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Câu 1a : tự kết luận nhé
\(2\left(x+3\right)=5x-4\Leftrightarrow2x+6=5x-4\Leftrightarrow-3x=-10\Leftrightarrow x=\frac{10}{3}\)
Câu 1b : \(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow x+3-2x+6=5-2x\Leftrightarrow-x+9=5-2x\Leftrightarrow x=-4\)
c, \(\frac{x+1}{2}\ge\frac{2x-2}{3}\Leftrightarrow\frac{x+1}{2}-\frac{2x-2}{3}\ge0\)
\(\Leftrightarrow\frac{3x+3-4x+8}{6}\ge0\Rightarrow-x+11\ge0\Leftrightarrow x\le11\)vì 6 >= 0
1) 2(x + 3) = 5x - 4
<=> 2x + 6 = 5x - 4
<=> 3x = 10
<=> x = 10/3
Vậy x = 10/3 là nghiệm phương trình
b) ĐKXĐ : \(x\ne\pm3\)
\(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)
=> \(\frac{x+3-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{5-2x}{\left(x-3\right)\left(x+3\right)}\)
=> x + 3 - 2(x - 3) = 5 - 2x
<=> -x + 9 = 5 - 2x
<=> x = -4 (tm)
Vậy x = -4 là nghiệm phương trình
c) \(\frac{x+1}{2}\ge\frac{2x-2}{3}\)
<=> \(6.\frac{x+1}{2}\ge6.\frac{2x-2}{3}\)
<=> 3(x + 1) \(\ge\)2(2x - 2)
<=> 3x + 3 \(\ge\)4x - 4
<=> 7 \(\ge\)x
<=> x \(\le7\)
Vậy x \(\le\)7 là nghiệm của bất phương trình
Biểu diễn
-----------------------|-----------]|-/-/-/-/-/-/>
0 7
\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x\left(1-\dfrac{x-1}{x+1}\right)\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x.\dfrac{x+1-\left(x-1\right)}{x+1}\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x.\dfrac{2}{x+1}\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{6x}{x+1}\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}-\dfrac{6x}{x+1}=0\)
\(\Rightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2-6x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow\dfrac{4x-6x^2+6x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow\dfrac{10x-6x^2}{\left(x-1\right)\left(x+1\right)=0}\)
\(\Rightarrow10x-6x^2=0\)
\(\Rightarrow x-6x^2=0\)
\(\Rightarrow2x\left(5-3x\right)=0\)
\(\Rightarrow x\left(5-3x\right)=0\)
\(\Rightarrow5-3x=0\)
\(\Rightarrow3x=5\)
\(\Rightarrow x=\dfrac{5}{3}\)
a) \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x\left(1-\dfrac{x-1}{x+1}\right)\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}3\left(1-\dfrac{x-1}{x+1}\right),\left(đk:x\ne1;x\ne-1\right)\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3-\dfrac{3\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3-\dfrac{3x-3}{x+1}\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{3x-3}{x+1}=3\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x-1\right)\cdot\left(3x-3\right)}{\left(x-1\right)\left(x+1\right)}=3\)
\(\Leftrightarrow\dfrac{2\cdot2x+3x^2-3x-3x+3}{\left(x-1\right)\left(x+1\right)}=3\)
\(\Leftrightarrow\dfrac{4x+3x^2-3x-3x+3}{\left(x-1\right)\left(x+1\right)}=3\)
\(\Leftrightarrow\dfrac{-2x+3x^2+3}{\left(x-1\right)\left(x+1\right)}=3\)
\(\Leftrightarrow-2x+3x^2+3=3\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow-2x+3x^2+3=3\left(x^2-1\right)\)
\(\Leftrightarrow-2x+3x^2+3=3x^2-3\)
\(\Leftrightarrow-2x+3=-3\)
\(\Leftrightarrow-2x=-3-3\)
\(\Leftrightarrow-2x=-6\)
\(\Rightarrow x=3\left(đk:x\ne1,x\ne-1\right)\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
a) \(x^3+x^2+2x-16\ge0\)
\(\Leftrightarrow x^3-2x^2+3x^2-6x+8x-16\ge0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+8\right)\ge0\)
Mà \(x^2+3x+8>x^2+3x+2,25=\left(x+1,5\right)^2\ge0\)
Cho nên \(x-2\ge0\)
\(\Leftrightarrow x\ge2\)
a,x^3-2x^2+3x^2-6x+8x-16>=0
(x^2+3x+8)(x-2)>=0
x^2+3x+8>0
=> để lớn hơn hoac bang 0 thì x-2 phải>=0
=>x>=2
b,hình như là vô nghiệm ko chắc chắn lắm
ĐK : 2x - 1 \(\ge0\)=> \(x\ge\frac{1}{2}\)
Khi đó |2x - 1| = 2x - 1
<=> \(\orbr{\begin{cases}2x-1=2x-1\\2x-1=-2x+1\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x=0\\4x=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}\forall x\\x=\frac{1}{2}\end{cases}}\Leftrightarrow\forall x\)
Kết hợp điều kiện => \(x\ge\frac{1}{2}\)là giá trị phải tìm
Vậy \(x\ge\frac{1}{2}\)là nghiệm phương trình
=> Chọn B
ĐKXĐ: \(x\ne1\)
- Với \(a=\pm1\) pt vô nghiệm
- Với \(a\ne1\)
\(\Rightarrow1-x=\dfrac{1+a}{1-a}\)
\(\Leftrightarrow x=1-\dfrac{1+a}{1-a}=\dfrac{-2a}{1-a}\)
Vậy: \(a=\pm1\) hệ vô nghiệm
\(a\ne\pm1\) hệ có nghiệm duy nhất \(x=\dfrac{2a}{a-1}\)