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KV
31 tháng 10 2023
a) ∆ABC vuông tại A
⇒ BC² = AC² + AB² (Pytago)
= 10² + 5²
= 125
⇒ BC = 55 (cm)
AM là đường trung tuyến ứng với cạnh huyền BC
⇒ AM = BC : 2 = 5√5/2 (cm)
b) ∆ABC vuông tại A
⇒ BC² = AB² + AC² (Pytago)
= 24² + 7²
= 625
⇒ BC = 25 (cm)
AM là đường trung tuyến ứng với cạnh huyền BC
⇒ AM = BC : 2 = 25/2 (cm)
c) ∆ABC vuông tại A
⇒ BC² = AB² + AC² (Pytago)
= 4² + 3²
= 25
⇒ BC = 5 (cm)
AM là đường trung tuyến ứng với cạnh huyền BC
⇒ AM = BC : 2 = 5/2 (cm)
19 tháng 12 2023
a: \(VP=a^3+b^3+c^3-3bac\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=VT\)
b: \(VT=\left(3a+2b-1\right)\left(a+5\right)-2b\left(a-2\right)\)
\(=3a^2+15a+2ab+10b-a-5-2ab+4b\)
\(=3a^2+14a+14b-5\)
\(VP=\left(3a+5\right)\left(a+3\right)+2\left(7b-10\right)\)
\(=3a^2+9a+5a+15+14b-20\)
\(=3a^2+14a+14b-5\)
=>VT=VP
c: \(VT=a\left(b-x\right)+x\left(a+b\right)\)
\(=ab-ax+ax+bx\)
\(=ab+bx=b\left(a+x\right)=VP\)
d: \(VT=a\left(b-c\right)-b\left(a+c\right)+c\left(a-b\right)\)
\(=ab-ac-ab-bc+ca-cb\)
\(=-2bc\)
=VP
4 tháng 1
\(4x^4+1\)
\(=4x^4+4x^2+1-4x^2\)
\(=\left(2x^2+1\right)^2-\left(2x\right)^2\)
\(=\left(2x^2+1+2x\right)\left(2x^2+1-2x\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 1
\(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2\)
\(=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)
14 tháng 11 2023
e: \(\dfrac{x^2+3x+9}{x^3+4x^2+4x}\cdot\dfrac{x^2+2x}{x^3-27x}\)
\(=\dfrac{x^2+3x+9}{x\left(x^2+4x+4\right)}\cdot\dfrac{x\left(x+2\right)}{x\left(x^2-27\right)}\)
\(=\dfrac{x^2+3x+9}{\left(x+2\right)^2}\cdot\dfrac{x+2}{x\left(x^2-27\right)}\)
\(=\dfrac{\left(x^2+3x+9\right)}{\left(x+2\right)\cdot x\left(x^2-27\right)}\)
f: \(\dfrac{2x^2+4xy+2y^2}{5x-5y}\cdot\dfrac{15x-15y}{2x^3+2y^3}\)
\(=\dfrac{2\left(x^2+2xy+y^2\right)}{5\left(x-y\right)}\cdot\dfrac{15\left(x-y\right)}{2\left(x^3+y^3\right)}\)
\(=\dfrac{\left(x+y\right)^2}{1}\cdot\dfrac{3}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)
\(=\dfrac{3\left(x+y\right)}{x^2-xy+y^2}\)
g: \(\dfrac{x^3-4x}{x^2-7x+12}\cdot\dfrac{x-4}{x^2-2x}\)
\(=\dfrac{x\left(x^2-4\right)}{\left(x-3\right)\left(x-4\right)}\cdot\dfrac{x-4}{x\left(x-2\right)}\)
\(=\dfrac{x^2-4}{\left(x-3\right)\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x-2\right)}=\dfrac{x+2}{x-3}\)
BT
2
29 tháng 8 2023
o: x^4+x^3+x^2-1
=x^3(x+1)+(x-1)(x+1)
=(x+1)(x^3+x-1)
q: \(=\left(x^3-y^3\right)+xy\left(x-y\right)\)
=(x-y)(x^2+xy+y^2)+xy(x-y)
=(x-y)(x^2+2xy+y^2)
=(x-y)(x+y)^2
s: =(2xy)^2-(x^2+y^2-1)^2
=(2xy-x^2-y^2+1)(2xy+x^2+y^2-1)
=[1-(x^2-2xy+y^2]+[(x+y)^2-1]
=(1-x+y)(1+x-y)(x+y-1)(x+y+1)
u: =(x^2-y^2)-4(x+y)
=(x+y)(x-y)-4(x+y)
=(x+y)(x-y-4)
x: =(x^3-y^3)-(3x-3y)
=(x-y)(x^2+xy+y^2)-3(x-y)
=(x-y)(x^2+xy+y^2-3)
z: =3(x-y)+(x^2-2xy+y^2)
=3(x-y)+(x-y)^2
=(x-y)(x-y+3)
H9
HT.Phong (9A5)
CTVHS
29 tháng 8 2023
o) \(x^4+x^3+x^2-1\)
\(=\left(x^4+x^3\right)+\left(x^2-1\right)\)
\(=x^3\left(x+1\right)+\left(x+1\right)\left(x-1\right)\)
\(=\left(x+1\right)\left(x^3+x-1\right)\)
q) \(x^3+x^2y-xy^2-y^3\)
\(=\left(x^3+x^2y\right)-\left(xy^2+y^3\right)\)
\(=x^2\left(x+y\right)-y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)\)
\(=\left(x+y\right)^2\left(x-y\right)\)
s) \(4x^2y^2-\left(x^2+y^2-1\right)^2\)
\(=\left(2xy\right)^2-\left(x^2+y^2-1\right)^2\)
\(=\left(2xy-x^2-y^2+1\right)\left(2xy+x^2+y^2-1\right)\)
\(=-\left(x^2-2xy+y^2-1\right)\left(x^2+2xy+y^2-1\right)\)
\(=-\left(x-y-1\right)\left(x-y+1\right)\left(x+y+1\right)\left(x+y-1\right)\)
u) \(x^2-y^2-4x-4y\)
\(=\left(x^2-y^2\right)-\left(4x+4y\right)\)
\(=\left(x+y\right)\left(x-y\right)-4\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-4\right)\)
x) \(x^3-y^3-3x+3y\)
\(=\left(x^3-y^3\right)-\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2-3\right)\)
z) \(3x-3y+x^2-2xy+y^2\)
\(=\left(3x-3y\right)+\left(x^2-2xy+y^2\right)\)
\(=3\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3+x-y\right)\)
5 tháng 8 2023
m: \(=x^m\cdot x^2-x^m=x^m\left(x^2-1\right)=x^m\left(x-1\right)\left(x+1\right)\)
n: \(=5\cdot x^m\cdot x^2+10x^2\)
\(=5x^2\left(x^m+2\right)\)
o: \(=5x\left(x-2y\right)+2\left(x-2y\right)^2\)
\(=\left(x-2y\right)\left(5x+2x-4y\right)\)
=(x-2y)(7x-4y)
p: \(=7x\left(y-4\right)^2+\left(y-4\right)^3\)
\(=\left(y-4\right)^2\cdot\left(7x+y-4\right)\)
q: \(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
\(=\left(4x-8\right)\left(x^2+6-x-7\right)-9\left(4x-8\right)\)
\(=\left(4x-8\right)\left(x^2-x-10\right)\)
\(=4\left(x-2\right)\left(x^2-x-10\right)\)
H9
HT.Phong (9A5)
CTVHS
5 tháng 8 2023
m) \(x^{m+2}-x^m\)
\(=x^m\cdot x^2-x^m\)
\(=x^m\left(x^2-1\right)\)
\(=x^m\left(x^2-1^2\right)\)
\(=x^m\left(x-1\right)\left(x+1\right)\)
n) \(5x^{m+2}+10x^2\)
\(=5x^m\cdot x^2+10x^2\)
\(=5x^2\cdot x^m+10x^2\)
\(=5x^2\left(x^m+2\right)\)
o) \(5x\left(x-2y\right)+2\left(2y-x\right)^2\)
\(=5x\left(x-2y\right)+2\left(x-2y\right)^2\)
\(=\left(x-2y\right)\left[5x+2\left(x-2y\right)\right]\)
\(=\left(x-2y\right)\left(5x+2x-4y\right)\)
\(=\left(x-2y\right)\left(7x-4y\right)\)
p) \(7x\left(y-4\right)^2-\left(4-y\right)^3\)
\(=7x\left(4-y\right)^2-\left(4-y\right)^3\)
\(=\left(4-y\right)^2\left[7x-\left(4-y\right)\right]\)
\(=\left(4-y\right)^2\left(7x-4+y\right)\)
q) \(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
\(=4\left(x-2\right)\left(x^2+6\right)-4\left(x-2\right)\left(x+7\right)-36\left(x-2\right)\)
\(=4\left(x-2\right)\left[\left(x^2+6\right)-\left(x+7\right)-9\right]\)
\(=4\left(x-2\right)\left(x^2+6-x-7-9\right)\)
\(=4\left(x-2\right)\left(x^2-x-10\right)\)
Y
27 tháng 6 2023
\(\dfrac{x-1}{x+2}+\dfrac{6x}{x^2-4}=\dfrac{x+1}{2-x}\left(dkxd:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{x-1}{x+2}+\dfrac{6x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{x+1}{x-2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)+6x+\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x^2-2x-x+2+6x+x^2+2x+x+2=0\)
\(\Leftrightarrow2x^2+6x+4=0\)
\(\Leftrightarrow2x^2+2x+4x+4=0\)
\(\Leftrightarrow2x\left(x+1\right)+4\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-1\right\}\)
29 tháng 4 2022
1. Có sẵn kết quả kìa:))
2.\(B=\dfrac{2x-1}{x+1}-\dfrac{x+1}{x-1}-\dfrac{6}{\left(x-1\right)\left(x+1\right)}\)
\(B=\dfrac{\left(2x-1\right)\left(x-1\right)-\left(x+1\right)\left(x+1\right)-6}{\left(x-1\right)\left(x+1\right)}\)
\(B=\dfrac{2x^2-2x-x+1-x^2-2x-1-6}{\left(x-1\right)\left(x+1\right)}\)
\(B=\dfrac{x^2-5x-6}{\left(x-1\right)\left(x+1\right)}\)
\(B=\dfrac{\left(x-6\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(B=\dfrac{x-6}{x-1}\left(đpcm\right)\)
14:
a: Xét ΔHNM vuông tại H và ΔMNP vuông tại M có
góc N chung
=>ΔHNM đồng dạng với ΔMNP
b: NP=căn 3^2+4^2=5cm
MH=3*4/5=2,4cm
NH=3^2/5=1,8cm
13:
a: 3x+5=x-5
=>2x=-10
=>x=-5
b: (x-2)(2x+5)=0
=>x-2=0 hoặc 2x+5=0
=>x=2 hoặc x=-5/2
c: =>2(5x-2)=3(3x+1)
=>10x-4=9x+3
=>x=7
d: =>(3x+6-x+1)/(x+2)(x-1)=17-3x/(x+2)(x-1)
=>2x+7=17-3x
=>5x=10
=>x=2