Ta có: \(\left(\frac23x-\frac12\right)\cdot\left(\frac{x}{4}-\frac18\right)=0\)

=>\(\left[\begin{array}{l}\frac23x-\frac12=0\\ \frac{x}{4}-\frac18=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac23x=\frac12\\ \frac{x}{4}=\frac18\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12:\frac23=\frac12\cdot\frac32=\frac34\\ x=\frac48=\frac12\end{array}\right.\)

Bài 3:

a:

ĐKXĐ: x>=-5

\(x^2-7x=6\sqrt{x+5}-30\)

=>\(x^2-4x-3x+12=6\sqrt{x+5}-18\)

=>\(\left(x-4\right)\left(x-3\right)=6\left(\sqrt{x+5}-3\right)\)

=>\(\left(x-4\right)\left(x-3\right)=6\cdot\frac{x+5-9}{\sqrt{x+5}+3}\)

=>\(\left(x-4\right)\left(x-3-\frac{6}{\sqrt{x+5}+3}\right)=0\)

=>x-4=0

=>x=4(nhận)

Bài 2:

a: ĐKXĐ: x>=0

\(\sqrt{x+4\sqrt{x}+4}=5x+2\)

=>\(\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)

=>\(5x+2=\sqrt{x}+2\)

=>\(5x-\sqrt{x}=0\)

=>\(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{x}=0\\ 5\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ \sqrt{x}=\frac15\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\frac{1}{25}\left(nhận\right)\end{array}\right.\)

b: \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)

=>\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\) =4

=>|x-1|+|x+2|=4(1)

TH1: x<-2

(1) sẽ trở thành: -x-2+1-x=4

=>-2x-1=4

=>-2x=5

=>x=-5/2(nhận)

TH2: -2<=x<1

(1) sẽ trở thành: x+2+1-x=4

=>3=4(vô lý)

TH3: x>=1

(1) sẽ trở thành: x+2+x-1=4

=>2x+1=4

=>2x=3

=>x=3/2(nhận)

c: ĐKXĐ: x>=1

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)

=>\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)

=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)

=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\)

=>\(\left|\sqrt{x-1}-1\right|=2-\sqrt{x-1}-1=1-\sqrt{x-1}\)

=>\(\sqrt{x-1}-1\le0\)

=>\(\sqrt{x-1}\le1\)

=>0<=x-1<=1

=>1<=x<=2

\(a,=\dfrac{x^3-\left(x-1\right)\left(x^2+x+1\right)}{1-x}=\dfrac{x^3-x^3+1}{1-x}=\dfrac{1}{1-x}\\ b,=\dfrac{2x+x^2+3x+2+2-x}{\left(x+2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x+2\right)^2}=1\)

Thank bạn <3

<>?/[;b[]rwel;u];53pjkjnlgkljtreylkeuro;uwqr[i5uiwehhwwejokejoiyufljukneghnmknbfvhdbg.elkgiwr;iewqirluoyeiwhtgo

tớ chịu.

hi hi.

a) Ta có: \(A=\left(\dfrac{3}{2x+4}+\dfrac{x}{2-x}+\dfrac{2x^2+3}{x^2-4}\right):\dfrac{2x-1}{4x-8}\)

\(=\left(\dfrac{3\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}-\dfrac{2x\left(x+2\right)}{2\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right):\dfrac{2x-1}{4x-8}\)

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

\(=\dfrac{2x^2-x}{x+2}\cdot\dfrac{2}{2x-1}\)

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

\(=\dfrac{2x}{x+2}\)

rút gọn à banj

đúng rồi á

`2)x^4+2x^3-x^2-2x+1=0`

`<=>x^4+2x^3+x^2-2x^2-2x+1=0`

`<=>(x^2+x)^2-2(x^2+x)+1=0`

`<=>(x^2+x-1)^2=0`

`<=>x^2+x-1=0`

`\Delta=1+4=5`

`=>x_{1,2}=(-1+-sqrt5)/2`

Vậy `S={(-1+sqrt5)/2,(-1+sqrt5)/2`

`3)x^4-4x^3-9x^2+8x+4=0`

`<=>x^4-x^3-3x^3+3x^2-12x^2+12x-4x+4=0`

`<=>(x-1)(x^3-3x^2-12x-4)=0`

`<=>(x-1)(x^3+2x^2-5x^2-10x-2x-4)=0`

`<=>(x-1)(x+2)(x^2-5x-10)=0`

`+)x=1`

`+)x=-2`

`+)x^2-5x-10=0`

`Delta=25+40=65`

`=>x_{12}=(5+sqrt{65})/2`

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

sao câu 1 hoài v ạ.Còn câu 2,3 nữa á.