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a) Tớ làm luôn nhé , không chép lại đề đâu
P = \(\left[\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right].\dfrac{x\left(x+6\right)}{2x-6}\)
ĐKXĐ : x # -6 ; x # 6 ; x # 0 ; x # 3 . Khi đó , ta có :
P = \(\left[\dfrac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right]\).\(\dfrac{x\left(x+6\right)}{2x-6}\)
P = \(\dfrac{x^2-x^2+12x-36}{x-6}.\dfrac{1}{2x-6}\)
P = \(\dfrac{6\left(2x-6\right)}{x-6}.\dfrac{1}{2x-6}=\dfrac{6}{x-6}\)
b) Tương tự
a, Rút gọn Biểu thức:
A=\(\left(\dfrac{x+2}{2x-4}-\dfrac{x-2}{2x+4}\right):\dfrac{2x}{x2+2x}\)
= \(\left(\dfrac{x+2}{2x-4}+\dfrac{-x-2}{2x+4}\right):\dfrac{2x}{x2+2x}\)
= \(\left(\dfrac{x+2+-x-2}{2x-4+2x+4}\right):\dfrac{2x}{x2+2x}\)
= 0 \(:\dfrac{2x}{x2+2x}\)
b, \(\left(\dfrac{x+2}{2x-4}-\dfrac{x-2}{2x+4}\right):\dfrac{2x}{x2+2x}\)
Thay tất cả x= -4
=> \(\left(\dfrac{-4+2}{2-4-4}-\dfrac{-4-2}{2-4+4}\right):\dfrac{2.-4}{-4.2+2.-4}\)
= -16 : \(\dfrac{1}{3}\)
= -18
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..
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)
\(A=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3-3x}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2+x-3-3x+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{1}{x^3+1}\)
a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
\(Câu\text{ }1:\)
\(\text{ a) }A=\dfrac{4}{x^2+2}+\dfrac{3}{2-x^2}-\dfrac{12}{4-x^4}\\ A=\dfrac{4\left(2-x^2\right)}{\left(x^2+2\right)\left(2-x^2\right)}+\dfrac{3\left(2+x^2\right)}{\left(2-x^2\right)\left(2+x^2\right)}-\dfrac{12}{\left(2+x^2\right)\left(2-x^2\right)}\\ A=\dfrac{4\left(2-x^2\right)+3\left(2+x^2\right)-12}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{8-4x^2+6+3x^2-12}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-x^2-2}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-\left(x^2+2\right)}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-1}{2-x^2}\)
\(\text{b) }Để\text{ }A=-3\\ thì\Rightarrow\dfrac{-1}{2-x^2}=-3\\ \Leftrightarrow2-x^2=3\\ \Leftrightarrow x^2=-1\\ \Leftrightarrow x\text{ }không\text{ }có\text{ }giá\text{ }trị\left(vì\text{ }x^2\ge0\forall x\right)\\ \text{ }Vậy\text{ }để\text{ }A=-3\text{ }thì\text{ }x\text{ }không\text{ }có\text{ }giá\text{ }trị.\)
\(\text{c) }Ta\text{ }có:\text{ }A=\dfrac{-1}{2-x^2}\\ A=\dfrac{1}{x^2-2}\\ x^2\ge0\forall x\\ \Rightarrow x^2-2\ge-2\forall x\\ \Rightarrow A=\dfrac{1}{x^2-2}\le-\dfrac{1}{2}\\ Dấu\text{ }"="\text{ }xảy\text{ }khi:\\ x^2=0\\ \Leftrightarrow x=0\\\text{ }Vậy\text{ }A_{\left(Max\right)}=-\dfrac{1}{2}\text{ }khi\text{ }x=0\)
\(Câu\text{ }2:\)
\(\text{a) }B=\dfrac{1}{x}+\dfrac{1}{x+5}+\dfrac{x-5}{x\left(x+5\right)}\\ B=\dfrac{x+5}{x\left(x+5\right)}+\dfrac{x}{\left(x+5\right)x}+\dfrac{x-5}{x\left(x+5\right)}\\ B=\dfrac{x+5+x+x-5}{x\left(x+5\right)}\\ B=\dfrac{3x}{x\left(x+5\right)}\\ B=\dfrac{3}{x+5}\left(\text{*}\right)\)
\(\text{b) }Ta\text{ }có:\text{ }\left|x-1\right|=6\\ \Leftrightarrow\left[{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\\ Ta\text{ }lại\text{ }có:\text{ }B=\dfrac{3}{x+5}\\ \RightarrowĐKCĐ:x+5\ne0\\ \Rightarrow x\ne-5\\ \Rightarrow x=7\text{ }thỏa\text{ }mãn\text{ }với\text{ }điều\text{ }kiện\text{ }của\text{ }biến.\\ x=-5\text{ }không\text{ }thỏa\text{ }mãn\text{ }với\text{ }điều\text{ }kiện\text{ }của\text{ }biến.\\ Thay\text{ }x=7\text{ }vào\text{ }\left(\text{*}\right),ta\text{ }được:\text{ }B=\dfrac{3}{7+5}=\dfrac{3}{12}=\dfrac{1}{4}\\ \text{ }Vậy\text{ }với\text{ }x=7\text{ }thì\text{ }B=\dfrac{1}{4}\\ với\text{ }x=-5\text{ }thì\text{ }B\text{ }không\text{ }có\text{ }giá\text{ }trị.\)
\(\text{c) }Ta\text{ }có:B=\dfrac{3}{x+5}\\ \RightarrowĐể\text{ }B\in Z\\ thì\Rightarrow3⋮x+5\\ \Rightarrow x+5\inƯ_{\left(3\right)}\\ Mà\text{ }Ư_{\left(3\right)}=\left\{\pm1;\pm3\right\}\\ Ta\text{ }lập\text{ }bảng\text{ }xét\text{ }giá\text{ }trị:\)
| \(x+5\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
| \(x\) | \(-8\) | \(-6\) | \(-4\) | \(-2\) |
\(\Rightarrow x\in\left\{-8;-6;-4;-2\right\}\\ Vậy\text{ }để\text{ }B\in Z\\ thì x\in\left\{-8;-6;-4;-2\right\}\)
Câu 3:
\(\Leftrightarrow3x^3-2x^2+6x^2-4x+9x-6>0\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2+2x+3\right)>0\)
=>3x-2>0
=>x>2/3
Câu 1:
a: \(A=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{x+1+2x-2}{\left(x^2-1\right)}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{3x-1}{x^2-1}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{3x^2-x-3x^2+3}{x\left(x^2-1\right)}\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{-\left(x-3\right)}{x\left(x+2\right)}\)
\(=x-2+\dfrac{6x-3-x^2+3x}{x\left(x+2\right)}\)
\(=x-2+\dfrac{-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x\left(x^2-4\right)-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-4x-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-x^2+5x-3}{x\left(x+2\right)}\)
b: TH1: \(\left\{{}\begin{matrix}x^3-x^2+5x-3>0\\x\left(x+2\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2< x< 2\\x>0.63\end{matrix}\right.\Leftrightarrow0.63< x< 2\)
TH2: \(\left\{{}\begin{matrix}x^3-x^2+5x-3< 0\\x\left(x+2\right)>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0.63\\\left[{}\begin{matrix}x>0\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x< 0.63\\x< -2\end{matrix}\right.\)
\(=\dfrac{-3\left(x-2\right)-2\left(x+2\right)+4x}{x^2-4}\)
\(=\dfrac{-3x+6-2x-4+4x}{x^2-4}\)
\(=\dfrac{-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=-\dfrac{1}{x+2}\left(x\ne2;x\ne-2\right)\)
\(\dfrac{-3}{x+2}-\dfrac{2}{x-2}+\dfrac{4x}{x^2-4}\left(x\ne\pm2\right)\)
\(=\dfrac{-3\left(x-2\right)-2\left(x+2\right)+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-3x+6-2x-4+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=-\dfrac{1}{x+2}\)