a) \(\left(2x+1\right)^3=27\)

\(\Leftrightarrow2x+1=3\)

\(\Leftrightarrow x=1\)

b) \(\left(2x-1\right)^3=125\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow x=3\)

c) \(\left(x+1\right)^4=\left(2x\right)^4\)

\(\Leftrightarrow x+1=2x\)

\(\Leftrightarrow x=1\)

d) \(\left(2x-1\right)^5=x^5\)

\(\Leftrightarrow2x-1=x\)

\(\Leftrightarrow x=1\)

a. ( 2x + 1 )³ = 27

<=> ( 2x + 1 )³ = 3³

<=> 2x + 1 = 3

<=> 2x = 2

<=> x = 1

b. ( 2x - 1 )³ = 125

<=> ( 2x - 1 )³ = 5³

<=> 2x - 1 = 5

<=> 2x = 6

<=> x = 3

c. ( x + 1 )⁴ = 2x⁴

<=> x + 1 = 2x

<=> x = 1

d. ( 2x - 1 )⁵ = x⁵

<=> 2x - 1 = x

<=> x = 1

a) \(\left(2x+1\right)^3=27\)

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

\(\Rightarrow2x+1=3\)

\(2x=3-1\)

\(2x=2\)

\(x=2\div2\)

\(x=1\)

Vậy \(x=1\).

b) \(\left(2x-1\right)^3=125\)

\(\left(2x-1\right)^3=5^3\)

\(\Rightarrow2x-1=5\)

\(2x=5+1\)

\(2x=6\)

\(x=6\div2\)

\(x=3\)

Vậy \(x=3\).

c) \(\left(x+1\right)^4=\left(2x\right)^4\)

\(\Rightarrow x+1=2x\)

\(x-2x=-1\)

\(-x=-1\)

\(x=1\)

Vậy \(x=1\).

d) \(\left(2x-1\right)^5=x^5\)

\(\Rightarrow2x-1=x\)

\(2x-x=1\)

\(x=1\)

Vậy \(x=1\).

Cho đa thức

P(x)= x mũ 2 + 2x mũ 2 +1 (1)

Thay P(-1) vào đa thức (1) , ta có :

P= \((-1)^2 +2.(-1) ^3\)

P= \(1+ (-2)\)

P= \(-1\)

Thay P(\(\dfrac{1}{2}\)) vào đa thức (1) , ta có :

\(P= (\dfrac{1}{2})^2 +2.(\dfrac{1}{2})^3\)

\(P= \dfrac{1}{4} + \dfrac{1}{4}\)

\(P=\dfrac{1}{2}\)

Q(x)=x mũ 4 +4x mũ 3 +2x mũ 2 trừ 4x+ 1. (2)

Thay Q(-2) vào đa thức (2) , ta có :

Q =\((-2)^4 +4.(-2)^3 +2.(-2)^2-4(-2)+1\)

\(Q = 16-32+8+8+1\)

\(Q= 1\)

Thay Q(1) vào đa thức (2) , ta có:

\(Q= \) \(1^4+4.1^3+2.1^2-4.1+1\)

\(Q= 1+ 4+2-4+1\)

\(Q= 4\)

Tính P(-1) ; P(1/2) ; Q(-2) ; Q(1)

mng giúp em với tối em nộp bài rồi a

cức + điên= lan ngọc cức điên

lớp 8 sao giống lớp 6 bọn em vậy

Đề bài bn ghi thek thì ai làm nổi cho bn :V ?

ko có biết

Giải:

1) \(9x^2-12xy+4y^2-3\)

\(=\left(3x-2y\right)^2-3\)

\(=\left(3x-2y-\sqrt{3}\right)\left(3x-2y+\sqrt{3}\right)\) (Bước này chắc không cần)

2) \(x^2+4x+1\)

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

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

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

(Bước này chắc không cần)

3) \(x^2-4x+7\)

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

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

4) \(x^2+6x+15\)

\(=x^2+6x+9+6\)

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

5) \(x^2-x+\dfrac{1}{3}\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{1}{12}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{12}\)

6) \(\dfrac{1}{4}x^2+x\)

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

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

7) \(3x^2+2x+1\)

\(=x^2+2x+1+2x^2\)

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

8) \(2x^2-2x+1\)

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

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

9) \(10a^2+5b^2+12ab+4a-6b+15\)

\(=4a^2+6a^2+4b^2+b^2+12ab+4a-6b+15\)

\(=\left(6a^2+b^2+12ab\right)+4a+4a^2-6b+4b^2+15\)

\(=\left(6a+b\right)^2+4a\left(1+a\right)-2b\left(3+2b\right)+15\)

Giải:

1) \(9x^2-12xy+4y^2-3\)

\(=\left(9x^2-12xy+4y^2\right)-3\)

\(=\left(3x-2y\right)^2-3\)

2) \(x^2+4x+1\)

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

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

3) \(x^2-4x+7\)

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

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

4) \(x^2+6x+15\)

\(=x^2+6x+9+6\)

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

5) \(x^2-x+\dfrac{1}{3}\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{1}{12}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{12}\)

6) \(\dfrac{1}{4}x^2+x\)

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

7) \(3x^2+2x+1\)

\(=x^2+2x+1+2x^2\)

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

8) \(2x^2-2x+1\)

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

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

9) \(10a^2+5b^2+12ab+4a-6b+15\)

\(=a^2+b^2+9a^2+12ab+4b^2+4a-6b+15\)

\(=9a^2+12ab+4b^2+a^2+4a-6b+b^2+15\)

\(=\left(3a+2b\right)^2+a\left(a+4\right)-b\left(6-b\right)+15\)

Vậy ...