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2:
a: BC=căn 15^2+20^2=25cm
AH=15*20/25=12cm
góc ADH=góc AEH=góc DAE=90 độ
=>ADHE là hình chữ nhật
=>DE=AH=12cm
b: ΔAHB vuông tại H có HD vuông góc AB
nên AD*AB=AH^2
ΔAHC vuông tại H có HE vuông góc AC
nên AE*AC=AH^2
=>AD*AB=AE*AC
c: góc IAC+góc AED
=góc ICA+góc AHD
=góc ACB+góc ABC=90 độ
=>AI vuông góc ED
4:
a: góc BDH=góc BEH=góc DBE=90 độ
=>BDHE là hình chữ nhật
b: BDHE là hình chữ nhật
=>góc BED=góc BHD=góc A
Xét ΔBED và ΔBAC có
góc BED=góc A
góc EBD chung
=>ΔBED đồng dạng với ΔBAC
=>BE/BA=BD/BC
=>BE*BC=BA*BD
c: góc MBC+góc BED
=góc C+góc BHD
=góc C+góc A=90 độ
=>BM vuông góc ED
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gọi dài=x , rộng=2x -->2x^2=S (1)
lại có (x+4)(2x+5)=S+111 -->2x^2+13x+20=S+111 (2)
thay (1) cho (2) -->13x=91 -->x=7 -->dài=7 rộng=3,5
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Phân tích đa thức thành nhân tử à?
1) \(\left(x+y\right)^3-x^3-y^3\)
\(=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-x^2+xy-y^2\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)
\(=3xy\left(x+y\right)\)
2) \(x^3+1-x^2-x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left[x^2-x+1-x\right]\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
( x + y )3 - x3 - y3
= ( x + y )3 - ( x3 + y3 )
= ( x + y )3 - ( x + y )( x2 - xy + y2 )
= ( x + y )[ ( x + y )2 - ( x2 - xy + y2 ) ]
= ( x + y )( x2 + 2xy + y2 - x2 + xy - y2 )
= 3xy( x + y )
x3 + 1 - x2 - x
= ( x3 + 1 ) - ( x2 + x )
= ( x + 1 )( x2 - x + 1 ) - x( x + 1 )
= ( x + 1 )( x2 - x + 1 - x )
= ( x + 1 )( x2 - 2x + 1 )
= ( x + 1 )( x - 1 )2
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a: \(2x^2\left(3xy+x^2-2y^2\right)\)
\(=6x^3y+2x^4-4x^2y^2\)
b: \(\dfrac{1}{3}x^2y^3\left(2x-3y+1\right)\)
\(=\dfrac{2}{3}x^3y^3-x^2y^4+\dfrac{1}{3}x^2y^3\)
h: \(\left(x-1\right)\left(x+1\right)\left(2x-3\right)\)
\(=\left(x^2-1\right)\left(2x-3\right)\)
\(=2x^3-3x^2-2x+3\)
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Ta có:\(x+y=1\)\(\Rightarrow x=1-y\)
Khi đó: \(P=\left(1-y\right)^3+y^3+\left(1-y\right)y\)
\(=1-3y+3y^2-y^3+y^3+y-y^2\)
\(=2y^2-2y+1\)
\(=2\left(y^2-y+\frac{1}{4}\right)-\frac{1}{2}+1\)
\(=2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)
Dấu ''='' xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(5x^2y-10xy^2=5xy\left(x-y\right)\)
b) \(4x\left(2y-z\right)+7y\left(z-2y\right)=\left(4x-7y\right)\left(2y-z\right)\)
c) \(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
d) \(12x^2y-18xy^2-30y^2=6y\left(2x^2-3xy-5y\right)\)
e) \(y\left(a-b\right)-7y^2\left(b-a\right)=\left(y+7y^2\right)\left(a-b\right)\)
f) \(27x^2\left(y-1\right)-9x^3\left(1-y\right)=\left(27x^2+9x^3\right)\left(y-1\right)=9x^2\left(3+x\right)\left(y-1\right)\)
Bài 4:
Ta có: \(\left(n+2\right)^2-\left(n-2\right)^2\)
\(=n^2+4n+4-n^2+4n-4\)
\(=8n⋮8\)
Bài 1:
a: \(x^2-12x+36=\left(x-6\right)^2\)
b: \(4x^2+12x+9=\left(2x+3\right)^2\)
c: \(\dfrac{1}{4}x^2-5xy+25y^2=\left(\dfrac{1}{2}x-5y\right)^2\)
d: \(\left(x-5\right)^2-16=\left(x-5-4\right)\left(x-5+4\right)=\left(x-9\right)\left(x-1\right)\)
e: \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(x+2\right)\left(8-x\right)\)
g: \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)
\(=\left(5x-5\right)\left(9x-3\right)\)
\(=15\left(x-1\right)\left(3x-1\right)\)
f: \(8x^3+\dfrac{1}{27}=\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)\)
g: \(49\left(x-4\right)^2-9\left(x+2\right)^2\)
\(=\left(7x-28-3x-6\right)\left(7x-28+3x+6\right)\)
\(=\left(4x-34\right)\left(10x-24\right)\)
\(=4\left(2x-17\right)\left(5x-12\right)\)