Bài 1:
1: \(\left(x-3\right)^2-\left(x+2\right)^2=5\)
=>(x-3-x-2)(x-3+x+2)=5
=>-5(2x-1)=5
=>2x-1=-1
=>2x=0
=>x=0
2: \(\left(x-1\right)\left(x+1\right)-\left(5-x\right)^2=4\)
=>\(x^2-1-\left(x^2-10x+25\right)-4=0\)
=>\(x^2-5-x^2+10x-25=0\)
=>10x-30=0
=>x=3
3: \(2\cdot\left(x+3\right)\left(x-4\right)+\left(x+3\right)^2+\left(x-4\right)^2=4x^2\)
=>\(2\left(x^2-x-12\right)+x^2+6x+9+x^2-8x+16=4x^2\)
=>\(2x^2-2x-24+2x^2-2x+25-4x^2=0\)
=>-4x+1=0
=>-4x=-1
=>\(x=\dfrac{1}{4}\)
4: \(x\left(x-5\right)^2-x^2\left(x-1\right)-x^2=4\)
=>\(x\left(x^2-10x+25\right)-x^3+x^2-x^2=4\)
=>\(x^3-10x^2+25x-x^3=4\)
=>\(-10x^2+25x-4=0\)
=>\(x=\dfrac{25\pm\sqrt{465}}{20}\)
5: \(\left(2x-3\right)^2-\left(4-2x\right)^2=0\)
=>\(\left(2x-3-4+2x\right)\left(2x-3+4-2x\right)=0\)
=>4x-7=0
=>4x=7
=>\(x=\dfrac{7}{4}\)
6: \(\left(x-5\right)^2+2x+3-2\left(5-x\right)\left(3+2x\right)-9x\left(x-1\right)=0\)
=>\(x^2-10x+25+2x+3+2\left(2x+3\right)\left(x-5\right)-9x^2+9x=0\)
=>\(10x^2+x+28+2\left(2x^2-7x-15\right)=0\)
=>\(10x^2+x+28+4x^2-14x-30=0\)
=>\(14x^2-13x-2=0\)
=>\(x=\dfrac{13\pm\sqrt{281}}{28}\)
Bài 2:
a: \(501^2=\left(500+1\right)^2=500^2+2\cdot500\cdot1+1^2\)
=250000+1000+1
=251001
b: \(999^2=\left(1000-1\right)^2\)
\(=1000^2-2\cdot1000\cdot1+1^2\)
=1000000-2000+1
=998001
c: \(48\cdot52=\left(50-2\right)\left(50+2\right)\)
\(=50^2-2^2=2500-4=2496\)
d: \(64\cdot56=\left(60+4\right)\left(60-4\right)\)
\(=60^2-4^2=3600-16=3584\)
