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

Giải:

ĐK: \(x\ge-1\)

PT tương đương với: \(\sqrt{x+1}=1-x^2\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-x^2\ge0\\x+1=1-2x^2+x^4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-1\le x^2\le1\\x^4-2x^2-x=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-1\le x\le1\\\left\{{}\begin{matrix}x=0\left(TM\right)\\x^3-2x-1=0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-1\le x\le1\\\left[{}\begin{matrix}x=0\\x=-1\\x^2-x-1=0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left(2x+5\right)\left(x-3\right)-\left(x-3\right)\left(x+3\right)=0\)

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

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

=>x=3 hoặc x=-2

     x²-9=(x-3)(2x-5)
(=) (x-3)(x+3)=(x-3)(2x-5)

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

(=) (x-3)(x+3-2x+5)=0
(=) (x-3)(8-x)=0

(=)x-3=0 hoặc 8-x=0

(=)x=0 hoặc x=8

Vậy S=\(\left\{0;8\right\}\)

=>\(\left(\dfrac{x^2-8}{2008}-1\right)+\left(\dfrac{x^2-7}{2009}-1\right)=\left(\dfrac{x^2-6}{2010}-1\right)+\left(\dfrac{x^2-5}{2011}-1\right)\)

=>x^2-2016=0

=>x^2=2016

=>\(x=\pm\sqrt{2016}\)

\(đk:4x^2-12x+44\ge0\left(luôn-đúng\right)\)

\(x^2+\sqrt{4x^2-12x+44}=3x+4\)

\(\Leftrightarrow x^2-3x-4+2\sqrt{x^2-3x+11}=0\)

\(\Leftrightarrow x^2-3x+11+2\sqrt{x^2-3x+11}-15=0\)

\(đặt:\sqrt{x^2-3x+11}=t\left(t\ge0\right)\)

\(\Rightarrow t^2+2t-15=0\Leftrightarrow\left[{}\begin{matrix}t=3\left(tm\right)\\t=-5\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-3x+11}=3\Leftrightarrow x^2-3x+2=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

ĐKXĐ: ...

\(x^2+\sqrt{4x^2-12x+44}=3x+4\)

\(\Leftrightarrow\sqrt{4x^2-12x+44}=3x+4-x^2\)

\(\Leftrightarrow4x^2-12x+44=\left(3x+4-x^2\right)^2\)

\(\Leftrightarrow4x^2-12x+44=x^4-6x^3+x^2+24x+16\)

\(\Leftrightarrow x^4-6x^3-3x^2+36x-28=0\)

...........

`|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`

lx

lỗi r bn

lx

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

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

=>(x+2)(x-1)=0

=>x=-2 hoặc x=1

b: ĐKXĐ: x<>4; x<>1

PT =>\(\dfrac{x+3+3x-12}{x-4}=\dfrac{6}{1-x}\)

=>(4x-9)(1-x)=6(x-4)

=>4x-4x^2-9+9x=6x-24

=>-4x^2+13x-9-6x+24=0

=>-4x^2+7x+15=0

=>x=3(nhận) hoặc x=-5/4(nhận)

Đặt \(y=x^2-2x+3=\left(x-1\right)^2+2\ge2\), ta có:

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

\(\Leftrightarrow\left(y+3\right)\left(y-2\right)=0\Leftrightarrow\orbr{\begin{cases}y=2\\y=-3\end{cases}\Rightarrow y=2\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1}\)

Vậy \(S=\left\{1\right\}\)