Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2+2x+1=x^2+2\cdot1x+1^2=\left(x+1\right)^2\)
\(4x^2+12x+9=\left(2x\right)^2+2\cdot3\cdot2x+3^2=\left(2x+3\right)^2\)
\(\dfrac{4}{9}a^2-\dfrac{4}{3}a+1=\left(\dfrac{2}{3}a\right)^2-2\cdot\dfrac{2}{3}\cdot1a+1^2=\left(\dfrac{2}{3}a-1\right)^2\)
\(a^2+5a+\dfrac{25}{4}=a^2+2\cdot2,5a+2,5^2=\left(2,5+a\right)^2\)
Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .
1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)
\(=>A=-12x+16\)
2) \(=>B=8x^3+27-8x^3+2=29\)
3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)
4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)
5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)
\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)
\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)
6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)
k cho mik nha ,
Bài 3:
4.
$(x-3)(2x^2-x-4)=x(2x^2-x-4)-3(2x^2-x-4)$
$=(2x^3-x^2-4x)-(6x^2-3x-12)$
$=2x^3-7x^2-x+12$
5.
$(2x^2-x+3)(1-2x+2x^2)=2x^2(2x^2-2x+1)-x(2x^2-2x+1)+3(2x^2-2x+1)$
$=4x^4-6x^3+10x^2-7x+3$
6.
$(x-2)(2x-5)=2x^2-9x+10$
7.
$(x^2-x)(2x-3x^2)=x^2(x-1)(2-3x)$
$=x^2(5x-3x^2-2)=5x^3-3x^4-2x^2$
Bài 3:
8.
$(x+2)(x^2+3x)(4-x)$
$=(x+2)(4-x)(x^2+3x)=(2x-x^2+8)(x^2+3x)$
$=-x^4-x^3+14x^2+24x$
9. Giống câu 8
10.
$(3x+5)(-x^2+4x-2)=-(3x+5)(x^2-4x+2)$
$=-[3x(x^2-4x+2)+5(x^2-4x+2)]$
$=-(3x^3-7x^2-14x+10)$
Bài 13:
a: Ta có: \(AE=EB=\dfrac{AB}{2}\)
\(AD=DC=\dfrac{AC}{2}\)
mà AB=AC
nên AE=EB=AD=DC
Xét ΔAED có AE=AD
nên ΔADE cân tại A
b: Xét ΔABD và ΔACE có
AB=AC
\(\widehat{BAD}\) chung
AD=AE
Do đó: ΔABD=ΔACE
c: Xét ΔABC có
\(\dfrac{AE}{EB}=\dfrac{AD}{DC}\left(=1\right)\)
Do đó: DE//BC
Xét tứ giác BEDC có DE//BC
nên BEDC là hình thang
mà BD=CE
nên BEDC là hình thang cân
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)-4x^2\\ P=\left(x-y-x-y\right)^2-4x^2\\ P=4y^2-4x^2=4\left(y-x\right)\left(x+y\right)\)
Bài 1:
a) (2x+5)(x-6)=2x2+5x-12x-30=2x2-7x-30
b) (2x-1)(x2-4x+3)=2x3-8x2+6x-x2+4x-3=2x3-9x2+10x-3
c) x2-2x-(x-7)(x+2)=x2-2x-x2+7x-2x+14=3x+14
d) 3x-(x+2)(x+4)=3x-x2-2x-4x-8=-x2-3x-8
Bài 2:
a) 2(x+1)=x-1
⇒2x+2=x-1
⇒2x+2-x+1=0
⇒x+3=0
⇒x=-3
b) x(x+2)-x2=1
⇒x2+2x-x2=1
⇒2x=1
⇒x=0,5
c) 3x(x-2)=(3x-1)(x-1)-5
⇒3x2-6x=3x2-x-3x+1-5
⇒3x2-6x-3x2+x+3x-1+5=0
⇒-2x+4=0
⇒-2x=-4
⇒x=2
d) 6(x-1)(x-2)-6x(x+3)=2x
⇒6(x2-x-2x+2)-6x2-18x-2x=0
⇒6x2-6x-12x+12-6x2-18x-2x=0
⇒-38x+12=0
⇒-38x=-12
⇒x=\(\dfrac{6}{19}\)
\(\Leftrightarrow\left(x+3\right)^2\cdot\left(x-3\right)^2-6\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[\left(x^2-9\right)\left(x-3\right)-6\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3-3x^2-9x+21\right)=0\)
=>x+3=0
hay x=-3
Bài 5:
1) Ta có: \(2x\left(x+1\right)-2x^2-2x\)
\(=2x^2+2x-2x^2-2x\)
=0
2) Ta có: \(3x\left(x-2\right)-3\left(x^2-2x\right)+4\)
\(=3x^2-6x-3x^2+6x+4\)
=4
3) Ta có: \(\left(x-1\right)\left(x-5\right)-x^2+6x-5\)
\(=x^2-6x+5-x^2+6x-5\)
=0
4) Ta có: \(\left(2x+1\right)\left(x-1\right)-2x^2+x-5\)
\(=2x^2-2x+x-1-2x^2+x-5\)
=-6
5) Ta có: \(\left(3x-2\right)\left(x-1\right)-3x^2+5x-4\)
\(=3x^2-3x-2x+2-3x^2+5x-4\)
=-2
6) Ta có: \(2x\left(x+1\right)-x\left(x+3\right)-x^2+x+5\)
\(=2x^2+2x-x^2-3x-x^2+x+5\)
=5
cảm ơn bạn