Ta có:
\(\left|x-1\right|+\left|x-5\right|=\left|x-1\right|+\left|5-x\right|\)
≥ \(\left|x-1+5-x\right|=4\)
mà \(4-\left|x-3\right|\)≤ 4
Dấu "="⇔ \(x=3\)
Giải các PT sau
a)\(\left(3\left(x+1\right)-2\left(x+3\right)\right)^3+\left(2\left(x+3\right)-x+5\right)^3+\left(x-5-3\left(x+1\right)\right)^3=0\)
b)\(\left(x-2\right)^3+\left(x-4\right)^3+\left(x-7\right)^3+3\left(x-2\right)\left(x-4\right)\left(x-7\right)=0\)
Giải các pt sau
a, \(\left(x-1\right)\left(2x+5\right)\left(x^2+2\right)\)=0
b,\(\left(2x-1\right)\left(x-5\right)\left(x^2+3\right)\)=0
c,\(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
d,\(\left(2x+3\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
Bằng cách phân tích vế trái thành nhân tử, giải các PT sau:
a) \(2x.\left(x-3\right)+5\left(x-3\right)\)
b) \(\left(x^2-4\right)+\left(x-2\right).\left(3-2x\right)=0\)
c) \(x^3-3x^2+3x-1=0\)
Giải PT: \(48x\left(x+1\right)\left(x^3-4\right)=\left(x^4+8x+12\right)^2\)
Giải pt
\(\left(x+2\right)\left(x-3\right)+3=\left(x-4\right)\left(x+2\right)-7\)
giải pt
\(a,\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(b,\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)=1680\)
\(c,\left(x+1\right)^4+\left(x-1\right)^4=82\)
d,\(\left(4-x\right)^5+\left(x+2\right)^5=32\)
\(x^2+\dfrac{1}{x^2}=x+\dfrac{1}{x}\)
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
e,\(\left(x^2+x+1\right)^2-2x^2-2x=5\)
Giải pt
giải pt sau: \(\dfrac{1}{x-5}\)-\(\dfrac{4}{\left(x-5\right)\left(x-1\right)}\)=\(\dfrac{5}{x-1}\)
Giải các pt sau:
a, \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
b,\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
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