\right)=\left(2x-3\right)\left(5x+7\right)=0\)

\(\Leftrightarrow\) \(\left[\begin{array}{nghiempt}2x-3=0\\5x+7=0\end{array}\right.\) \(\Leftrightarrow\) \(\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{7}{5}\end{array}\right.\)

Vậy \(x=\frac{3}{2}\) và \(x=-\frac{7}{5}\) là nghiệm của đa thức C(x)

\(b,D\left(x\right)=\left(15x^5+4x^2-8\right)-\left(15x^5-x-8\right)=0\)

\(\Leftrightarrow15x^5+4x^2-8-15x^5+x+8=0\)

\(\Leftrightarrow4x^2+x=0\) \(\Leftrightarrow x\left(4x+1\right)=0\)  \(\Leftrightarrow\) \(\left[\begin{array}{nghiempt}x=0\\4x+1=0\end{array}\right.\)  \(\Leftrightarrow\) \(\left[\begin{array}{nghiempt}x=0\\x=-\frac{1}{4}\end{array}\right.\)

Vậy \(x=0\) và \(x=-\frac{1}{4}\) là nghiệm đa thức D(x)

\(c,E\left(x\right)=\left(5x^7-8x^2\right)-\left(4x^7+4x^4\right)-\left(x^7+4\right)=0\)

\(\Leftrightarrow5x^7-8x^2-4x^7-4x^4-x^7-4=0\)

\(\Leftrightarrow-8x^2-4x^4-4=0\)

\(\Leftrightarrow-4\left(2x^2+x^4+1\right)=0\)

\(\Leftrightarrow2x^2+x^4+1=0\) \(\Leftrightarrow x^4+x^2+x^2+1=0\) 

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

\(\Leftrightarrow\left(x^2+1\right)^2=0\) \(\Leftrightarrow x^2+1=0\) \(\Leftrightarrow x^2=-1\) \(\Rightarrow x\in\varnothing\)

Vậy E(x) vô nghiệm