a) 21.7²-11.7²+90.7²+49.125.16

=21.7²-11.7²+90.7²+7².2000

=7².(21-11+90+2000)

=7².2100

=49.2100

=102900

a) \(21\cdot7^2-11\cdot7^2+90\cdot7^2+7^2\cdot125\cdot16=7^2\cdot\left(21-11+90+125\cdot16\right)=7^2\cdot2100=102900\)

\(21.7^2-11.7^2+90.7^2+49.125.16\)

=\(21.7^2-11.7^2+90.7^2+7^2.2000\)

=\(7^2.\left(21-11+90+2000\right)\)

=\(7^2.2100\)

=\(102900\)

\(a,12\left(x-1\right)=0\\ x-1=0\\ x=1\\ b,45+5\left(x-3\right)=70\\ 5\left(x-3\right)=25\\ x-3=5\\ x=8\\ c,3.x-18:2=12\\ 3.x-9=12\\ 3.x=21\\ x=7\)

12(x-1)=0

     (x-1)=0:12

      x-1=0

      x=0+1

      x= 1

Vậy x= 1

12 ( x - 1 ) = 0
       x - 1   = 0 : 12
       x - 1   = 0
=>   x        = 1

Bài 1: 

\(=\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}+\dfrac{1}{\left(x+13\right)\left(x+16\right)}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}+\dfrac{3}{\left(x+10\right)\left(x+13\right)}+\dfrac{3}{\left(x+13\right)\cdot\left(x+16\right)}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+13}+\dfrac{1}{x+13}-\dfrac{1}{x+16}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{x+1}-\dfrac{1}{x+16}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{x+16-x-1}{\left(x+1\right)\left(x+16\right)}=\dfrac{5}{\left(x+1\right)\left(x+16\right)}\)

Bài 2: 

\(\Leftrightarrow a^2-2a+1+b^2+4b+4+4c^2-4c+1=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b+4\right)^2+\left(2c-1\right)^2=0\)

Dấu '=' xảy ra khi a=1; b=-4; c=1/2

1) \(2x^2+5x-3=0\)

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

\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{2}\end{cases}}}\)

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

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

\(\Leftrightarrow\left(x-\frac{1}{2}\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)

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

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

\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)

\(\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)

(5x-1)3-2=70

<=>(5x-1).3=70+2=72

<=>5x-1=72:3

<=>5x-1=24

<=>5x=24+1=25

<=>5x=25:5=5

de vay sao ban k lam

2x-1=5x

2x-1= -5x

tu lam tiep

l2x-1l=5x

TH1: 2x-1=5x <=> 2x-5x=1<=>-3x=1<=>x=-1/3

TH2:2x-1=-5x<=>2x+5x=1<=>7x=1<=>x=1/7

2. -x² + x - 33 = -x² + x - 1/4 - 131/4 = -( x² - x + 1/4 ) - 131/4 = -( x - 1/2 )² - 131/4

-( x - 1/2 )² ≤ 0 ∀ x => -( x - 1/2 )² - 131/4 ≤ -131/4 < 0 ∀ x ( đpcm )

3. x² + 4x + 33 = x² + 4x + 4 + 29 = ( x + 2 )² + 29

( x + 2 )² ≥ 0 ∀ x => ( x + 2 )² + 29 ≥ 29 > 0 ∀ x ( đpcm )

4. x² + 8x = x² + 8x + 16 - 16 = ( x + 4 )² - 16

( x + 4 )²  ≥ 0 ∀ x =>  ( x + 4 )² - 16 ≥ -16 ∀ x

Đẳng thức xảy ra <=> x + 4 = 0 => x = -4

Vậy GTNN của biểu thức = -16, đạt được khi x = -4