Câu hỏi của Phương Thảo

Question

1. khai triển HĐT :

(3x+4)^2=

(x-1)^2=

(x-3)^2=

(1/2x-5)^2 =

x^2-1=

x^2-y^2 =

x^2-2=

4x-1/9 =

3. viết cac biểu thức sau dươi dang HĐT

x^2 -2x+1

x^2 -10x +25

x^2+x +1/4

(2x-3) (2x+3)

(2/3...

Dung Nguyễn Thị Xuân · Accepted Answer

Bài 1:

$\left(3x+4\right)^2=9x^2+24x+16$

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

$\left(x-3\right)^2=x^2-6x+9$

$\left(\dfrac{1}{2}x-5\right)^2=\dfrac{1}{4}x^2-5x+25$

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

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

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

$4x-\dfrac{1}{9}=\left(2\sqrt{x}+\dfrac{1}{3}\right)\left(2\sqrt{x}-\dfrac{1}{3}\right)$Câu trả lời: Bài 1:

$\left(3x+4\right)^2=9x^2+24x+16$

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

$\left(x-3\right)^2=x^2-6x+9$

$\left(\dfrac{1}{2}x-5\right)^2=\dfrac{1}{4}x^2-5x+25$

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

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

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

$4x-\dfrac{1}{9}=\left(2\sqrt{x}+\dfrac{1}{3}\right)\left(2\sqrt{x}-\dfrac{1}{3}\right)$

Dung Nguyễn Thị Xuân · Answer

Bài 3:

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

$x^2-10x+25=\left(x-5\right)^2$

$x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2$

$\left(2x-3\right)\left(2x+3\right)=4x^2-9$

$\left(\dfrac{2}{3}x+5\right)\left(\dfrac{2}{3}x-5\right)=\dfrac{4}{9}x^2-25$Câu trả lời: Bài 3:

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

$x^2-10x+25=\left(x-5\right)^2$

$x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2$

$\left(2x-3\right)\left(2x+3\right)=4x^2-9$

$\left(\dfrac{2}{3}x+5\right)\left(\dfrac{2}{3}x-5\right)=\dfrac{4}{9}x^2-25$

Phương Thảo · Answer

help meCâu trả lời: help me

Bài 3: Những hằng đẳng thức đáng nhớ