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a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\frac{x+1}{2\cdot\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2x+1-3x-1+x^2}{3x}\)
\(A=\frac{x^2-x}{3x}\)
\(A=\frac{x\left(x-1\right)}{3x}\)
\(A=\frac{x-1}{3}\)
b) Thay x = 4 ta có :
\(A=\frac{4-1}{3}=\frac{3}{3}=1\)
c) Để A thuộc Z thì \(x-1⋮3\)
\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)
\(\Rightarrow x\in\left\{1;4;7;...\right\}\)
Vậy.....
a, ĐKXĐ: \(x\ne\pm3\)
\(A=\frac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(=\frac{3x-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}=\frac{3x-12}{3x+9}\)
b, \(x=-4\Rightarrow A=\frac{3.\left(-4\right)-12}{3.\left(-4\right)+9}=8\)
c, \(A\in Z\Rightarrow3x-12⋮\left(3x+9\right)\Rightarrow3x+9-21⋮\left(3x+9\right)\Rightarrow21⋮\left(3x+9\right)\)
\(\Rightarrow3x+9\inƯ\left(21\right)=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)
Mà \(3x+9⋮3\Rightarrow3x+9\in\left\{-21;-3;3;21\right\}\Rightarrow x\in\left\{-10;-4;-2;4\right\}\) (thỏa mãn điều kiện)
a, ĐỂ A xác định :
\(\Rightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{cases}}\Rightarrow x\ne\pm3.\)
\(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x+3\right)\left(x-3\right)}\right):\frac{3}{x-3}\)
\(A=\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}:\frac{3}{x-3}\)
\(A=\frac{x^2-3x+2x^2+6x-3x^2+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(A=\frac{3x+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(A=\frac{x-4}{x+3}\)
b
a) Điều kiện xác định của phân thức A là x#+-5
\(A=\frac{2\left(x+15\right)}{x^2-25}-\frac{x+3}{x+5}+\frac{x}{x-5}
\)
\(A=\frac{2\left(x+15\right)}{\left(x+5\right)\left(x-5\right)}-\frac{x+3}{x+5}+\frac{x}{x-5}\)
\(A=\frac{2\left(x+15\right)}{\left(x+5\right)\left(x-5\right)}-\frac{\left(x+3\right)\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\frac{x\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(A=\frac{2x+30-\left(x^2-5x+3x-15\right)+x^2+5x}{\left(x+5\right)\left(x-5\right)}\)
\(A=\frac{2x+30-x^2+5x+3x-15+x^2+5x}{\left(x+5\right)\left(x-5\right)}=\frac{15x+15}{\left(x+5\right)\left(x-5\right)}=\frac{15\left(x+1\right)}{\left(x+5\right)\left(x-5\right)}\)
tick đúng nha, ý b tí mình giải nhé
a, \(ĐKXĐ\hept{\begin{cases}2-x\ne0\\2+x\ne0\end{cases}\Leftrightarrow x\ne\pm2}\)
b, Ta có: \(A=\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\)
\(=\frac{\left(2+x\right)^2}{\left(2-x\right)\left(2+x\right)}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}-\frac{\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\)
\(=\frac{4+4x+x^2+4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\)
\(=\frac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}\)
\(=\frac{4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x}{x-2}\)
a) ĐKXĐ: \(\hept{\begin{cases}2-x\ne0\\x^2-4\ne0\\2+x\ne0\end{cases}}\)<=>\(\hept{\begin{cases}2-x\ne0\\2+x\ne0\\\left(x-2\right)\left(x+2\right)\ne0\end{cases}}\)<=>\(x\ne\pm2\)
b)\(A=\frac{2+x}{2-x}-\frac{4x}{x^2-4}-\frac{2-x}{2+x}\)
\(\Leftrightarrow A=\frac{2+x}{2-x}+\frac{4x}{4-x^2}-\frac{2-x}{2+x}\)
\(\Leftrightarrow A=\frac{\left(2+x\right)\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}+\frac{4x}{\left(2-x\right)\left(2+x\right)}-\frac{\left(2-x\right)\left(2-x\right)}{\left(2+x\right)\left(2-x\right)}\)
\(\Leftrightarrow A=\frac{x^2+4x+4+4x-x^2+4x-4}{\left(2+x\right)\left(2-x\right)}\)
\(\Leftrightarrow A=\frac{12x}{\left(2+x\right)\left(2-x\right)}\)
a) \(ĐKXĐ:x\ne\pm1\)
\(A=\frac{x^3-2x^2+x}{x^2-1}\)
\(\Leftrightarrow A=\frac{x\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow A=\frac{x^2-x}{x+1}\)
b) Để A có giá trị nguyên
\(\Leftrightarrow\frac{x^2-x}{x+1}\inℤ\)
\(\Leftrightarrow x^2-x⋮x+1\)
\(\Leftrightarrow x^2-x-2+2⋮x+1\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)+2⋮x+1\)
\(\Leftrightarrow2⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)
Ta sẽ loại các giá trị ktm
\(\Leftrightarrow x\in\left\{-2;0;-3\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{-2;0;-3\right\}\)
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
a) ĐKXD: \(x+2\ne0\)và \(x^2+4x+4\ne0\)và \(x^2-4\ne0\)và \(2-x\ne0\)
\(\Leftrightarrow x\ne-2\)và \(\left(x+2\right)^2\ne0\)và \(\left(x-2\right)\left(x+2\right)\ne0\)và \(x\ne2\)
\(\Leftrightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
+) \(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(=\left[\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right]:\left[\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x+2\right)^2}:\frac{-x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x+2\right)^2}.\frac{\left(x-2\right)\left(x+2\right)}{-x}\)
\(=\frac{-2x+4}{x+2}\)
b) Ta có: x-1=3 <=> x=4 Thay vào A ta được:
\(\frac{-2.4-4}{4+2}=-2\)
c)
Để \(A\in Z\Leftrightarrow8⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(8\right)=\left\{\pm1;\pm4;\pm8\right\}\)
Bạn làm nốt nha
a) Biểu thức A xác định khi \(\hept{\begin{cases}x+1\ne0\\x^2-1\ne0\end{cases}\Leftrightarrow}\)\(\begin{cases}x\ne1\\x\ne\pm1\end{cases}\)(bạn thông cảm chỗ này mình ko viết được ngoặc nhọn)
Vậy biểu thức A xác định khi \(x\ne\pm1\)
b)\(A=\frac{2x}{x+1}+\frac{1+2x}{x^2-1}=\frac{2x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{1+2x}{x^2-1}=\frac{2x^2-2x}{x^2-1}+\frac{1+2x}{x^2-1}\)
\(=\frac{2x^2+1}{x^2-1}=\frac{2x^2-2+3}{x^2-1}=\frac{2\left(x^2-1\right)+3}{x^2-1}=\frac{2\left(x^2-1\right)}{x^2-1}+\frac{3}{x^2-1}=2+\frac{3}{x^2-1}\)
c) A nguyên khi và chỉ khi \(\frac{3}{x^2-1}\) nguyên
<=>3 chia hết cho x2-1
<=>\(x^2-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
<=>\(x^2\in\left\{-2;0;2;4\right\}\)
Vì \(x^2\ge0\Rightarrow x^2\in\left\{0;2;4\right\}\)<=>\(x\in\left\{-2;0;\sqrt{2};2\right\}\)
Vì \(x\in Z\Rightarrow x\in\left\{-2;0;2\right\}\)
Vậy A nguyên khi \(x\in\left\{-2;0;2\right\}\)
a)A xác khi \(\hept{\begin{cases}x+1\ne0\\x^2-1\ne0\end{cases}\Rightarrow x\ne\left\{-1,1\right\}}\)
b) \(A=\frac{2x}{x+1}+\frac{1+2x}{\left(x-1\right)\left(x+1\right)}=\frac{2x\left(x-1\right)+1+2x}{\left(x-1\right)\left(x+1\right)}=\frac{2x^2+1}{x^2-1}=2+\frac{3}{\left(x^2\right)-1}\)
c)x^2-1=U(3)={-3,-1,1,3}
x^2={-2,0,2,4}
x={-2,0,2}
a, \(A=\left(\frac{2}{x-2}-\frac{2}{x+2}\right)\frac{x^2+4x+4}{8}\)ĐK : \(x\ne\pm2\)
\(=\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right)\frac{\left(x+2\right)^2}{8}\)
\(=\frac{2x+2-2x+2}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}=\frac{4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}\)
\(=\frac{x+2}{2\left(x-2\right)}=\frac{x+2}{2x-4}\)
b, A = x hay
\(\frac{x+2}{2x-4}=x\Leftrightarrow x+2=2x^2-4x\)
\(\Leftrightarrow5x+2-2x^2=0\)vô nghiệm
tương tự với A = x/2 nhé !