d: Ta có: \(D=x^2-x+\dfrac{1}{2}\)

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

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

D= (x^2 -2x1/2 +1/4) +1/4

   =(x-1/2)^2 +1/4

MinD=1/4 khi x=1/2

Ta có:

x² + 7x + 10 = 0

<=> x^2 + 5x + 2x + 10 = 0

<=> x(x + 5) + 2(x + 5) = 0

<=> (x+2)(x+5) = 0

<=> x+2=0 hoặc x+5=0

<=> x= -2 hoặc x= -5

Vậy x = -2; -5.

\(x^2+7x+10=0\)

\(\Leftrightarrow\left(x^2+5x\right)+\left(2x+10\right)=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x+5=0\\x+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\x=-2\end{cases}}}\)

(X+2)(X+5)=0

=>x=-2 hoặc x=-5

x^2+7x+10=0

x(x+7)=-10

=>x>0                 x<0

    x+7<0             x+7<0

Mà x+7>x

=>x<0      =>x<0

    x+7>0      x>-7

=>x thuộc -1;-2;-3;-4;-5;-6

\(x^2+7x+10=0\)

\(x^2+2x+5x+10=0\)

\(x\left(x+2\right)+5\left(x+2\right)=0\)

\(\left(x+2\right)\left(x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-2\\x=-5\end{cases}}\)

Tim x do giup minh voi

|x+1|+|x+2|+....+|x+10|=7x

= | 10x | + (1+2+3....+10)=7x

=> | 10x | + 55 =7x 

=> 55 = -3x

=> x= 55 : (-3)=-18,333..

hình như đề sai 

Vì |x+1| ≥ 0; |x+2|≥0 ;...; |x+10| ≥ 0 => |x+1|+|x+2|+...+|x+10| ≥ 0 => 7x ≥ 0 => x ≥ 0

=>x+1+x+2+...+x+10=7x

=>10x+55 = 7x

=>10x-7x=-55

=>3x=-55

=>x=-55/3

\(C=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x-10\right)\)

Đặt \(x^2-7x=t\),khi đó:

\(C=\left(t+10\right).\left(t-10\right)=t^2-10^2=t^2-100\)

Vì \(t^2\ge0=>t^2-100\ge-100\) (với mọi t)

Dấu "=" xảy ra\(< =>t=0< =>x^2-7x=0< =>x\left(x-7\right)=0< =>\orbr{\begin{cases}x=0\\x=7\end{cases}}\)

Vậy minC=-100 khi x=0 hoặc x=7

\(3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\\ 2x\left(x-1\right)-\left(1+2x\right)=-34\\ \Leftrightarrow2x^2-2x-1-2x=-34\\ \Leftrightarrow2x^2-4x+33=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+30=0\\ \Leftrightarrow2\left(x-1\right)^2+30=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-1\right)^2+30\ge30>0\right]\\ x^2+9x-10=0\\ \Leftrightarrow x^2-x+10x-10=0\\ \Leftrightarrow\left(x-1\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-10\end{matrix}\right.\\ \left(7x-1\right)\left(2+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x-1=0\\2+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

a) Đặt x^2+2x+2=t

\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)

\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)

Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

      B = (x-2)(x-5)(x²-7x-10)

    =(x²-7x+10)(x²-7x-10)

    =(x²-7x)²-10²

     ⁼(x²-7x)²-100

=>GTNN của B là 100 <=>x²-7x=0

             x(x-7)=0

        =>x=0 hoặc x=7

Vậy GTNN của B là 100 khi x=0 hoặc x=7

A=x^2+2x.3/2+3/2^2+11/2

=(x+3/2)^2+11/2>=11/2

hình như linh chi làm sai r