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a: \(A=\left(\dfrac{2\left(2x+1\right)}{2\left(2x+4\right)}-\dfrac{x}{3x-6}-\dfrac{2x^3}{3x^3-12x}\right):\dfrac{6x+13x^2}{24x-12x^2}\)
\(=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^3}{3x\left(x^2-4\right)}\right):\dfrac{x\left(13x+6\right)}{x\left(24-12x\right)}\)
\(=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^2}{3\left(x-2\right)\left(x+2\right)}\right):\dfrac{13x+6}{-12\left(x-2\right)}\)
\(=\dfrac{3\left(2x+1\right)\left(x-2\right)-2x\left(x+2\right)-4x^2}{6\left(x+2\right)\left(x-2\right)}\cdot\dfrac{-12\left(x-2\right)}{13x+6}\)
\(=\dfrac{3\left(2x^2-3x-2\right)-2x^2-4x-4x^2}{x-2}\cdot\dfrac{-2}{13x+6}\)
\(=\dfrac{6x^2-9x-6-6x^2-4x}{x-2}\cdot\dfrac{-2}{13x+6}\)
\(=\dfrac{-\left(13x+6\right)\cdot\left(-2\right)}{\left(13x+6\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
b: Để A>0 thì x-2>0
hay x>2
Để A>-1 thì A+1>0
\(\Leftrightarrow\dfrac{2+x-2}{x-2}>0\)
=>x/x-2>0
=>x>2 hoặc x<0
2x3 + 3x2 + 6x + 5 = 02
<=> 2x3 + x2 + 5x + 2x2 + x + 5 = 0
<=> x(2x2 + x + 5) + (2x2 + x + 5) = 0
<=> (2x2 + x + 5)(x + 1) = 0
<=> x + 1 = 0 (vì 2x2 + x + 5 \(\ge\) 4,875 > 0 \(\forall\) x)
<=> x = - 1
Vậy tập nghiệm của pt là \(S=\left\{-1\right\}\)
b) 4x4 + 12x3 + 5x2 - 6x - 15 = 0
<=> 4x4 + 10x3 + 2x3 + 5x2 - 6x - 15 = 0
<=> 2x3(2x + 5) + x2(2x + 5) - 3(2x + 5) = 0
<=> (2x + 5)(2x3 + x2 - 3) = 0
<=> (2x + 5)(2x3 - 2x2 + 3x2 - 3) = 0
<=> (2x + 5)(x - 1)(2x2 + 3x + 3) = 0
<=> (2x + 5)(x - 1)[x2 + (x + 3/2)2 + 3/4]= 0
Mà x2 + (x + 3/2)2 + 3/4 > 0\(\forall x\)
\(\Rightarrow\left[\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-\frac{5}{2}\\x=1\end{matrix}\right.\)
Vậy ...
a) \(ĐKXĐ:\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(=\left[\frac{\left(x+1\right)\left(x+2\right)}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x\left(x+1\right)}{3x\left(x+1\right)}\right]:\frac{2\left(1-2x\right)}{x+1}-\frac{3x+1-x^2}{3x}\)
\(=\frac{\left(x+1\right)\left(x+2\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(=\frac{2-8x^2}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(=\frac{1+2x-3x-1+x^2}{3x}\)
\(=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
b)\(\text{Với }x\ne0,x\ne-1,x\ne\frac{1}{2}\text{ ta có:}\)
\(\text{Để A< 0\Leftrightarrow}\frac{x-1}{3}< 0\Rightarrow x-1< 0\Leftrightarrow x< 1\)
a) ĐKXĐ : \(x\ne0;x\ne\pm2;x\ne3\)
\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)
Đặt \(B=\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\)
\(B=\frac{\left(x+2\right)\left(x+2\right)}{-\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(2-x\right)\left(x-2\right)}{\left(2+x\right)\left(x-2\right)}\)
\(B=\frac{-\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(B=\frac{-\left(x+2\right)^2-4x^2--\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(B=\frac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(B=\frac{-4x^2-8x}{\left(x-2\right)\left(x+2\right)}\)
\(B=\frac{-4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(B=\frac{-4x}{x-2}\)
\(\Rightarrow A=\frac{-4x}{x-2}:\left(\frac{x^2-3x}{2x^2-x^3}\right)\)
\(\Leftrightarrow A=\frac{-4x\cdot x^2\cdot\left(2-x\right)}{\left(x-2\right)\cdot x\cdot\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{4x^2\cdot x\cdot\left(x-2\right)}{\left(x-3\right)\cdot x\cdot\left(x-2\right)}\)
\(\Leftrightarrow A=\frac{4x^2}{x-3}\)
b) \(\left|x-7\right|=4\)
\(\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=3\end{cases}}}\)
Mà ĐKXĐ x khác 3 => x = 11
\(\Leftrightarrow A=\frac{4\cdot11^2}{11-3}=\frac{121}{2}\)
c) \(A=\frac{4x^2}{x-3}\)
Để A dương thì hoặc cả tử và mẫu âm hoặc cả tử và mẫu dương
Dễ thấy \(4x^2\ge0\forall x\)
=> Để A dương thì x - 3 dương
hay x - 3 > 0
<=> x > 3
Vậy x > 3 thì A > 0
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm3\\x\ne0\end{cases}}\)
a) \(B=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}\right):\frac{3x^2}{x+3}\)
\(\Leftrightarrow B=\left(\frac{3-x}{x+3}\cdot\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(\Leftrightarrow B=\frac{\left(3-x\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{3x^2}\)
\(\Leftrightarrow B=-\frac{x+3}{3x^2}\)
b) Khi \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=3\left(ktm\right)\end{cases}}\)
\(\Leftrightarrow x=1\)
\(\Leftrightarrow B=-\frac{1+3}{3.1^2}=-\frac{4}{3.}\)
c) Để B > 0
\(\Leftrightarrow-\frac{x+3}{3x^2}>0\)
\(\Leftrightarrow\frac{x+3}{3x^2}< 0\)
\(\Leftrightarrow x+3< 0\) (Do 3x2 > 0; loại giá trị = 0)
\(\Leftrightarrow x< -3\)
Vậy để \(B>0\Leftrightarrow x< -3\)
a) ta có :x2+2x+2=(x+1)2+1>0,với mọi x
x2+2x+3=(x+1)2+2>0,với mọi x
ĐKXĐ:x\(\in\)R.Đặt x2+2x+2=a (a>0),ta có:\(\dfrac{a-1}{a}+\dfrac{a}{a+1}=\dfrac{7}{6}\)
<=>\(\dfrac{6\left(a-1\right)\left(a+1\right)}{6a\left(a+1\right)}+\dfrac{6a^2}{6a\left(a+1\right)}=\dfrac{7a\left(a+1\right)}{6a\left(a+1\right)}\)
=>6(a2-1)+6a2=7a2+7a<=>6a2-6+6a2=7a2+7a<=>12a2-7a2-7a-6=0
<=>5a2-7a-6=0<=>(a-2)(5a+3)=0<=>a-2=0(vì a>0,nên 5a+3>0)
<=>a=2=>x2+2x+2=2<=>x(x+2)=0<=>\(|^{x=0}_{x+2=0< =>x=-2}\)
Vậy tặp nghiệm của PT là S\(=\left\{0;-2\right\}\)
a) ĐKXĐ: \(x\ne-2;x\ne2\), rút gọn:
\(A=\left[\frac{3\left(x-2\right)-2x\left(x+2\right)+2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right]\div\frac{2x-1}{4\left(x-2\right)}\)
\(A=\frac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}=\frac{4\left(2x^2-x\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4x\left(2x-1\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4}{x+2}\)
b) Ta có: \(\left|x-1\right|=3\Leftrightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(n\right)\\x=-2\left(l\right)\end{cases}}}\)
=> Khi \(x=4\)thì \(A=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3}\)
c) \(A< 2\Leftrightarrow\frac{4}{x+2}< 2\Leftrightarrow4< 2x+4\Leftrightarrow0< 2x\Leftrightarrow x>0\)Vậy \(A< 2,\forall x>0\)
d) \(\left|A\right|=1\Leftrightarrow\left|\frac{4}{x+2}\right|=1\Leftrightarrow\hept{\begin{cases}\frac{4}{x+2}=1\\\frac{4}{x+2}=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\left(l\right)\\x=-6\left(n\right)\end{cases}}}\)Vậy \(\left|A\right|=1\)khi và chỉ khi x = -6
