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Bài 1:
a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)
\(\Leftrightarrow6-8x-10+2x-5=0\)
\(\Leftrightarrow-6x+11=0\)
\(\Leftrightarrow-6x=-11\)
hay \(x=\dfrac{11}{6}\)
b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)
\(\Leftrightarrow6-12x-11+3x-1=0\)
\(\Leftrightarrow-9x-6=0\)
\(\Leftrightarrow-9x=6\)
hay \(x=-\dfrac{2}{3}\)
=>5(4x-1)-2+x<=3(10x-3)
=>20x-5+x-2<=30x-9
=>21x-7<=30x-9
=>-9x<=-2
=>x>=2/9
Câu 1:
a) Ta có: 7x+21=0
\(\Leftrightarrow7x=-21\)
hay x=-3
Vậy: S={-3}
b) Ta có: 3x-2=2x-3
\(\Leftrightarrow3x-2-2x+3=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
Vậy: S={-1}
c) Ta có: 5x-2x-24=0
\(\Leftrightarrow3x=24\)
hay x=8
Vậy: S={8}
Câu 2:
a) Ta có: \(\left(2x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{2};1\right\}\)
b) Ta có: \(\left(2x-3\right)\left(-x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\-x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\-x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=7\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};7\right\}\)
c) Ta có: \(\left(x+3\right)^3-9\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2-9\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+3-3\right)\left(x+3+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-6\end{matrix}\right.\)
Vậy: S={0;-3;-6}
câu 1
a) 5x(x-2)=0 =>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b)(x+5)(2x-7)=0 =>\(\left[{}\begin{matrix}x+5=0\\2x-7=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-5\\x=\dfrac{7}{2}\end{matrix}\right.\)
c) \(\dfrac{5x}{x+2}\)=4 Đk x\(\ne\)-2
=> 5x=4(x+2)
=>5x-4x=8
=>x=8(tmđk)
`|5x| = - 3x + 2`
Nếu `5x>=0<=> x>=0` thì phương trình trên trở thành :
`5x =-3x+2`
`<=> 5x +3x=2`
`<=> 8x=2`
`<=> x= 2/8=1/4` ( thỏa mãn )
Nếu `5x<0<=>x<0` thì phương trình trên trở thành :
`-5x = -3x+2`
`<=>-5x+3x=2`
`<=> 2x=2`
`<=>x=1` ( không thỏa mãn )
Vậy pt đã cho có nghiệm `x=1/4`
__
`6x-2<5x+3`
`<=> 6x-5x<3+2`
`<=>x<5`
Vậy bpt đã cho có tập nghiệm `x<5`
a: 2x-3>3(x-2)
=>2x-3>3x-6
=>-x>-3
hay x<3
b: \(\dfrac{12x+1}{12}< =\dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)
=>12x+1<=36x+4-24x-3
=>12x+1<=12x+1(luôn đúng)
\(\dfrac{x+2}{x-1}=\dfrac{x-1}{x-3}\) (1)
ĐKXĐ: \(x\ne1;x\ne3\)
(1) \(\Leftrightarrow\left(x+2\right)\left(x-3\right)=\left(x-1\right)^2\)
\(\Leftrightarrow x^2-3x+2x-6=x^2-2x+1\)
\(\Leftrightarrow-3x+2x+2x=1+6\)
\(\Leftrightarrow x=7\) (nhận)
Vậy S = {7}
\(\Rightarrow\dfrac{1}{\left(2x-1\right)\left(2x-2\right)}+\dfrac{1}{\left(2x-2\right)\left(2x-3\right)}+\dfrac{1}{\left(2x-3\right)\left(2x-4\right)}+\dfrac{1}{\left(2x-4\right)\left(2x-5\right)}=\dfrac{4}{21}\)
\(\Rightarrow\dfrac{1}{2x-2}-\dfrac{1}{2x-1}+\dfrac{1}{2x-3}-\dfrac{1}{2x-2}+\dfrac{1}{2x-4}-\dfrac{1}{2x-3}+\dfrac{1}{2x-5}-\dfrac{1}{2x-4}=\dfrac{4}{21}\)
\(\Rightarrow\dfrac{1}{2x-5}-\dfrac{1}{2x-1}=\dfrac{4}{21}\)
\(\Rightarrow\dfrac{4}{\left(2x-1\right)\left(2x-5\right)}=\dfrac{4}{21}\)
\(\Rightarrow\left(2x-1\right)\left(2x-5\right)=21\)
\(\Rightarrow4x^2-12x-16=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
f: =>(x-3)(x-1)>=0
=>x>=3 hoặc x<=1
g: =>(x-2)(x^2+3)<0
=>x-2<0
=>x<2
h: =>x+2>=0
=>x>=-2
i: =>(x-1-x+3)/(x-3)>0
=>x-3>0
=>x>3