a) \(x^2+1\ge1\)

\(\Rightarrow x^2+1< 1\)( Vô lí )

=> BPT vô nghiệm 

b) \(x^2+2x< 2x\)

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

\(\Leftrightarrow x^2< 0\)( vô lí )

Vậy BPT vô nghiệm

c) \(x^2-2x+3< -2x+3\)

\(\Leftrightarrow x^2-2x+3+2x-3< 0\)

\(\Leftrightarrow x^2< 0\)

Vậy,,,,,,,,,,,,,,,,,,,

a, \(x^2+1< 1\)(*)

Ta có : \(x^2\ge0< =>x^2+1\ge1\)

Nên không thể bé hơn 1 

Nên (*) vô lí 

b, \(x^2+2x< 2x\)(**)

Ta có : \(x^2\ge0< =>x^2+2x\ge2x\)

Nên không thể bé hơn 2x

Nên (**) vô lí 

c, \(x^2-2x+3< -2x+3\)

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

\(< =>x^2< 0\)( vô lí )

a) \(-x^2+3x+4>0\)

\(\Leftrightarrow-\left(x^2-3x-4\right)>0\)

\(\Leftrightarrow x^2-3x-4< 0\)

\(\Leftrightarrow x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{25}{4}< 0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2-\frac{25}{4}< 0\)

\(\Leftrightarrow\left(x-\frac{3}{2}-\frac{5}{2}\right)\left(x-\frac{3}{2}+\frac{5}{2}\right)< 0\)

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

\(\Leftrightarrow1< x< 4\)

b) \(x^2-6x+5\ge0\)

\(\Leftrightarrow x^2-2.3x+9-4\ge0\)

\(\Leftrightarrow\left(x-3\right)^2-4\ge0\)

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

\(\Leftrightarrow x\left(x-5\right)\ge0\)

Còn lại tự làm

\(-A=2x^2-0,5x+8\)

=> \(-2A=4x^2-x+16\)

=> \(-2A=\left(2x-\frac{1}{4}\right)^2+\frac{255}{16}\)

Có: \(\left(2x-\frac{1}{4}\right)^2\ge0\)

=> \(-2A\ge\frac{255}{16}\)

=> \(A\le-\frac{255}{16}:2\)

=>  \(A\le-\frac{255}{32}\)

DẤU "=" XẢY RA <=> \(\left(2x-\frac{1}{4}\right)^2=0\)

<=> \(x=\frac{1}{8}\)

sửa đề câu a đi 

\(x^2-6x-7=x^2+x-7\left(x+1\right)=\left(x-7\right)\left(x+1\right)\)

+)\(x^3+2x^2+xy^2-4x\)

\(=x^3+xy^2+2x^2-4x\)

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

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

+) \(x^2-6x-7\)

\(=x^2-6x+9-16\)

\(=\left(x-3\right)^2-16\)

\(=\left(x-3-4\right)\left(x-3+4\right)=\left(x-7\right)\left(x+1\right)\)

Mọi người ơi mk đang cần gấp giúp mk với ạ

\(-B=x^2-2xy+4y^2-2x-10y-5\)

=> \(-B=\left(x-y-1\right)^2+3y^2-12y+12-18\)

=> \(-B=\left(x-y-1\right)^2+3\left(y-2\right)^2-18\)

CÓ: \(\left(x-y-1\right)^2;3\left(y-2\right)^2\ge0\forall x;y\)

=> \(B\ge-18\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x-y-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=3\\y=2\end{cases}}\)

a) 

<=> \(3x-12x^2+12x^2-6x=9\)

<=> \(-3x=9\)

<=> \(x=-3\)

b)

<=> \(6x-24x^2-12x+24x^2=6\)

<=> \(-6x=6\)

<=> \(x=-1\)

c) 

<=> \(6x-4-3x+6=1\)

<=> \(3x+2=1\)

<=> \(x=-\frac{1}{3}\)

d) 

<=> \(9-6x^2+6x^2-3x=9\)

<=> \(-3x=0\)

<=> \(x=0\)

e) KO HIỂU ĐỀ

f) 

<=> \(4x^2-8x+3-\left(4x^2+9x+2\right)=8\)

<=> \(-17x+1=8\)

<=> \(x=-\frac{7}{17}\)

g) 

<=> \(-6x^2+x+1+6x^2-3x=9\)

<=> \(-2x=8\)

<=> \(x=-4\)

h)

<=> \(x^2-x+2x^2+5x-3=4\)

<=> \(3x^2+4x=7\)

<=> \(\orbr{\begin{cases}x=1\\x=-\frac{7}{3}\end{cases}}\)

a. \(3x\left(1-4x\right)+6x\left(2x-1\right)=9\)

\(\Rightarrow3x-12x^2+12x^2-6x=9\)

\(\Rightarrow-3x=9\)

\(\Rightarrow x=-3\)

b. \(3x\left(2-8x\right)-12x\left(1-2x\right)=6\)

\(\Rightarrow6x-24x^2-12x+24x^2=6\)

\(\Rightarrow-6x=6\)

\(\Rightarrow x=-1\)

c. \(2\left(3x-2\right)-3\left(x-2\right)=1\)

\(\Rightarrow6x-4-3x+6=1\)

\(\Rightarrow3x+2=1\)

\(\Rightarrow3x=-1\)

\(\Rightarrow x=-\frac{1}{3}\)

\(=\left(a^2-1\right)^3-\left(a^6-1\right)\)

\(=a^6-3a^4+3a^2-1-a^6+1\)

\(=-3a^4+3a^2\)

\(=-3a^2\left(a^2-1\right)\)

\(\left(a^2-1\right)^3-\left(a^4+a^2+1\right)\left(a^2-1\right)\)

\(=a^6-3a^4+3a^2-1-\left(a^6-a^4+a^4-a^2+a^2-1\right)\)

\(=a^6-3a^4+3a^2-1-a^6+1\)

\(=-3a^4+3a^2\)

\(=-3a^2\left(a^2-1\right)\)

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

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-4\left(a^3+b^3+c^3\right)-12abc\)

\(=-3\left(a^3+b^3+c^3\right)+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-12abc\)

\(=-3\left(\left(a^3+b^3+c^3\right)-\left(a+b\right)\left(b+c\right)\left(c+a\right)+4abc\right)\)

XONG NHAAAAA :3333333