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\(ĐKXĐ:x\ne-3;2\)
\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{1}{x-2}\)
\(=\frac{x^2+4x+4}{\left(x+3\right)\left(x+2\right)}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{x+3}{\left(x+2\right)\left(x+3\right)}\)
\(=\frac{x^2+4x+4-5-x-3}{\left(x+2\right)\left(x+3\right)}=\frac{x^2+3x-4}{\left(x+3\right)\left(x+2\right)}=\frac{\left(x+4\right)\left(x-1\right)}{\left(x+3\right)\left(x+2\right)}\)
\(x^2-9=0\Leftrightarrow x=3\left(vì:x\ne-3\right)\)
\(\Rightarrow P=\frac{7}{15}\)
\(P\inℤ\Leftrightarrow x^2+3x-4⋮x^2+5x+6\Leftrightarrow2x+10⋮x^2+5x+6\Leftrightarrow12⋮x^2+5xx+6\)
\(................\left(dễ\right)\)
P/s: shitbo sai rồi nha bạn!Nếu không tin thì thay x = 3 vào P ban đầu và giá trị P sau khi rút gọn sẽ thấy sự khác biệt =)
ĐK: \(x\ne-3;x\ne2\)
a) \(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}\)
\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)
Thay vào điều kiện,tìm loại x = -3 .Tìm được x =3
Ta có: \(P=\frac{x-4}{x-2}=\frac{3-4}{3-2}=-1\)
c)Ta có: \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)
Để P có giá trị nguyên thì \(\frac{2}{x-2}\) nguyên hay \(x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Suy ra \(x=\left\{0;1;3;4\right\}\)
mk nghỉ bài này đề sai
a) điều kiện : \(x\ne0;x\ne-1;x\ne2\)
ta có : \(A=1+\left(\dfrac{x+1}{x^3+1}-\dfrac{1}{x-x^2-1}+\dfrac{2}{x+1}\right):\dfrac{x^3-2x^2}{x^3-x^2+x}\)
\(\Leftrightarrow A=1+\left(\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}+\dfrac{2}{x+1}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{x+1+x+1+2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{2x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x^2-x+1}{x\left(x-2\right)}\) \(\Leftrightarrow A=1+\dfrac{2x^2+4}{x\left(x+1\right)\left(x-2\right)}=\dfrac{2x^2+4+x\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}\)\(\Leftrightarrow A=\dfrac{x^3+x^2-2x+4}{x\left(x+1\right)\left(x-2\right)}\)
b) ta có : \(\left|x-\dfrac{3}{4}\right|=\dfrac{5}{4}\) \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{4}\\x-\dfrac{3}{4}=\dfrac{-5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=\dfrac{-1}{2}\end{matrix}\right.\)
thế vào \(A\) ta có : \(A=\dfrac{41}{5}\)
vậy ...............................................................................................................
2) \(\dfrac{2x}{x^2-25}+\dfrac{5}{5-x}-\dfrac{1}{x+5}\)
\(=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\) MTC: \(\left(x-5\right)\left(x+5\right)\)
\(=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5x+25-x+5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4x+30}{\left(x-5\right)\left(x+5\right)}\)
3)
Thay \(x=\dfrac{4}{5}\) vào biểu thức A ta được:
\(\dfrac{-4.\left(\dfrac{4}{5}\right)+30}{\left[\left(\dfrac{4}{5}\right)-5\right]\left[\left(\dfrac{4}{5}\right)+5\right]}\)
\(=\dfrac{-3,2+30}{-4,2.5,8}\)
\(=\dfrac{26,8}{-24,36}\)
\(=\dfrac{-670}{609}\)
Vậy giá trị của biểu thức A tại \(x=\dfrac{4}{5}\) là \(\dfrac{-670}{609}\)
1/ a, \(A=\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
Vậy \(A=x\)
b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)
Vậy...
2/a,
\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)
\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)
\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)
\(=\dfrac{3x+2}{x\left(3x+2\right)}\)
\(=\dfrac{1}{x}\)
Vậy....
b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy..
Bài 3:
a: DKDXĐ: x<>1
b: \(=\dfrac{x^2+2+x^2-x-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{2}{x-1}=\dfrac{x^2-2x+1}{\left(x-1\right)^2}\cdot\dfrac{2}{x^2+x+1}=\dfrac{2}{x^2+x+1}\)
c: Để C lớn nhất thì \(A=x^2+x+1_{MIN}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)
Dấu = xảy ra khi x=-1/2
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
\(Câu\text{ }1:\)
\(\text{ a) }A=\dfrac{4}{x^2+2}+\dfrac{3}{2-x^2}-\dfrac{12}{4-x^4}\\ A=\dfrac{4\left(2-x^2\right)}{\left(x^2+2\right)\left(2-x^2\right)}+\dfrac{3\left(2+x^2\right)}{\left(2-x^2\right)\left(2+x^2\right)}-\dfrac{12}{\left(2+x^2\right)\left(2-x^2\right)}\\ A=\dfrac{4\left(2-x^2\right)+3\left(2+x^2\right)-12}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{8-4x^2+6+3x^2-12}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-x^2-2}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-\left(x^2+2\right)}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-1}{2-x^2}\)
\(\text{b) }Để\text{ }A=-3\\ thì\Rightarrow\dfrac{-1}{2-x^2}=-3\\ \Leftrightarrow2-x^2=3\\ \Leftrightarrow x^2=-1\\ \Leftrightarrow x\text{ }không\text{ }có\text{ }giá\text{ }trị\left(vì\text{ }x^2\ge0\forall x\right)\\ \text{ }Vậy\text{ }để\text{ }A=-3\text{ }thì\text{ }x\text{ }không\text{ }có\text{ }giá\text{ }trị.\)
\(\text{c) }Ta\text{ }có:\text{ }A=\dfrac{-1}{2-x^2}\\ A=\dfrac{1}{x^2-2}\\ x^2\ge0\forall x\\ \Rightarrow x^2-2\ge-2\forall x\\ \Rightarrow A=\dfrac{1}{x^2-2}\le-\dfrac{1}{2}\\ Dấu\text{ }"="\text{ }xảy\text{ }khi:\\ x^2=0\\ \Leftrightarrow x=0\\\text{ }Vậy\text{ }A_{\left(Max\right)}=-\dfrac{1}{2}\text{ }khi\text{ }x=0\)
\(Câu\text{ }2:\)
\(\text{a) }B=\dfrac{1}{x}+\dfrac{1}{x+5}+\dfrac{x-5}{x\left(x+5\right)}\\ B=\dfrac{x+5}{x\left(x+5\right)}+\dfrac{x}{\left(x+5\right)x}+\dfrac{x-5}{x\left(x+5\right)}\\ B=\dfrac{x+5+x+x-5}{x\left(x+5\right)}\\ B=\dfrac{3x}{x\left(x+5\right)}\\ B=\dfrac{3}{x+5}\left(\text{*}\right)\)
\(\text{b) }Ta\text{ }có:\text{ }\left|x-1\right|=6\\ \Leftrightarrow\left[{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\\ Ta\text{ }lại\text{ }có:\text{ }B=\dfrac{3}{x+5}\\ \RightarrowĐKCĐ:x+5\ne0\\ \Rightarrow x\ne-5\\ \Rightarrow x=7\text{ }thỏa\text{ }mãn\text{ }với\text{ }điều\text{ }kiện\text{ }của\text{ }biến.\\ x=-5\text{ }không\text{ }thỏa\text{ }mãn\text{ }với\text{ }điều\text{ }kiện\text{ }của\text{ }biến.\\ Thay\text{ }x=7\text{ }vào\text{ }\left(\text{*}\right),ta\text{ }được:\text{ }B=\dfrac{3}{7+5}=\dfrac{3}{12}=\dfrac{1}{4}\\ \text{ }Vậy\text{ }với\text{ }x=7\text{ }thì\text{ }B=\dfrac{1}{4}\\ với\text{ }x=-5\text{ }thì\text{ }B\text{ }không\text{ }có\text{ }giá\text{ }trị.\)
\(\text{c) }Ta\text{ }có:B=\dfrac{3}{x+5}\\ \RightarrowĐể\text{ }B\in Z\\ thì\Rightarrow3⋮x+5\\ \Rightarrow x+5\inƯ_{\left(3\right)}\\ Mà\text{ }Ư_{\left(3\right)}=\left\{\pm1;\pm3\right\}\\ Ta\text{ }lập\text{ }bảng\text{ }xét\text{ }giá\text{ }trị:\)
| \(x+5\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
| \(x\) | \(-8\) | \(-6\) | \(-4\) | \(-2\) |
\(\Rightarrow x\in\left\{-8;-6;-4;-2\right\}\\ Vậy\text{ }để\text{ }B\in Z\\ thì x\in\left\{-8;-6;-4;-2\right\}\)
1,
\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)
\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)
2.
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
3.
Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)
4.
\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)
\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)
5.
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)