a/ \(2x^2-3x+1>0\Rightarrow\left[{}\begin{matrix}x>1\\x< \frac{1}{2}\end{matrix}\right.\)

b/ \(-3x^2+2x+1< 0\Rightarrow-\frac{1}{3}< x< 1\)

c/ \(\frac{x+3}{x-2}\ge0\Rightarrow\left[{}\begin{matrix}x>2\\x\le-3\end{matrix}\right.\)

d/ \(\frac{2x+1}{x+2}\ge1\Leftrightarrow\frac{2x+1}{x+2}-1\ge0\Leftrightarrow\frac{x-1}{x+2}\ge0\Rightarrow\left[{}\begin{matrix}x\ge1\\x< -2\end{matrix}\right.\)

e/ \(\frac{\sqrt{x}+3}{2-\sqrt{x}}\le0\Rightarrow\left\{{}\begin{matrix}x\ge0\\2-\sqrt{x}< 0\end{matrix}\right.\) \(\Rightarrow x>4\)

g/\(\frac{\sqrt{x}-3}{\sqrt{x}-2}\ge0\Rightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x\ge9\\x< 4\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.\)

h/ \(\frac{\sqrt{x}-3}{\sqrt{x}-1}-\frac{1}{3}< 0\Rightarrow\frac{2\left(\sqrt{x}-4\right)}{3\left(\sqrt{x}-1\right)}< 0\Rightarrow1< x< 16\)

a: \(-3x^2\ge0\)

\(\Leftrightarrow x^2< =0\)

=>x=0

b: \(\dfrac{-5}{4x^2}\ge0\)

\(\Leftrightarrow4x^2< 0\)(vô lý)

c: \(\dfrac{4}{x+3}>=0\)

=>x+3>0

hay x>-3

d: \(\dfrac{-5}{2x-1}>=0\)

=>2x-1<0

hay x<1/2

e: \(\dfrac{-2}{x^2+1}>=0\)

=>x²+1<0(vô lý)

f: \(\dfrac{10}{x^2+9}>=0\)

=>x²+9>0(luôn đúng)

sai đề hết??

a: \(-3x^2\ge0\)

\(\Leftrightarrow x^2< =0\)

=>x=0

b: \(\dfrac{-5}{4x^2}\ge0\)

\(\Leftrightarrow4x^2< 0\)(vô lý)

c: \(\dfrac{4}{x+3}>=0\)

=>x+3>0

hay x>-3

d: \(\dfrac{-5}{2x-1}>=0\)

=>2x-1<0

hay x<1/2

e: \(\dfrac{-2}{x^2+1}>=0\)

=>x²+1<0(vô lý)

f: \(\dfrac{10}{x^2+9}>=0\)

=>x²+9>0(luôn đúng)

b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18

4x 2 -4x+1-4x 2+25=18

26-4x=18

4x=8

x=2

a,27x-18=2x-3x^2

<=> 3x^2-2x+27-18x=0

<=> 3x^2-20x+27=0

\(\Delta\)= 20^2-4-12.27

tính \(\Delta\)rồi tìm x1 ,x2

a) \(x^3+x^2+2x-16\ge0\)

\(\Leftrightarrow x^3-2x^2+3x^2-6x+8x-16\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+8\right)\ge0\)

Mà \(x^2+3x+8>x^2+3x+2,25=\left(x+1,5\right)^2\ge0\)

Cho nên \(x-2\ge0\)

\(\Leftrightarrow x\ge2\)

a,x^3-2x^2+3x^2-6x+8x-16>=0

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

x^2+3x+8>0

=> để lớn hơn hoac bang 0 thì x-2 phải>=0

=>x>=2

b,hình như là vô nghiệm ko chắc chắn lắm

\(b, (2x^2 + 3x-1) - 5(2x^2 + 3x + 2) + 24 =0 \)

Đặt \(2x^2 + 3x + 1 = a \)

\(=> (a-2) - 5(a+2) + 24 = 0\)\(\)

\(=> a - 2 - 5a - 10 + 24 = 0\)

\(=> a = 3=> 2x^2 + 3x + 1 = 3\)

\(<=> 2x^2 + 3x - 2 = 0\)

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

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

\(<=> 2x - 1 = 0 hoặc x+2 =0\)

\(<=> x = 1/2 hoặc x = -2\)

~~

\(\dfrac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)

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

\(=\dfrac{\left(x-1\right)^2}{x^2+2}\ge0\forall x\) (đpcm)

Dấu "=" xảy ra khi x = 1

Bn kia giải bài 1 r nên mk giải bài 2 nha!

Sửa lại:\(\dfrac{x^7+x^2+1}{x^8+x+1}\)

\(\dfrac{x^7+x^2+1}{x^8+x+1}=\dfrac{x^7-x+x^2+x+1}{x^8-x^2+x^2+x+1}\)

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

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

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

\(=\dfrac{\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)}{\left(x^2+x+1\right)(x^6-x^5+x^3-x^2+1)}\)

Cả tử và mẫu đều có nhân tử:\(x^2+x+1>1\Rightarrowđpcm\)

Mk viết sai đề câu d

\(\frac{X^2-3x+3}{x^2-2x+1}\)

\(c=\frac{x^2+y^2}{x^2+2xy+y^2}\)

\(c=x^2+y^2:x^2+2xy+y^2\)

\(c=x^2:x^2+y^2+y^2+2xy\)

\(c=x+2y^2+2xy\)

\(d=\frac{x^2-3x+3}{x^2-2x+1}\)

\(d=x^2-3x+3:x^2-2x+1\)

\(d=x^2:x^2-3x-2x+3+1\)

\(d=x-5x+4\)

a)\(x^2+4y^2-2x+4y+2\)

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

\(=\left(x-1\right)^2+\left(2y+1\right)^2\ge0\)(đúng)

b) Sửa đề

\(3y^2+x^2+2xy+2x+6y+3\)

\(=\left(x^2+y^2+2xy\right)+2y^2+2x+6y+3\)

\(=\left(x+y\right)^2+2\left(x+y\right)+1+2y^2+4y+2\)

\(=\left(x+y+1\right)^2+2\left(y+1\right)^2\ge0\) (đúng)

Làm cái này thử đi:

Cho \(x,y\ge0\)giải phương trình.

\(9^x-8^x=19y\)

Giải được thì nói tiếp :3