\(25a^2-25x^2+10x-1\)

\(=25a^2-\left(5x-1\right)^2\)

\(=\left(5a-5x+1\right)\left(5a+5x-1\right)\)

\(25a^2-25x^2+10x-1=25a^2-\left(25x^2-10x+1\right)=\left(5a\right)^2-\left(5x-1\right)^2=\left(5a+5x-1\right)\left(5a-5x+1\right)\)

a) \(\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Rightarrow x-\dfrac{1}{2}=0\)

\(\Rightarrow x=\dfrac{1}{2}\)

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

\(\Rightarrow x-2=1\)

\(\Rightarrow x=3\)

c) \(\left(2x-1\right)^3=-8\)

\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Rightarrow2x-1=-2\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=\dfrac{-1}{2}\)

d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

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

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\).

a , \(\left(x-\dfrac{1}{2}\right)^2=0\)

<=> \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

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

c , \(\left(2x-1\right)^3=-8\Rightarrow2x-1=-2\Rightarrow x=\dfrac{-1}{2}\)

d , \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4^2}\)

<=> \(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)

a) \(\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0^2\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\)

\(\Leftrightarrow x=0+\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{2}\left(TM\right)\)

Vậy \(x=\dfrac{1}{2}\) là giá trị cần tìm

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

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

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

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

Vậy \(x\in\left\{3;1\right\}\)

c) \(\left(2x-1\right)^3=-8\)

\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Rightarrow2x-1=-2\)

\(\Rightarrow2x=-2+1\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=\dfrac{-1}{2}\left(TM\right)\)

Vậy \(x=\dfrac{-1}{2}\)

d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\\left(x+\dfrac{1}{2}\right)^2=\dfrac{-1}{4}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{-1}{4};\dfrac{-3}{4}\right\}\) là giá trị cần tìm

Ta có : \(\left(3x-1\right)^2-\left(x+7\right)^2=0\)

\(\left(3x-1+x+7\right)\left(3x-1-x-7\right)=0\)

\(\left(4x+6\right)\left(2x-8\right)=0\)

Nên : 4x + 6 = 0 hoặc 2x - 8 = 0

4x = -6 hoặc 2x = 8

x = \(\frac{-3}{2}\) hoặc x = 4

Vậy x = \(\frac{-3}{2}\) hoặc x = 4

A=-(11+5x+5x+x^2)

A=-((5x+x^2)+(5x+20)-9)

A=-(x(5+x)+5(5+x)-9)

A=-((5+x)^2-9)

A=(5+x)^2+9

Max A= 9 Khi x=-5

3S=1x2x3+2x3x3+...+39x40x3

3S=1x2x(3-0)+2x3x(4-1)+...+39x40x(41-38)

3S=1x2x4-1x2x0+2x3x4-2x3x1+.....+39x40x41-38x39x40

3S=39x40x41

S=39x40x41:3

S=13x40x41

S=21320

S = 1 x 2 + 2 x 3 + 3 x 4 + ... + 39 x 40

S = 2 + 6 + 12 + ... + 1560

Sau đó tự làm nhé !

khong phai the

C=[(x+1)(x-6)][(x-2)(x-3)]

=(x²-5x-6)(x²-5x+6)

=(x²-5x)²-36>=-36

GTNN cua C=-36 tai x²-5x=0=>x(x-5)=0=>x=0 hoac x=5

B=(x-3)²+(x-11)²

  =x²-6x+9+x²-22x+121

  =2x²-28x+130

  =2(x²-14x+65)

  =2(x²-2.7x+7²-7²+65)

  =2[(x-7)²-49+65]

  =2(x-7)²+32

=> vì 2(x-7)² >= 0 

=>2(x-7)²+32 >= 32

=> GTNN của B=32. Khi x=7

A=25x²+3y²-10x+11

  =25x²-10x+1+3y²+10

  =(5x-1)²+3y²+10

 =>vì (5x-1)²+3y² >= 0

 =>(5x-1)²+3y²+10>= 10

 => GTNN của A=10 . Khi x=1/5 và y=0