Lời giải:

a. $9x^2-16-(3x-4)(2x+5)=0$

$\Leftrightarrow [(3x)^2-4^2]-(3x-4)(2x+5)=0$

$\Leftrightarrow (3x-4)(3x+4)-(3x-4)(2x+5)=0$

$\Leftrightarrow (3x-4)(3x+4-2x-5)=0$

$\Leftrightarrow (3x-4)(x-1)=0$

$\Leftrightarrow 3x-4=0$ hoặc $x-1=0$

$\Leftrightarrow x=\frac{4}{3}$ hoặc $x=1$.

b.

$x^2+4x=12$

$\Leftrightarrow x^2+4x-12=0$

$\Leftrightarrow (x^2-2x)+(6x-12)=0$

$\Leftrightarrow x(x-2)+6(x-2)=0$

$\Leftrightarrow (x-2)(x+6)=0$

$\Leftrightarrow x-2=0$ hoặc $x+6=0$

$\Leftrightarrow x=2$ hoặc $x=-6$

c.

$x^2-2x=35$

$\Leftrightarrow x^2-2x-35=0$

$\Leftrightarrow (x^2+5x)-(7x+35)=0$

$\Leftrightarrow x(x+5)-7(x+5)=0$

$\Leftrightarrow (x+5)(x-7)=0$

$\Leftrightarrow x+5=0$ hoặc $x-7=0$

$\Leftrightarrow x=-5$ hoặc $x=7$

\(A=-x^2+4x=-\left(x^2-4x+4\right)+4=4-\left(x-2\right)^2\)

Ta có: \(\left(x-2\right)^2\ge0\Rightarrow4-\left(x-2\right)^2\le4\)

\(\Rightarrow A_{max}=4\Leftrightarrow x=2\)

\(\left(4x-1\right)^3=\frac{-1}{27}\)

\(\Rightarrow\left(4x-1\right)^3=\left(\frac{-1}{3}\right)^3\)

\(\Rightarrow4x-1=\frac{-1}{3}\)

\(\Rightarrow4x=\frac{-1}{3}+1\)

\(\Rightarrow4x=\frac{2}{3}\)

\(\Rightarrow x=\frac{2}{3}:4\)

\(\Rightarrow x=\frac{1}{6}\)

\(x+x\cdot3:\dfrac{2}{9}+x:\dfrac{2}{7}=252\)

\(\Leftrightarrow x+x\cdot3\cdot\dfrac{9}{2}+x\cdot\dfrac{7}{2}=252\)

\(\Leftrightarrow x\cdot18=252\)

hay x=14

Bài 1.

a) -2x( -3x + 2 ) - ( x + 2 )²

= 6x² - 4x - ( x² + 4x + 4 )

= 6x² - 4x - x² - 4x - 4

= 5x² - 8x - 4

b) ( x + 2 )( x² - 2x + 4 ) - 2( x + 1 )( 1 - x )

= x³ + 8 + 2( x + 1 )( x - 1 )

= x³ + 8 + 2( x² - 1 )

= x³ + 8 + 2x² - 2

= x³ + 2x² + 6

c) ( 2x - 1 )² - 2( 4x² - 1 ) + ( 2x + 1 )²

= 4x² - 4x + 1 - 8x² + 2 + 4x² + 4x + 1

= 4

d) x² - 3x + xy - 3y

= x( x - 3 ) + y( x - 3 )

= ( x - 3 )( x + y )

Bài 2.

a) 4x² - 4xy + y² = ( 2x - y )²

b) 9x³ - 9x²y - 4x + 4y

= 9x²( x - y ) - 4( x - y )

= ( x - y )( 9x² - 4 )

= ( x - y )( 3x - 2 )( 3x + 2 )

c) x³ + 2 + 3( x³ - 2 )

= x³ + 2 + 3x³ - 6

= 4x³ - 4

= 4( x³ - 1 )

= 4( x - 1 )( x² + x + 1 )

Bài 3.

2( x - 2 ) = x² - 4x + 4

⇔ ( x - 2 )² - 2( x - 2 ) = 0

⇔ ( x - 2 )( x - 2 - 2 ) = 0

⇔ ( x - 2 )( x - 4 ) = 0

⇔ x = 2 hoặc x = 4

\(4x^2-28=0\)

\(4x^2=28\)

\(x^2=7\)

\(\)

\(4x^2-28=0\)

\(\Leftrightarrow4\left(x^2-7\right)=0\)

\(\Leftrightarrow x^2-7=0\)

\(\Leftrightarrow x^2=7\)

\(\Leftrightarrow x=\pm\sqrt{7}\)

\(x^2+36=12x\)

\(\Rightarrow x^2+36-12x=0\)

\(\Rightarrow\left(x-6\right)^2=0\)

\(\Rightarrow x-6=0\)

\(\Rightarrow x=6\)

A) \(X+15=4^2\)

=>  \(X=16-15\)

=>  \(X=1\)

b)  100 : ( 35 - X ) = 5

=>   35 - X = 100 : 5

=>  35 - X  = 20

=>   X =  35 -  20

=>  X =  15 

C) \(3X-17=2^2\cdot2^4\)

=> \(3X-17=2^6=64\)

=>  \(3X=64+17\)

=>  \(3X=81\)

=>   \(X=27\)

d) 120  -  20 ( 50 - 4X )  = 0

=>  20 ( 50 - 4X )   =  120

=>   50 - 4X  = 120 : 20

=>   50 - 4X  =  6

=>  4X  =  50 -   6

=>   4X  =  44

=>  X  = 11

a) \(A=x^2-2.10x+100+1\)

\(A=\left(x-10\right)^2+1>=1\)với mọi x

Dấu = xảy ra khi x-10 =0

                           =>x=10

Min A=1 khi x=10

b) Câu b bạn viết sai đề rồi B= -x^2 +4x -3  mới làm dc

a)A= \(\left(x^2-2.x.10+100\right)+1\)

=\(\left(x-10\right)^2+1>=1\)

Dấu "=" xảy ra <=> \(\left(x-10\right)^2=0\)<=> \(x-10=0\)<=>\(x=10\)

Vậy MinA = 1 khi x=10

a) \(A=x^2-20x+101\)

\(\Rightarrow A=x^2-20x+100+1\)

\(\Rightarrow A=\left(x-10\right)^2+1\)

Ta có : \(\left(x-10\right)^2\ge0\forall x\)

\(\Rightarrow A=\left(x-10\right)^2+1\ge1\forall x\)

\(A=1\Leftrightarrow\left(x-10\right)^2=0\)

            \(\Leftrightarrow x-10=0\Leftrightarrow x=10\)

Vậy Min A = 1 <=> x = 10

b) \(B=-x^2+4x+3\)

\(\Rightarrow B=-\left(x^2-4x+4\right)+7\)

\(\Rightarrow B=-\left(x-2\right)^2+7\)

Ta có : \(\left(x-2\right)^2\ge0\forall x\)\(\Rightarrow-\left(x-2\right)^2\le0\forall x\)

\(\Rightarrow B=-\left(x-2\right)^2+7\le7\forall x\)

\(B=7\Leftrightarrow-\left(x-2\right)^2=0\Leftrightarrow x=2\)

Vậy Max B = 7 <=> x = 2