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\(\frac{4x^2-6x+5}{2x-1}=2x-2+\frac{3}{2x-1}\)
Để biểu thức có giá trị nguyên thì \(\left(2x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
Với 2x - 1 = 1 => 2x = 2 => x = 1
2x - 1 = -1 => 2x = 0 => x = 0
2x - 1 = 3 => 2x = 4 => x = 2
2x - 1 = -3 => 2x = -2 => x = -1
Vậy x = {1;0;2;-1}
a: Thay x=5 vào B, ta được:
\(B=\dfrac{5-1}{5-3}=\dfrac{4}{2}=2\)
b: \(A=\dfrac{2x^2+6x-2x^2-3x-1}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-1}{\left(x+3\right)\left(x-3\right)}\)
\(\left(\text{*}\right)\) Tìm giá trị lớn nhất của biểu thức sau:
Ta có:
\(A=\frac{x^2+1}{x^2-x+1}=\frac{2\left(x^2-x+1\right)-\left(x^2-2x+1\right)}{x^2-x+1}=2-\frac{\left(x-1\right)^2}{x^2-x+1}\le2\) với mọi \(x\)
Dấu \("="\) xảy ra \(\Leftrightarrow\) \(\left(x-1\right)^2=0\) \(\Leftrightarrow\) \(x-1=0\) \(\Leftrightarrow\) \(x=1\)
Vậy, \(A_{max}=2\) \(\Leftrightarrow\) \(x=1\)
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\(B=\frac{3-4x}{x^2+1}=\frac{4\left(x^2+1\right)-\left(4x^2+4x+1\right)}{x^2+1}=4-\frac{\left(2x+1\right)^2}{x^2+1}\le4\) với mọi \(x\)
Dấu \("="\) xảy ra \(\Leftrightarrow\) \(\left(2x+1\right)^2=0\) \(\Leftrightarrow\) \(2x+1=0\) \(\Leftrightarrow\) \(x=-\frac{1}{2}\)
Vậy, \(B_{max}=4\) \(\Leftrightarrow\) \(x=-\frac{1}{2}\)
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\(\left(\text{*}\text{*}\right)\) Tìm giá trị nhỏ nhất của biểu thức sau:
Từ \(A=\frac{x^2+1}{x^2-x+1}\)
\(\Rightarrow\) \(3A=\frac{3x^2+3}{x^2-x+1}=\frac{\left(x^2+2x+1\right)+2\left(x^2-x+1\right)}{x^2-x+1}=\frac{\left(x+1\right)^2}{x^2-x+1}+2\ge2\) với mọi \(x\)
Vì \(3A\ge2\) nên \(A\ge\frac{2}{3}\)
Dấu \("="\) xảy ra \(\Leftrightarrow\) \(\left(x+1\right)^2=0\) \(\Leftrightarrow\) \(x+1=0\) \(\Leftrightarrow\) \(x=-1\)
Vậy, \(A_{min}=\frac{2}{3}\) \(\Leftrightarrow\) \(x=-1\)
Câu b) tự giải
A = \(\frac{x^2+6x+5}{x^2+2x-15}=\frac{x^2+x+5x+5}{x^2-3x+5x-15}=\frac{x.\left(x+1\right)+5.\left(x+1\right)}{x.\left(x-3\right)+5.\left(x-3\right)}=\frac{\left(x+1\right)\left(x+5\right)}{\left(x-3\right)\left(x+5\right)}\)
\(=\frac{x+1}{x-3}=\frac{x-3}{x-3}+\frac{4}{x-3}=1+\frac{4}{x-3}\)
Để A nguyên thì \(1+\frac{4}{x-3}\text{ nguyên }\Rightarrow\frac{4}{x-3}\text{ nguyên }\Rightarrow x-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau:
| x-3 | 1 | -1 | 2 | -2 | 4 | -4 |
| x | 4 | 2 | 5 | 1 | 7 | -1 |
Vậy x={-1;1;2;4;5;7} thì A nguyên
a: \(A=\left(\dfrac{2x^2+2}{x^3-1}+\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+3}{x^3-x^2+3x-3}\right):\dfrac{1}{x-1}\)
\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{x^4+2x^2+1-x^2}-\dfrac{x^2+3}{x^2\left(x-1\right)+3\left(x-1\right)}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x^2-x+1\right)}{\left(x^2+1\right)^2-x^2}-\dfrac{x^2+3}{\left(x-1\right)\left(x^2+3\right)}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{\left(x^2+1+x\right)\left(x^2+1-x\right)}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x^2+x+1}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{1}\)
\(=\dfrac{2x^2+3+x-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-1}{1}\)
\(=\dfrac{x^2+1}{x^2+x+1}\)
b: Để A là số nguyên thì \(x^2+1⋮x^2+x+1\)
=>\(x^2+x+1-x⋮x^2+x+1\)
=>\(x⋮x^2+x+1\)
=>\(x^2+x⋮x^2+x+1\)
=>\(x^2+x+1-1⋮x^2+x+1\)
=>\(-1⋮x^2+x+1\)
=>\(x^2+x+1\in\left\{1;-1\right\}\)
=>\(x^2+x+1=1\)
=>x2+x=0
=>x(x+1)=0
=>\(x\in\left\{0;-1\right\}\)
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
đề như thế này hả B=\(\frac{x-3}{2x-1}\) hayB=\(1-\frac{3}{2x-1}\)
câu 1 nếu theo đề thì để B nguyên khi 2x-1 thuộc ước của 3 thay vào là xong