a)Ta có: 5x+7y=112

\(\Rightarrow x=\frac{112-7y}{5}=22-y+\frac{2-2y}{5}\)

Do x,y nguyên \(\Rightarrow\frac{2-2y}{5}\)nguyên hay (2-2y) chia hết 5 <=>2(1-y) chia hết 5;(2,5)=1

=>(1-y) chia hết 5 hay (y-1) chia hết 5.Đặt y-1=5t \(\left(t\in Z\right)\)

\(\Rightarrow y=5t+1\)

Thay y vào x ta có:x=21-7t

Lại có x>0;y>0 \(\Rightarrow\hept{\begin{cases}5t+1>0\\21-7t>0\end{cases}\Rightarrow}\hept{\begin{cases}t>-\frac{1}{5}\\t< 3\end{cases}}\)

\(\Rightarrow t=\left\{0;1;2\right\}\)

a)5x+7y=112

tách ra các giá trị nguyên

tìm 1 nghiệm riêng

\(\text{a)Ta có: 5x+7y=112 }\)

\(\Rightarrow x=\frac{112-7y}{5}=22-y+\frac{2-2y}{5}\)

\(6x^4+5x^3-38x^2+5x+6=0\\ \Leftrightarrow6x^4+20x^3+6x^2-15x^3-50x^2-15x+6x^2+20x+6=0\\ \Leftrightarrow2x^2\left(3x^2+10x+3\right)-5x\left(3x^2+10x+3\right)+2\left(3x^2+10x+3\right)=0\\ \Leftrightarrow\left(3x^2+10x+3\right)\left(2x^2-5x+2\right)=0\\ \Leftrightarrow\left(3x^2+x+9x+3\right)\left(2x^2-x-4x+2\right)=0\\ \Leftrightarrow\left[x\left(3x+1\right)+3\left(3x+1\right)\right]\left[x\left(2x-1\right)-2\left(2x-1\right)\right]=0\\ \Leftrightarrow\left(3x+1\right)\left(x+3\right)\left(2x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+3=0\\2x-1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=-3\\x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

                    \(6x^4+5x^3-38x^2+5x+6=0\)

\(\Leftrightarrow\)\(6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6=0\)

\(\Leftrightarrow\)\(\left(x-2\right)\left(6x^3+17x^2-4x-3\right)=0\)

\(\Leftrightarrow\)\(\left(x-2\right)\left(6x^3+18x^2-x^2-3x-x-3\right)=0\)

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

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

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

P/s: lm tiếp nhé

c: =>(x+2)(x+3)(x-5)(x-6)=180

=>(x^2-3x-10)(x^2-3x-18)=180

=>(x^2-3x)^2-28(x^2-3x)=0

=>x(x-3)(x-7)(x+4)=0

=>\(x\in\left\{0;3;7;-4\right\}\)

c: =>(x-3)(x+2)(2x+1)(3x-1)=0

=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)

a) Ta có: \(\left(x+1\right)^4+\left(x-3\right)^4=0\)

Nhận thấy: \(\hept{\begin{cases}\left(x+1\right)^4\ge0\left(\forall x\right)\\\left(x-3\right)^4\ge0\left(\forall x\right)\end{cases}\Rightarrow}\left(x+1\right)^4+\left(x-3\right)^4\ge0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\x=3\end{cases}}\) (mâu thuẫn)

=> pt vô nghiệm

b) \(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow\left(x^4-2x^3\right)+\left(4x^3-8x^2\right)+\left(4x^2-8x\right)+\left(3x-6\right)=0\)

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

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

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

Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\left(\forall x\right)\)

=> \(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

a,\(\left(x+1\right)^4+\left(x-3\right)^4=0\)

\(x^4-1+x^4-81=0\)

\(2x^4-82=0\)

\(2x^4=82\)

\(x^4=41\)

\(x=\sqrt[4]{41}\)

\(\Rightarrow\)vô nghiệm

pt bậc 4 => có 4 nghiệm.

bấm máy tính tìm nghiệm đẹp (-2 và 3). Chia sơ đồ hoocne.

2 nghiệm đẹp (-2 và 3) được rồi, còn 2 nghiệm còn lại thì giải pt bậc 2 là ra.

kq: x=-2, x=3, x=1/3 , x=-1/2

Ta có \(6x^4-5x^3-38x^2-5x+6=0\Leftrightarrow6x^4+12x^3-17x^3-34x^2-4x^2-8x+3x+6=0\Leftrightarrow6x^3\left(x+2\right)-17x^2\left(x+2\right)-4x\left(x+2\right)+3\left(x+2\right)=0\Leftrightarrow\left(x+2\right)\left(6x^3-17x^2-4x+3\right)=0\Leftrightarrow\left(x+2\right)\left(6x^3-18x^2+x^2-3x-x+3\right)=0\Leftrightarrow\left(x+2\right)\left[6x^2\left(x-3\right)+x\left(x-3\right)-\left(x-3\right)\right]=0\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(6x^2+x-1\right)=0\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(6x^2-2x+3x-1\right)=0\Leftrightarrow\left(x+2\right)\left(x-3\right)\left[2x\left(3x-1\right)+\left(3x-1\right)\right]=0\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(3x-1\right)\left(2x+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x+2=0\\x-3=0\\3x-1=0\\2x+1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-2\\x=3\\x=\dfrac{1}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy S={\(-\dfrac{1}{2};-2;\dfrac{1}{3};3\)}

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

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

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

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

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

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

- Khi x - 1 = 0 thì x = 1

- Khi x + 1 = 0 thì x = -1

- Khi \(x^2-4x+1=0\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow\orbr{\begin{cases}x=\sqrt{3}+2\\x=-\sqrt{3}+2\end{cases}}\)

Pt có tậo nghiệm là: \(S=\left\{1;-1;\sqrt{3}+2;-\sqrt{3}+2\right\}\)

Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath tích mình nha

Ta có : \(6x^4+5x^3-38x^2+5x+6=0\)

\(\Leftrightarrow6x^4+20x^3+6x^2-15x^3-50x^2-15x+6x^2+20x+6=0\)

\(\Leftrightarrow2x^2\left(3x^2+10x+3\right)-5x\left(3x^2+10x+3\right)+2\left(3x^2+10x+3\right)=0\)

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

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

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

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

\(\Leftrightarrow\)\(3x+1=0\)

hoặc    \(x+3=0\)

hoặc   \(2x-1=0\)

hoặc    \(x-2=0\)

\(\Leftrightarrow\)\(x=-\frac{1}{3}\)

hoặc   \(x=-3\)

hoặc   \(x=\frac{1}{2}\)

hoặc   \(x=2\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{1}{3};-3;\frac{1}{2};2\right\}\)