tìm x,y,z
a, \(\dfrac{\left(x-2\right)^2}{-1}\) = \(\dfrac{1}{2-x}\) (x ≠ 2)
b, \(\dfrac{x-2}{3.2-x}\) = \(\dfrac{1}{2}\) (x ≠ 3.2)
c, \(\dfrac{x-1}{3}\) = \(\dfrac{x+3}{5}\)
d, 4x(x-5)-(x-1)(4x-3)=5
e, (x-5)(-x+4)-(x-1)(x+3)=\(-2x^2\)
f, (x-3)(\(x^2\)+3x+9)+x(5-\(x^2\))=6x
h, (x-5)(x-4)-(x+1)(x-2)=7
l, \(\dfrac{x}{2}\)=\(\dfrac{y}{3}\);\(\dfrac{y}{4}\)=\(\dfrac{z}{5}\) và x+y+z=5
m, \(\dfrac{y}{3}\)=\(\dfrac{z}{5}\); 4x=3y và 3x - z = 21
a: \(\dfrac{\left(x-2\right)^2}{-1}=\dfrac{1}{2-x}\)
=>\(\dfrac{\left(x-2\right)^2}{1}=\dfrac{1}{x-2}\)
=>\(\left(x-2\right)^3=1\)
=>x-2=1
=>x=3(nhận)
b: \(\dfrac{x-2}{3\cdot2-x}=\dfrac{1}{2}\)
=>\(2\left(x-2\right)=6-x\)
=>2x-4=6-x
=>3x=10
=>x=10/3(nhận)
c: \(\dfrac{x-1}{3}=\dfrac{x+3}{5}\)
=>5(x-1)=3(x+3)
=>5x-5=3x+9
=>2x=14
=>x=7
d: \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
=>\(4x^2-20x-4x^2+3x+4x-3=5\)
=>-13x=8
=>\(x=-\dfrac{8}{13}\)
e: \(\left(x-5\right)\left(-x+4\right)-\left(x-1\right)\left(x+3\right)=-2x^2\)
=>\(-x^2+4x+5x-20-x^2-2x+3=-2x^2\)
=>7x-17=0
=>\(x=\dfrac{17}{7}\)
f: \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)
=>\(x^3-27+5x-x^3=6x\)
=>6x=5x-27
=>x=-27
h: \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)
=>\(x^2-9x+20-\left(x^2-x-2\right)=7\)
=>-8x+22=7
=>-8x=-15
=>x=15/8
từ câu a đến h chỉ là nhân phá và rút gọn thôi nên mình xin phép ko làm nhé
câu l,
có \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y+z}{8+12+15}=\dfrac{1}{7}\) đến đây dễ rồi nhé
câu m,
đổi 4x=3y thành x/3 = y/4 rồi làm tương tự câu trên
