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a: \(M=\dfrac{18+5x+15+3x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{8x+24}{\left(x+3\right)\left(x-3\right)}=\dfrac{8}{x-3}\)
b: Thay x=11 vào M, ta được:
\(M=\dfrac{8}{11-3}=1\)
a) \(M=\dfrac{18}{x^2-9}+\dfrac{5}{x-3}+\dfrac{3}{x+3}.\left(x\ne\pm3\right).\)
\(M=\dfrac{18}{\left(x-3\right)\left(x+3\right)}+\dfrac{5}{x-3}+\dfrac{3}{x+3}=\dfrac{18+5\left(x+3\right)+3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{18+5x+15+3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{24+8x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{8\left(3+x\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{8}{x-3}.\)
b) Thay \(x=11\left(TM\right)\) vào biểu thức M:
\(\dfrac{8}{11-3}=\dfrac{8}{8}=1.\)
M xác định
\(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\x^2-x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\left(x-1\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne0;x\ne1\end{cases}}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)
Vậy ĐKXĐ của M là \(\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)
\(M=\frac{3}{x-1}+\frac{1}{x^2-x}=\frac{3}{x-1}+\frac{1}{x\left(x-1\right)}=\frac{3x}{x\left(x-1\right)}+\frac{1}{x\left(x-1\right)}=\frac{3x+1}{x\left(x-1\right)}\)
Thay x=5 ta có:
\(M=\frac{3.5+1}{5\left(5-1\right)}=\frac{15+1}{5.4}=\frac{16}{20}=\frac{4}{5}\)
Vậy \(M=5\)tại x=5
\(M=0\)
\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=0\Leftrightarrow3x+1=0\Leftrightarrow x=-\frac{1}{3}\)( thỏa mãn đkxđ)
Vậy với \(x=-\frac{1}{3}\)thì \(M=0\)
\(M=-1\)
\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=-1\Leftrightarrow3x+1=-x^2+x\Leftrightarrow x^2+2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)
Vậy với \(x=-1\)thì \(M=-1\)
a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right)\cdot\dfrac{x+2}{6}\)
\(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-6}{6}\cdot\dfrac{1}{x-2}=\dfrac{-1}{x-2}\)
b: x=2 ko thỏa mãn ĐKXĐ
=>Loại
Khi x=3 thì A=-1/(3-2)=-1
c: A=2
=>x-2=-1/2
=>x=3/2
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2x+4}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
Khi \(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)
a: \(M=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
a: \(M=\dfrac{2x^2-10x-x^2+x+30-x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{x+5}\)
b: Để M là số nguyên thì \(x+5\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
hay \(x\in\left\{-4;-6;-3;-7;0;-10;-15\right\}\)
a) Ta có M = (x + 3)2 - (x - 1)(x + 4) + 5
= x2 + 6x + 9 -(x2 + 3x - 4) + 5
= x2 + 6x + 9 - x2 - 3x + 4 + 5
= 3x + 18 (1)
b) Thay x = 2 vào (1)
=> M = 3.2 + 18 = 24
c) Ta có M = 15x2
=> 15x2 = 3x + 18
=> 15x2 - 3x - 18 = 0
=> 15x2 + 15x - 18x - 18 = 0
=> 15x(x + 1) - 18(x + 1) = 0
=> (15x - 18)(x + 1) = 0
=> 3(5x - 6)(x + 1) = 0
=> (5x - 6)(x + 1) = 0
=> \(\orbr{\begin{cases}5x-6=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1,2\\x=-1\end{cases}}\)
Vậy \(x\in\left\{1,2;-1\right\}\)là giá trị cần tìm
a, \(M=\left(x+3\right)^2-\left(x-1\right)\cdot\left(x+4\right)+5\)\(=x^2+6x+9-\left(x^2-x+4x-4\right)+5\)\(=3x+18\)
b, Thay x=2 vào M có \(M=3\cdot2+18=24\)
c, \(M=15x^2\Leftrightarrow15x^2=3x+18\Leftrightarrow15x^2-3x-18=0\Leftrightarrow3\cdot\left(x+1\right)\cdot\left(5x-6\right)=0\) \(\Leftrightarrow\hept{\begin{cases}x=-1\\x=\frac{6}{5}\end{cases}}\)
Vậy ....