a, 4x² - 49 = 0

⇔⇔ (2x)² - 7² = 0

⇔⇔ (2x - 7)(2x + 7) = 0

⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72

b, x² + 36 = 12x

⇔⇔ x² + 36 - 12x = 0

⇔⇔ x² - 2.x.6 + 6² = 0

⇔⇔ (x - 6)² = 0

⇔⇔ x = 6

e, (x - 2)² - 16 = 0

⇔⇔ (x - 2)² - 4² = 0

⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0

⇔⇔ (x - 6)(x + 2) = 0

⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2

f, x² - 5x -14 = 0

⇔⇔ x² + 2x - 7x -14 = 0

⇔⇔ x(x + 2) - 7(x + 2) = 0

⇔⇔ (x + 2)(x - 7) = 0

⇔{x+2=0x−7=0⇔{x=−2x=7

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

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

\(\Leftrightarrow x+1=1-x\)

\(\Leftrightarrow x=0\)

Trả lời:

( x + 1 )³ - ( 1 - x )³ = 0

<=> ( x + 1 )³ = ( 1 - x )³ 

<=> x + 1 = 1 - x

<=> x + x = 1 - 1

<=> 2x = 0

<=> x = 0

Vậy S = { 0 }

Đặt A =  x² + xy + y² + 1 

= \(x^2+2.x.\frac{1}{2}y+\frac{1}{4}y^2+\frac{3}{4}y^2+1=\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1\ge1\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+\frac{1}{2}y=0\\y=0\end{cases}}\Rightarrow x=y=0\)

Vậy Min A = 1 <=> x = y = 0

a) (x + y)² - 2(x + y) + 1 

= (x + y)² - 2.1.(x + y) + 1

= (x + y - 1)²

b) x³ + 1 - x² - x

= (x³ - x²) - (x - 1) 

= x²(x - 1) - (x - 1)

= (x² - 1)(x - 1) = (x - 1)(x + 1)(x - 1) = (x - 1)²(x + 1) 

c) 27x³ - 0,001 

= \(\left(3x\right)^3-\frac{1}{1000}=\left(3x\right)^3-\left(\frac{1}{10}\right)^3=\left(3x-\frac{1}{10}\right)\left(9x^2+\frac{3}{10}x+\frac{1}{100}\right)\)

d) 125x³ - 1 =(5x)³ - 1 = (5x - 1)(25x² + 5x + 1) 

e) (x² + 4)² - 16x² 

= (x² + 4)² - (4x)²

= (x² - 4x + 4)(x² + 4x + 4) 

= (x - 2)²(x + 2)²

= [(x - 2)(x + 2)]²

a.\(\left(x+y\right)^2-2\left(x+y\right)+1\)

\(=\left(x+y\right)^2-2.\left(x+y\right).1+1^2\)

\(=\left[\left(x+y\right)-1\right]^2\)

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

b.\(x^3+1-x^2-x\)

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

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

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

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

còn cái nịt

A B C 9 12 H I D

a, Xét tam giác ABH và tam giác CBA ta có : 

^AHB = ^CAB = 90⁰

^B _ chung 

Vậy tam giác ABH ~ tam giác CBA ( g.g )

A C B H D I

a) ta có:

AH\(\perp\)BC tại H (gt)

=>AHB=90^o

\(\Delta ABC\)vuông tại \(\widehat{A}\)(gt)=>\(\widehat{BAC}\)=90^o (tc)

=> AHB=BAC=90^o

xét \(\Delta ABC\)và \(\Delta HBA\)có:

\(\widehat{AHB}\)=\(\widehat{BAC}\)90^o (cmt)

B là góc chung(gt)

=> \(\Delta ABH\)đồng dạng \(\Delta CBA\)(trg hợp đồng dạng của \(\Delta\)vuông)

Câu 1: 

\(a^3=a^2.a=\left(b^2+c^2\right).a>b^2.b+c^2.c=b^3+c^3\)

Câu 2: 

\(\left|x-3y\right|^{2007}+\left|y+4\right|^{2008}=0\)

\(\Leftrightarrow\hept{\begin{cases}x-3y=0\\y+4=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-12\\y=-4\end{cases}}\)

\(\left(2x-3\right)\left(x+2\right)=3-\left(x-6\right)\left(3x-2\right)\)

\(\Leftrightarrow2x^2+x-6=3-\left(3x^2-20x+12\right)\)

\(\Leftrightarrow2x^2+x-6=-3x^2+20x-9\)

\(\Leftrightarrow5x^2-19x+3=0\Leftrightarrow x=\frac{19\pm\sqrt{301}}{10}\)