a: \(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)

=>x=0

b: \(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{x+1}{x-3}=\dfrac{1}{2}\)

\(\Leftrightarrow x^2-4x+3=2x\left(x-3\right)-2\left(x^2-1\right)\)

\(\Leftrightarrow x^2-4x+3=2x^2-6x-2x^2+2=-6x+2\)

\(\Leftrightarrow x^2+2x+1=0\)

=>x=-1(nhận)

\(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)

\(\Leftrightarrow x=0\) ( vì \(x^2+2x+5>0;\forall x\)

b.\(\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\)

\(ĐK:x\ne1;3;4\)

\(\Leftrightarrow\dfrac{x}{\left(x-1\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\)

\(\Leftrightarrow\dfrac{x\left(x-3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-1\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)\)

\(\Leftrightarrow x^2-3x-\left(x^2-3x-x+3\right)=x^2-1\)

\(\Leftrightarrow x^2-3x-x^2+4x-3=x^2-1\)

\(\Leftrightarrow x^2+x-2=0\)

\(\Leftrightarrow x^2-x+2x-2=0\)

\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{-2\right\}\)

\(a,2x^3+4x^2+10x=0\\ \Leftrightarrow2x\left(x^2+2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x^2+2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x^2+2x+1\right)+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2+4=0\left(vô..lí\right)\end{matrix}\right.\)

\(b,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne3\\x\ne4\end{matrix}\right.\\ \dfrac{x^2-4x}{x^2-5x+4}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x}{x-1}-\dfrac{1}{2}-\dfrac{x+1}{x-3}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{2\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)}-\dfrac{x^2-4x+3}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2x^2-2}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2-6x-x^2+4x-3-2x^2+2}{2\left(x-1\right)\left(x-3\right)}=0\)

\(\Rightarrow-x^2-2x-1=0\)

\(\Leftrightarrow x^2+2x+1=0\\ \Leftrightarrow\left(x+1\right)^2=0\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\left(tm\right)\)

a: =>5x-5+17x=1-12x-4

=>22x-5=-12x-3

=>34x=2

hay x=1/17

b: =>\(\left(x-3\right)^2-4x\left(x-3\right)=0\)

=>(x-3)(-3x-3)=0

=>x=3 hoặc x=-1

c: =>(x-4)(x-6)=0

=>x=4 hoặc x=6

a: =>2x^2+9x-6x-27=0

=>x(2x+9)-3(2x+9)=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

b: =>-10x^2+6x-5x+3=0

=>-2x(5x-3)-(5x-3)=0

=>(5x-3)(-2x-1)=0

=>x=-1/2 hoặc x=5/3

c: =>-x^3+2x^2-x^2+4=0

=>-x^2(x-2)-(x-2)(x+2)=0

=>(x-2)(-x^2-x-2)=0

=>x-2=0

=>x=2

d: =>(x^3+8)-4x(x+2)=0

=>(x+2)(x^2-2x+4)-4x(x+2)=0

=>(x+2)(x^2-6x+4)=0

=>x=-2 hoặc \(x=3\pm\sqrt{5}\)

a) Xét tam thức \(f\left( x \right) = 7{x^2} - 19x - 6\) có \(\Delta  = 529 > 0\), có hai nghiệm phân biệt \({x_1} =  - \frac{2}{7},{x_2} = 3\) và có \(a = 7 > 0\)

Ta có bảng xét dấu như sau

Vậy nghiệm của bất phương trình là đoạn \(\left[ { - \frac{2}{7};3} \right]\)

b) \( - 6{x^2} + 11x > 10 \Leftrightarrow  - 6{x^2} + 11x - 10 > 0\)

Xét tam thức \(f\left( x \right) =  - 6{x^2} + 11x - 10\) có \(\Delta  =  - 119 < 0\)và có \(a =  - 6 < 0\)

Ta có bảng xét dấu như sau

Vậy bất phương trình vô nghiệm

c) \(3{x^2} - 4x + 7 > {x^2} + 2x + 1 \Leftrightarrow 2{x^2} - 6x + 6 > 0\)

Xét tam thức \(f\left( x \right) = 2{x^2} - 6x + 6\) có \(\Delta  =  - 12 < 0\)và có \(a = 2 > 0\)

Ta có bảng xét dấu như sau

Vậy bất phương trình có vô số nghiệm

d) Xét tam thức \(f\left( x \right) = {x^2} - 10x + 25\) có \(\Delta  = 0\), có nghiệm kép \({x_1} = {x_2} = 5\) và có \(a = 1 > 0\)

Ta có bảng xét dấu như sau

Vậy nghiệm của bất phương trình là \(x = 5\)

a.\(\left(x^2+2x+5\right)\left(x^2+4x\right)=0\)

Ta có: \(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\ge4>0;\forall x\)

 \(\Rightarrow x^2+4x=0\)

\(\Leftrightarrow x\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

b.\(\left(x^2-4x+4\right)\left(x^2-3x\right)=0\)

\(\Leftrightarrow\left(x-2\right)^2x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=3\end{matrix}\right.\)

c.\(1,2x^3-x^2-0,2x=0\)

\(\Leftrightarrow x\left(1,2x^2-x-0,2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-\dfrac{1}{6}\end{matrix}\right.\)

Các câu na ná chắc nên mk làm mẫu 2 bài thui nha !

a, pt <=> x-23/24 + x-23/25 - x-23/26 - x-23/27 = 0

<=> (x-23).(1/24+1/25-1/26-1/27) = 0

<=> x-23=0 ( vì 1/24+1/25-1/26-1/27 > 0 )

<=> x=23

b, pt <=> (201-x/99 + 1)+(203-x/97 + 1)+(205-x/95 + 1) = 0

<=> 300-x/99 + 300-x/97 + 300-x/95 = 0

<=> (300-x).(1/99+1/97+1/95) = 0

<=> 300-x = 0 ( vì 1/99+1/97+1/95 > 0 )

<=> x=300

Tk mk nha

sory mình học lớp 7

a: ĐKXĐ: \(\left\{{}\begin{matrix}x+6>=0\\x-2>=0\end{matrix}\right.\Leftrightarrow x>=2\)

\(\sqrt{x+6}-\sqrt{x-2}=2\)

=>\(\left(\sqrt{x+6}-\sqrt{x-2}\right)^2=4\)

=>\(x+6+x-2-2\sqrt{\left(x+6\right)\left(x-2\right)}=4\)

=>\(2\sqrt{\left(x+6\right)\left(x-2\right)}=2x+4-4=2x\)

=>\(\sqrt{\left(x+6\right)\left(x-2\right)}=x\)

=>\(\left\{{}\begin{matrix}x>=0\\\left(x+6\right)\left(x-2\right)=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=2\\x^2+4x-12=x^2\end{matrix}\right.\)

=>x=3

b: ĐKXĐ: \(x-3>=0\)

=>x>=3

\(2\sqrt{x-3}-2x+3=0\)

=>\(\sqrt{4x-12}=2x-3\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\4x-12=4x^2-12x+9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=3\\4x^2-12x+9-4x+12=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=3\\4x^2-16x+21=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)