\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)

\(\Rightarrow x\left(1-2y\right)=40\)

\(\Rightarrow x;1-2y\in U\left(40\right)\)

\(U\left(40\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm8;\pm10;\pm20;\pm40\right\}\)

Mà 1-2y lẻ nên:

\(\left\{{}\begin{matrix}1-2y=1\Rightarrow2y=0\Rightarrow y=0\\x=40\\1-2y=-1\Rightarrow2y=2\Rightarrow y=1\\x=-40\end{matrix}\right.\)

\(\left\{{}\begin{matrix}1-2y=5\Rightarrow2y=-4\Rightarrow y=-2\\x=8\\1-2y=-5\Rightarrow2y=6\Rightarrow y=3\\x=-8\end{matrix}\right.\)

b tương tự.

c) \(\left(x+1\right)\left(x-2\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\Rightarrow x< -1\\x-2>0\Rightarrow x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\Rightarrow x>-1\\x-2< 0\Rightarrow x< 2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1< x< 2\Rightarrow x\in\left\{0;1\right\}\)

d tương tự

a) \(L=\left(x-1\right)^2+\left(x+5\right)^2\)

Ta có: \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left(x+5\right)^2\ge0\end{cases}}\)

\(\Rightarrow L=0\Leftrightarrow\)\(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(x+5\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\x=-5\end{cases}}\left(L\right)\)

Vậy đa thức L vô nghiệm

d) \(M=x^2-5x-6\)

\(\Leftrightarrow M=x^2-6x+x-6\)

\(\Leftrightarrow M=x\left(x-6\right)+\left(x-6\right)\)

\(\Leftrightarrow M=\left(x+1\right)\left(x-6\right)\)

M = 0 \(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=6\end{cases}}\)

Vậy đa thức M có hai nghiệm là -1 hoặc 6

Bài 1:

\(M\left(1\right)=a+b+6\)

Mà \(M\left(1\right)=0\)

\(\Rightarrow a+b+6=0\)

\(\Rightarrow a+b=-6\)( * )

\(\Rightarrow2a+2b=-12\) (1)

Ta có: \(M\left(-2\right)=4a-2b+6\)

Mà \(M\left(-2\right)=0\)

\(\Rightarrow4a-2b=-6\)(2)

Lấy (1) cộng (2) ta được:

\(6a=-18\)

\(a=-3\)

Thay a=-3 vào (* ) ta được:

\(b=-3\)

Vậy a=-3 ; b=-3

Bài 2:

a) \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{8}-\frac{y}{4}=\frac{5}{x}\)

\(\Leftrightarrow\frac{1}{8}-\frac{2y}{8}=\frac{5}{x}\)

\(\Leftrightarrow\frac{1-2y}{8}=\frac{5}{x}\)

\(\Leftrightarrow\left(1-2y\right).x=5.8\)

\(\Leftrightarrow\left(1-2y\right).x=40\)

Vì \(x,y\in Z\Rightarrow1-2y\in Z\)

mà \(40=1.40=40.1=5.8=8.5=\left(-1\right).\left(-40\right)=\left(-40\right).\left(-1\right)=\left(-5\right).\left(-8\right)=\left(-8\right).\left(-5\right)\)

Thử từng TH

Bài 2b

Thay x = -1; y = 1 vào N ta đc:

\(N=\left(-1\right).1+\left(-1\right)^2.1^2+\left(-1\right)^3.1^3+\left(-1\right)^4.1^4+\left(-1\right)^5.1^5\)

\(=\left(-1\right)+1+\left(-1\right)+1+\left(-1\right)\)

\(=-1\)

a) \(\left(\frac{1}{3}x^2y^2\right).\left(-\frac{4}{5}xy^3\right).\left(yz^2\right)\)

\(=\left(\frac{1}{3}.\frac{-4}{5}\right)\left(x^2.x\right)\left(y^2.y^3.y\right)z^2\)

\(=\frac{-4}{15}x^3y^6z^2\)

→ Phần hệ số: \(-\frac{4}{15}\)

→ Phần biến: \(x^3y^6z^2\)

b) \(5xy^2\left(-3x^2y\right)^2.\left(-\frac{1}{9}y^2\right)\)

\(=5xy^2.\left(9x^4y^2\right)\left(-\frac{1}{9}y^2\right)\)

\(=\left(5.9.\frac{-1}{9}\right)\left(x.x^4\right)\left(y^2.y^2.y^2\right)\)

\(=-5x^5y^6\)

→ Phần hệ số: -5

→ Phần biến: \(x^5y^6\)

c) \(x\left(-\frac{5}{2}y\right).\left(-\frac{1}{3}x^3\right)\)

\(=\left[\left(-\frac{5}{2}\right).\left(-\frac{1}{3}\right)\right]\left(x.x^3\right)y\)

\(=\frac{5}{6}x^4y\)

→ Phần hệ số: \(\frac{5}{6}\)

→ Phần biến: \(x^4y\)

d) \(-\frac{1}{2}x^3y^6.\frac{6}{5}x^2y^3\left(-5xy^2\right)\)

\(=\left[\left(-\frac{1}{2}\right).\frac{6}{5}.\left(-5\right)\right]\left(x^3.x^2.x\right)\left(y^6.y^3.y^2\right)\)

\(=3x^6y^{11}\)

→ Phần hệ số: 3

→ Phần biến: \(x^6y^{11}\)

e) \(3xy\left(-\frac{2}{9}y\right).\frac{1}{2}ax^2b\)

\(=\left(3.\frac{-2}{9}.\frac{1}{2}\right)\left(x.x^2\right)\left(y.y\right)a.b\)

\(=-\frac{1}{3}x^3y^2ab\)

→ Phần hệ số: \(-\frac{1}{3}\)

→ Phần biến: \(x^3y^2ab\)

Bài 2:

a) Thay x = -2 vào M ta được:

\(M\left(-2\right)=3\left(-2\right)^2-5\left(-2\right)-2\)

\(=3.4+10-2=12+10-2=20\)

Thay \(x=\frac{1}{3}\) vào M ta được:

\(M\left(\frac{1}{3}\right)=3.\left(\frac{1}{3}\right)^2-5.\frac{1}{3}-2\)

\(=3.\frac{1}{9}-\frac{5}{3}-2\)

\(=\frac{1}{3}-\frac{5}{3}-2\)

\(=-\frac{4}{3}-2=-\frac{10}{3}\)