a. y⁴ - 14y² + 49

Gọi y² là t, ta có:

t² - 14t + 49

<=> t² - 14t + 7²

<=> (t - 7)²

Thay x² = t

<=> (x² - 7)²

b. \(\dfrac{1}{4}-x^2\)

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

\(\Leftrightarrow\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)

c. x⁴ - 16

<=> (x²)² - 4²

<=> (x² - 4)(x² + 4)

d. x² - 9

<=> x² - 3²

<=> (x - 3)(x + 3)

Bài 1: 

a: \(y^2-14y^2+49=\left(y-7\right)^2\)

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

c: \(x^4-16=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)

d: \(x^2-9=\left(x-3\right)\left(x+3\right)\)

a) Ta có: \(3x-1=0\)

\(\Leftrightarrow3x=1\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

Vậy: \(S=\left\{\dfrac{1}{3}\right\}\)

b) Ta có: \(5x-2=x+4\)

\(\Leftrightarrow5x-x=4+2\)

\(\Leftrightarrow4x=6\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)

a ) 4x⁴ + 1 

= 4x⁴ + 2.2x².1 + 1 - 2.2x².1

= ( 2x² + 1 )² - 4x²

= ( 2x² - 2x + 1 ) ( 2x² + 2x + 1)

b) 4x⁸ + 1

= 4x⁸ + 2. 2x⁴.1 + 1 - 2.2x⁴.1

= ( 2x⁴ + 1 )² - 4x⁴

= ( 2x⁴ + 2x² + 1 ) ( 2x⁴ - 2x² + 1 )

c ) x⁴ + 64

= x⁴ + 2.x². 8 + 64 - 2.x².8

= ( x² + 8 )² - 16x²

= ( x² + 4x + 8 ) ( x² - 4x + 8 )

a) 4x⁴ + 1 = ( 2x )² + 1² = ( 2x + 1 )² = 4x² + 2x + 1

b) 4x⁸ + 1 = ( 2x⁴ )² + 1² = ( 2x⁴ + 1 )² = 4x⁸ + 2x⁴ + 1

c) x⁴ + 64 = ( x² )² + 8² = ( x² + 8 )² = x⁴ + 8x² + 64

Hk tốt

Cái này chưa học bt làm mấy câu

b. x^2 + 2x - 3

= x^2 + 3x - x - 3

= x ( x - 1 ) + 3 ( x - 1 )

= ( x + 3 ) ( x - 1 )

\(4x^2-3x-4\)

\(=\left(2x\right)^2-2.2x.\frac{3}{4}+\frac{9}{16}-\frac{73}{16}\)

\(=\left(2x-\frac{3}{4}\right)^2-\frac{73}{16}\)

\(=\left(2x-\frac{3}{4}\right)^2-\left(\frac{\sqrt{73}}{4}\right)^2\)

\(=\left(2x-\frac{3}{4}-\frac{\sqrt{73}}{4}\right)\left(2x-\frac{3}{4}+\frac{\sqrt{73}}{4}\right)\)

\(=\left(2x-\frac{3+\sqrt{73}}{4}\right)\left(2x+\frac{-3+\sqrt{73}}{4}\right)\)

\(x^2+2x-3\)

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

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

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

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\) \(\left(1\right)\)

đặt \(x^2+5x+5=t\)

\(\left(1\right)\)\(=\) \(\left(t-1\right)\left(t+1\right)-24\)

            \(=t^2-1-24\)

            \(=t^2-25\)

            \(=\left(t-5\right)\left(t+5\right)\)

hay \(\left(1\right)=\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)

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

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

học tốt

d) 

Hướng dẫn :

(x+1)(x+2)(x+3)(x+4)-24

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

P/s: có gì vào trang web mathway.com viết đa thức vào rồi nhấn " factor " là ra nhân tử nhé

B

Chọn B

Giúp tui với mấy bạn ơi

a: =>x-2=0 hoặc x+3=0

=>x=2 hoặc x=-3

b:=>x-7=0 hoặc x+2=0

=>x=7 hoặc x=-2

c: =>4x+2=0 hoặc 3x-4=0

=>x=4/3 hoặc x=-1/2

d: =>2x+1=0 hoặc x-3=0

=>x=3 hoặc x=-1/2

a)

`(x-2)(x+3)=0`

`<=> x-2=0` hoặc `x+3=0`

`<=>x=2` hoặc `x=-3`

b)

`(x-7)(2+x)=0`

`<=>x-7=0` hoặc `2+x=0`

`<=>x=7` hoặc `x=-2`

c)

`(4x+2)(3x-4)=0`

`<=>4x+2=0` hoặc `3x-4=0`

`<=>x=-1/2` hoặc `x=4/3`

d)

`(2x+1)(x-3)=0`

`<=>2x+1=0` hoặc `x-3=0`

`<=>x=-1/2` hoặc `x=3`

e)

`(0,1x-3)(x+0,5)=0`

`<=>0,1x-3=0` hoặc `x+0,5=0`

`<=>x=30` hoặc `x=-0,5`

f)

`(0,2x-0,4)(0,1x+0,7)=0`

`<=>0,2x-0,4=0` hoặc `0,1x+0,7=0`

`<=>x=2` hoặc `x=-7`

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

x⁴ + 64

= x⁴ + 16x² + 64 - 16x²

= (x² + 8)² - (4x)²

= (x² - 4x + 8)(x² + 4x + 8)

4x⁴ + 1

= 4x⁴ + 4x² + 1 - 4x²

= (2x² + 1) - (2x)²

= (2x² - 2x + 1)(2x² + 2x + 1)

64x⁴ + 1

= 64x⁴ + 16x² + 1 - 16x²

= (8x² + 1)² - (4x)²

= (8x² - 4x + 1)(8x² + 4x + 1)