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-2/5x-1/3=3/4
-2/5x=3/4+1/3
-2/5x = 13/12
x=13/12:2/5
x=-65/24
-13/20x+1/2=-2/5
-13/20x=-2/5-1/2
-13/20x=-9/10
x=-9/10:13/20
x=-18/13
4/3-|2x-1/2|=1
|2x-1/2|=1-4/3
|2x-1/2|=-1/3
<=> 2x-1/2=1/3 hoặc 2x-1/2=-1/3
TH1:
2x-1/2=1/3
2x=1/3+1/2
2x=5/6
x=5/6:2
x=5/12
TH2:
2x-1/2=-1/3
2x=-1/3+1/2
2x=1/6
x=1/6:2
x=1/12
\(\left|5x+13\right|=2x-7\)
khi \(x>\frac{7}{2}\), biểu thức có dạng:
\(\orbr{\begin{cases}5x+13=2x-7\\5x+13=7-2x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=-20\\7x=-6\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{20}{3}\\x=-\frac{6}{7}\end{cases}}}\)
Câu a :
\(x^2-2x-3=0\)
\(\Leftrightarrow x^2-x+3x-3=0\)
\(\Leftrightarrow x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\x+3=0\Rightarrow x=-3\end{matrix}\right.\)
Câu b :
\(2x^2+3=-5x\)
\(\Leftrightarrow2x^2+3+5x=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\Rightarrow x=-1\\2x+3=0\Rightarrow x=-\dfrac{3}{2}\end{matrix}\right.\)
Mấy câu sau khó quá ko bt làm :)
a) \(\left|3x-2\right|+5x=4x-10\)
\(\Leftrightarrow\left|3x-2\right|=-x-10\) \(\left(x\le-10\right)\)
\(\Leftrightarrow\left(3x-2\right)^2=\left(-x-10\right)^2\)
\(\Leftrightarrow8x^2-32x-96=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\) ( loại )
Vậy pt vô nghiệm.
b) \(3+\left|2x-5\right|>13\)
\(\Leftrightarrow\left|2x-5\right|>10\)
\(\Leftrightarrow\left(2x-5\right)^2>100\)
\(\Leftrightarrow4x^2-20x-75>0\)
\(\Leftrightarrow\left(2x-15\right)\left(2x+5\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\frac{15}{2}\\x< \frac{-5}{2}\end{matrix}\right.\)
Vậy...
b) Ta có: 3+|2x+5|>13
⇒|2x+5|>10
\(\Rightarrow\left\{{}\begin{matrix}2x+5>3\\2x+5< -3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2x>-2\\2x< -8\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-1\\x< -4\end{matrix}\right.\)
Vậy: -4<x<-1
a: \(=\dfrac{2x^4+x^3-5x^2-3x-3}{x^2-3}\)
\(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}\)
\(=2x^2+x+1\)
b: \(=\dfrac{x^5+x^2+x^3+1}{x^3+1}=x^2+1\)
c: \(=\dfrac{2x^3-x^2-x+6x^2-3x-3+2x+6}{2x^2-x-1}\)
\(=x+3+\dfrac{2x+6}{2x^2-x-1}\)
d: \(=\dfrac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5}{3x^2-2x+1}\)
\(=x^2-2x-5\)
A = 3 x | 1 - 2x | - 5
Ta co : | 1 - 2x | \(\ge\)0 nen 3 x | 1 - 2x | \(\ge\)0
A = 3 x | 1 - 2x | - 5 \(\ge\)- 5
Vậy min A = -5 \(\Leftrightarrow\)x = \(\frac{1}{2}\)
1 bài thôi . còn lại tương tự
bài cuối dùng BĐT : | a | + | b | \(\ge\)| a + b | nhé