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14: \(=\dfrac{4x+7+1}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\)
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
1. Có sẵn kết quả kìa:))
2.\(B=\dfrac{2x-1}{x+1}-\dfrac{x+1}{x-1}-\dfrac{6}{\left(x-1\right)\left(x+1\right)}\)
\(B=\dfrac{\left(2x-1\right)\left(x-1\right)-\left(x+1\right)\left(x+1\right)-6}{\left(x-1\right)\left(x+1\right)}\)
\(B=\dfrac{2x^2-2x-x+1-x^2-2x-1-6}{\left(x-1\right)\left(x+1\right)}\)
\(B=\dfrac{x^2-5x-6}{\left(x-1\right)\left(x+1\right)}\)
\(B=\dfrac{\left(x-6\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(B=\dfrac{x-6}{x-1}\left(đpcm\right)\)
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
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)
\(4x^4+1\)
\(=4x^4+4x^2+1-4x^2\)
\(=\left(2x^2+1\right)^2-\left(2x\right)^2\)
\(=\left(2x^2+1+2x\right)\left(2x^2+1-2x\right)\)
\(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2\)
\(=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)
a: \(\dfrac{2}{x+5}=\dfrac{2\cdot4\cdot\left(x-5\right)}{4\left(x-5\right)\left(x+5\right)}=\dfrac{8\left(x-5\right)}{4\left(x-5\right)\left(x+5\right)}\)
\(\dfrac{-3}{4x-20}=\dfrac{-3}{4\left(x-5\right)}=\dfrac{-3\left(x+5\right)}{4\left(x-5\right)\left(x+5\right)}=\dfrac{-3x-15}{4\left(x-5\right)\left(x+5\right)}\)
\(\dfrac{-x+2}{x^2-25}=\dfrac{-x+2}{\left(x-5\right)\left(x+5\right)}=\dfrac{4\left(-x+2\right)}{4\left(x-5\right)\left(x+5\right)}=\dfrac{-4x+8}{4\left(x-5\right)\left(x+5\right)}\)
b: \(\dfrac{1}{3x-6y}=\dfrac{1}{3\left(x-2y\right)}=\dfrac{\left(x-2y\right)\left(x+2y\right)}{3\left(x-2y\right)^2\cdot\left(x+2y\right)}\)
\(\dfrac{-x}{x^2-4y^2}=\dfrac{-x}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{-x\cdot3\cdot\left(x-2y\right)}{3\left(x-2y\right)^2\cdot\left(x+2y\right)}\)
\(\dfrac{-2y^2}{x^2-4xy+4y^2}=\dfrac{-2y^2}{\left(x-2y\right)^2}=\dfrac{-2y^2\cdot3\left(x+2y\right)}{3\left(x+2y\right)\left(x-2y\right)^2}\)
\(=\dfrac{-6y^2\left(x+2y\right)}{3\left(x+2y\right)\left(x-2y\right)^2}\)
\(\dfrac{x-1}{x+2}+\dfrac{6x}{x^2-4}=\dfrac{x+1}{2-x}\left(dkxd:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{x-1}{x+2}+\dfrac{6x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{x+1}{x-2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)+6x+\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x^2-2x-x+2+6x+x^2+2x+x+2=0\)
\(\Leftrightarrow2x^2+6x+4=0\)
\(\Leftrightarrow2x^2+2x+4x+4=0\)
\(\Leftrightarrow2x\left(x+1\right)+4\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-1\right\}\)