b: \(\Leftrightarrow48x^2-12x-20x+5-48x^2+36x=30\)

\(\Leftrightarrow4x=25\)

hay \(x=\dfrac{25}{4}\)

\(a,\Leftrightarrow\dfrac{3x^3+6x^2-3x-5x^2-10x+5}{x^2+2x-1}=10\\ \Leftrightarrow\dfrac{3x\left(x^2+2x-1\right)-5\left(x^2+2x-1\right)}{x^2+2x-1}=10\\ \Leftrightarrow3x-5=10\Leftrightarrow3x=15\Leftrightarrow x=5\\ b,\Leftrightarrow\left(x^4+2x^2-4x^2-8\right):\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-4\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x^2+2\right)=0\Leftrightarrow x=-2\left(x^2+2>0\right)\\ c,\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-4\right)^2}=0\Leftrightarrow\dfrac{x}{x-4}=0\Leftrightarrow x=0\)

b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)

hay x=-2

A = - 3\(x\).(\(x-5\)) + 3(\(x^2\) - 4\(x\)) - 3\(x\) - 10

A = - 3\(x^2\) + 15\(x\) + 3\(x^2\) - 12\(x\) - 3\(x\) - 10

A = (- 3\(x^2\) + 3\(x^2\)) + (15\(x\) - 12\(x\) - 3\(x\)) - 10

A = 0 + (3\(x-3x\)) - 10

A = 0  - 10

A = - 10 

b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

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: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)

\(\Leftrightarrow20x-5+2-6x-6x-30=10\)

\(\Leftrightarrow8x=43\)

hay \(x=\dfrac{43}{8}\)

b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)

\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)

\(\Leftrightarrow6x^2-3x-3=0\)

\(\Leftrightarrow2x^2-x-1=0\)

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

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

c: Ta có: \(4\left(x-1\right)\left(x+5\right)-\left(x+5\right)\left(x+2\right)-3\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow4\left(x^2+4x-5\right)-\left(x^2+7x+10\right)-3\left(x^2+x-2\right)=0\)

\(\Leftrightarrow4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0\)

\(\Leftrightarrow6x=24\)

hay x=4

d: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-5+1=-4\)

hay \(x=\dfrac{2}{7}\)

`A=-x^2+2x+10`

`=-(x^2-2x)+10`

`=-(x-1)^2+11<=11`

Dấu "=" xảy ra khi `x=1`.

`B=4x-2x^2+8`

`=-2(x^2-2x)+8`

`=-2(x^2-2x+1)+10`

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

Dấu "=" xảy ra khi `x=1`

`C=-x^2-x+1`

`=-(x^2+x)+1`

`=-(x^2+x+1/4)+1+1/4`

`=-(x+1/2)^2+5/4<=5/4`

Dấu "=" xảy ra khi `x=-1/2`

`D=-4x^2+6x+3`

`=-(4x^2-6x)+3`

`=-(4x^2-6x+9/4)+21/4`

`=-(2x-3/2)^2+21/4<=21/4`

Dấu "=' xảy ra khi `2x=3/2<=>x=3/4`

\(a,A=-x^2+2x+10=-x^2+2x-1+11=-\left(x^2-2x+1\right)+11\)

\(=11-\left(x-1\right)^2\)

- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)

\(\Rightarrow A=11-\left(x-1\right)^2\le11\)

Vậy MaxA = 11 <=> x = 1 .

\(b,B=-2x^2+4x-2+10=-2\left(x^2-2x+1\right)+10=10-2\left(x-1\right)^2\)

- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)

\(\Rightarrow B=10-2\left(x-1\right)^2\le10\)

Vậy MaxB = 10 <=> x = 1 .

\(c,C=-x^2-\dfrac{1}{2}.2.x-\dfrac{1}{4}+\dfrac{5}{4}=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\)

- Thấy : \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\in R\)

\(\Rightarrow C=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\le\dfrac{5}{4}\)

Vậy MaxC = 5/4 <=> x = -1/2 .

\(d,D=-4x^2+6x+3=-4x^2+2x.2.\dfrac{6}{4}-\dfrac{9}{4}+\dfrac{21}{4}=-\left(4x^2-6x+\dfrac{9}{4}\right)+\dfrac{21}{4}\)

\(=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\)

- Thấy : \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\in R\)

\(\Rightarrow A=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\le\dfrac{21}{4}\)

Vậy MaxD=21/4 <=> x = 3/4 .