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a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Rightarrow\left(2x+1\right)^2-\left[2\left(x+2\right)\right]^2=9\\ \Rightarrow\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\\ \Rightarrow-3\left(4x+5\right)=9\\ \Rightarrow-12x-15=9\\ \Rightarrow x=-2\)
b) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\\ \Rightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\\ \Rightarrow x^2+6x+9-x^2-4x+32=1\\ \Rightarrow2x=-40\\ \Rightarrow x=-20\)
\(a,\Rightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Rightarrow-12x=24\Rightarrow x=-2\\ b,\Rightarrow x^2+6x+9-x^2-4x+32=1\\ \Rightarrow2x=-40\Rightarrow x=-20\\ c,\Rightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\\ \Rightarrow8x=-40\Rightarrow x=-5\\ d,\Rightarrow x^3-27+4x-x^3=1\\ \Rightarrow4x=28\Rightarrow x=7\\ e,\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-19\\ \Rightarrow12x=-15\Rightarrow x=-\dfrac{5}{4}\)
a: \(x^2-9y^2=\left(x-3y\right)\left(x+3y\right)\)
c: \(\left(x+5\right)^2-16=\left(x+1\right)\left(x+9\right)\)
e: \(\left(2x+3\right)^2-\left(x-7\right)^2\)
\(=\left(2x+3+x-7\right)\left(2x+3-x+7\right)\)
\(=\left(3x-4\right)\left(x+10\right)\)
a) \(=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
b) \(=a^2\left(a-x\right)-y\left(a-x\right)=\left(a-x\right)\left(a^2-y\right)\)
c) \(=3\left(x^2+4x+4\right)=3\left(x+2\right)^2\)
d) \(=2\left(a^2-b^2\right)-5\left(a-b\right)=2\left(a-b\right)\left(a+b\right)-5\left(a-b\right)\)
\(=\left(a-b\right)\left(2a+2b+5\right)\)
e) \(=xy\left(x-y\right)-3\left(x^2-y^2\right)=xy\left(x-y\right)-3\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(xy-3x-3y\right)\)
f) \(=x^2\left(x+5\right)-4\left(x+5\right)=\left(x+5\right)\left(x^2-4\right)\)
\(=\left(x+5\right)\left(x-2\right)\left(x+2\right)\)
\(3x\left(x-y\right)+x-y\)
\(=3x\left(x-y\right)+1\left(x-y\right)\)
\(=\left(x-y\right)\left(3x+1\right)\)
a: Xét tứ giác ADHE có
\(\widehat{ADH}=\widehat{AEH}=\widehat{DAE}=90^0\)
Do đó: ADHE là hình chữ nhật
a: Xét ΔABC vuông tại A và ΔHBA vuông tại H có
góc B chung
=>ΔABC đồng dạng với ΔHBA
=>AB/HB=AC/HA
=>AB*HA=HB*AC
b: AH=căn 5^2-3^2=4cm
BI là phân giác
=>HI/HB=IA/AB
=>HI/3=IA/5=(HI+IA)/(3+5)=0,5
=>HI=1,5cm; IA=1,5cm
e: \(E=\dfrac{x^2-9-x^2+4-x^2+9}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x+2}{x+3}\)
a: \(A=\dfrac{4x^2+x^2-2x+1+x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{6x^2+2}{\left(x-1\right)\left(x+1\right)}\)
13-C
14-b
15-d
16-a
13: C
14: B
15: D
16: A