a,

\(\Leftrightarrow\left(\left(2x^2-4\right)-2\left(x+1\right)^2\right)< 0\)

\(\Leftrightarrow2x^2-4-2\left(x^2+2x+1\right)< 0\)

\(\Leftrightarrow2x^2-4-2x^2-4x-2< 0\)

\(\Leftrightarrow-4x-6< 0\)

\(\Rightarrow x+\dfrac{3}{2}>0\)

\(\Rightarrow x>-\dfrac{3}{2}\)

\(x\in\left\{-\dfrac{3}{2};\infty\right\}\)

b/

\(\Leftrightarrow\left(x-3\right)^2-5+6x< 0\)

\(\Leftrightarrow x^2-6x+9-5+6x< 0\)

\(\Leftrightarrow x^2+4< 0\) ( điều này vô lý vì không có giá trị nào của x khiến x^2+4<0)

từ trên suy ra:

không có giá trị nào của x để pt này đúng .

c, đưa các hệ số vào công thức bậc 2 ( áp dụng ct bậc 2):

có: \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2.a}\)

Với : a=1 ; b=-4 ; c =3

Ta có:

\(x=\dfrac{-1.-4\pm\sqrt{-4^2-4.1.3}}{2.1}\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{4}}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{4+2}{2}\\x_2=\dfrac{4-2}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=3\\x_2=1\end{matrix}\right.\)

tìm khoảng bđt bằng parabol :

 có một dấu bất đẳng thức

\(\Rightarrow1< x< 3\)

a) BPT \(\Leftrightarrow-2\left(x^2-\dfrac{3}{2}x+\dfrac{7}{2}\right)>0\)

            \(\Leftrightarrow-2\left(x^2-2x\cdot\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{47}{16}\right)>0\)

            \(\Leftrightarrow-\dfrac{47}{8}-2\left(x-\dfrac{3}{4}\right)^2>0\)  (Vô lý)

b) Bạn xem lại đề !

1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)

a: ĐKXĐ: x>=3

Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)

=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)

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

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

=>x-3=4

=>x=7(nhận)

b: ĐKXĐ: x>=0

\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)

=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)

=>\(7\sqrt{x}-5< =0\)

=>\(\sqrt{x}< =\dfrac{5}{7}\)

=>0<=x<=25/49

c: ĐKXĐ: x>=5

\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)

=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)

=>\(\dfrac{3}{2}\sqrt{x-5}=3\)

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

=>x-5=4

=>x=9(nhận)

a.

ĐKXĐ: \(x\le\dfrac{2}{3}\)

\(3x^2-7x+2-\left(1-\sqrt{2-3x}\right)=0\)

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

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

Do \(x\le\dfrac{2}{3}\Rightarrow x-2< 0\Rightarrow x-2-\dfrac{1}{1+\sqrt{2-3x}}< 0;\forall x\in TXĐ\)

Nên (1) tương đương:

\(3x-1=0\Leftrightarrow x=\dfrac{1}{3}\)

b.

ĐKXĐ: \(x\ge-\dfrac{1}{2}\)

\(18x^2+6x+3=9x\sqrt{6x+3}\)

Đặt \(\sqrt{6x+3}=y\ge0\) ta được:

\(18x^2+y^2=9xy\)

\(\Leftrightarrow18x^2-9xy+y^2=0\)

\(\Leftrightarrow\left(6x-y\right)\left(3x-y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=3x\\y=6x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{6x+3}=3x\\\sqrt{6x+3}=6x\end{matrix}\right.\) (\(x\ge0\))

\(\Leftrightarrow\left[{}\begin{matrix}6x+3=9x^2\\6x+3=36x^2\end{matrix}\right.\) (\(x\ge0\))

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1+\sqrt{13}}{12}\end{matrix}\right.\)

Bài 2:

a: =>2x^2-4x+1=x^2+x+5

=>x^2-5x-4=0

=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)

b: =>11x^2-14x-12=3x^2+4x-7

=>8x^2-18x-5=0

=>x=5/2 hoặc x=-1/4

a)2.(x+3)-(3+x).(1`+2x)=0\(\Leftrightarrow\)2x+6-3-6x-x-2x\(^2\)=0

\(\Leftrightarrow\)-2x\(^2\)-5x+3=0\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+3=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy PT đã cho có tập nghiệm S=\(\left\{-3;\dfrac{1}{2}\right\}\)

b)x\(^2\)-4x+4=9\(\Leftrightarrow\)x\(^2\)-4x+4-9=0\(\Leftrightarrow\)x\(^2\)-4x-5=0

\(\Leftrightarrow\left\{{}\begin{matrix}5-x=0\\1+x=0\end{matrix}\right.\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

Vậy PT đã cho có tập nghiệm S=\(\left\{-1;5\right\}\)

\(a,\Leftrightarrow\left(x+3\right)\left(2-1-2x\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\-2x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(b,\Leftrightarrow\left(x-2\right)^2=9\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

a) \(2\left(x+3\right)-\left(x+3\right)\left(1+2x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2-1-2x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(1-2x\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

-Vậy \(S=\left\{-3;\dfrac{1}{2}\right\}\)

b) \(x^2-4x+4=9\)

\(\Leftrightarrow\left(x-2\right)^2-9=0\)

\(\Leftrightarrow\left(x-2-3\right)\left(x-2+3\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
-Vậy \(S=\left\{5;-1\right\}\)