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a: =>-4x>16
=>x<-4
c: =>20x-25<=21-3x
=>23x<=46
=>x<=2
d: =>20(2x-5)-30(3x-1)<12(3-x)-15(2x-1)
=>40x-100-90x+30<36-12x-30x+15
=>-50x-70<-42x+51
=>-8x<121
=>x>-121/8
a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)
\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)
\(\Leftrightarrow6x+6+12x-8=x-7\)
\(\Leftrightarrow6x+12x-x=-7-6+8\)
\(\Leftrightarrow17x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy .........................
b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)
\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)
\(\Leftrightarrow2x^2-x^2+x+15-21=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)
Vậy .........................
P/s: các câu còn lại tương tự, bn tự giải nha
a) Đk : \(x\ne0;\ne1\)
\(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}\)
\(\Rightarrow\dfrac{x^2+3x}{x\left(x+1\right)}+\dfrac{x^2-x-2}{x\left(x+1\right)}-\dfrac{2x^2+2x-2}{x\left(x+1\right)}=0\)
\(\Rightarrow\dfrac{x^2+3x+x^2-x-2-2x^2-2x+2}{x\left(x-1\right)}=0\)
\(\Rightarrow\dfrac{0}{x-1}=0\)
=> Phương trình có vô số nghiệm x
b) Đk : \(x\ne2;x\ne3\)
\(\dfrac{2}{x-2}-\dfrac{x}{x+3}=\dfrac{5x}{\left(x-2\right)\left(x+3\right)}-1\)
\(\Rightarrow\dfrac{2x+6}{\left(x-2\right)\left(x+3\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+3\right)}-\dfrac{5x}{\left(x-2\right)\left(x+3\right)}+\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}\)
=0
\(\Rightarrow\dfrac{2x+6-x^2+2x-5x+x^2+x+6}{\left(x-2\right)\left(x+3\right)}=0\)
\(\Rightarrow\dfrac{12}{\left(x-2\right)\left(x+3\right)}=0\)
=> Phương trình vô nghiệm
c)
\(\Leftrightarrow\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+x+1}{x^4+x^2+1}-\dfrac{1-2x}{x^4+x^2+1}=0\)
\(\Rightarrow\dfrac{x^2-x+1-x^2-x-1-1+2x}{x^4+x^2+1}=0\)
\(\Rightarrow\dfrac{-1}{x^4+x^2+1}=0\)
=> PTVN
d) Thôi tự làm đi, câu này dễ :Vvv
e)
\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)\)=40
\(\Rightarrow\left[\left(x+1\right)\left(x+5\right)\right]\cdot\left[\left(x+2\right)\left(x+4\right)\right]=40\)
\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt
\(x^2+6x+7=t\)
Phương trình tương đương
\(\left(t-1\right)\left(t+1\right)=40\)
\(t^2=41\)
\(\)\(t=\pm\sqrt{41}\)
Thay vào tìm x.
Bài 1:
a. A = x^2 - 5x - 1
\(=x^2-5x+\frac{25}{4}-\frac{29}{4}\)
\(=x^2-5x+\left(\frac{5}{2}\right)^2-\frac{29}{4}\)
\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge0-\frac{29}{4}=-\frac{29}{4}\)
Dấu = khi x=5/2
Vậy MinC=-29/4 khi x=5/2
2. Tìm x:
a. ( 2x - 3 )^2 - ( 4x + 1 )( 4x - 1 ) = ( 2x - 1 ).( 3 - 7x )
=>4x2-12x+9+1-16x2=-14x2+13x-3
=>-12x2-12x+10=-14x2+13x-3
=>2x2-25x+13=0
\(\Rightarrow2\left(x-\frac{25}{4}\right)^2-\frac{521}{8}=0\)
\(\Rightarrow\left(x-\frac{25}{4}\right)^2=\frac{521}{16}\)
\(\Rightarrow x-\frac{25}{4}=\pm\sqrt{\frac{521}{16}}\)
\(\Rightarrow x=\frac{25}{4}\pm\frac{\sqrt{521}}{4}\)
c. 4.( x - 3 ) - ( x + 2 ) = 0
=>4x-12-x-2=0
=>3x-14=0
=>3x=14
=>x=14/3
Bài 2:
\(=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right)\cdot\dfrac{x\left(x+5\right)}{2x-5}-\dfrac{x}{x-5}\)
\(=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{2x-5}-\dfrac{x}{x-5}\)
\(=\dfrac{5\left(2x-5\right)}{\left(x-5\right)\left(2x-5\right)}-\dfrac{x}{x-5}\)
\(=\dfrac{5}{x-5}-\dfrac{x}{x-5}=-1\)
a) 3x-7>4x+2
\(\Leftrightarrow3x-4x>2+7\)
\(\Leftrightarrow-x>9\Leftrightarrow x< -9\)
Vậy S={x<9|x\(\in R\)}
b) 2(x-3)<3-5(2x-1)+4x
\(\Leftrightarrow2x-6< 3-10x+5+4x\)
\(\Leftrightarrow2x+10x-4x< 3+5+6\)
\(\Leftrightarrow8x< 14\Leftrightarrow x< \dfrac{7}{4}\)
Vậy S={x<\(\dfrac{7}{4}\)|x\(\in R\)}
c) (x-2)2+x(x-3)<2x(x-3)+1
\(\Leftrightarrow x^2-4x+4+x^2-3x< 2x^2-6x+1\)
\(\Leftrightarrow-x< -3\)
\(\Leftrightarrow x>3\)
Vậy S =\(\left\{x>3|x\in R\right\}\)
d) \(\dfrac{x-1}{3}-x+1>\dfrac{2x-3}{2}\)
\(\Leftrightarrow2x-2-6x+6>6x-9\)
\(\Leftrightarrow-10x>-13\Leftrightarrow x< \dfrac{13}{10}\)
Vậy S=\(\left\{x< \dfrac{13}{10}|x\in R\right\}\)
Biểu diễn tập nghiệm thì bạn tự làm
A : 5
- Chọn câu A