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a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)
=-4/3x^2+8/3-10/3
=-4/3x^2-2/3
d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)
\(=3x^2+9x+22+\dfrac{68}{x-3}\)
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
a, x=-505
b, x=35/8 hoac -37/8
nhung cau con lai thi tong tu
\(P\left(x\right)=5x^5+5x^4-2x^2+5x^2-x^5-4x^4+1-4x^5=x^4+3x^2+1\)
Mà \(x^4\ge0;3x^2\ge0=>x^4+3x^2+1\ge1>0\) nên \(P\left(x\right)\) vô nghiệm
Hok tốt nha !
P(x) = 5x5 + 5x4 - 2x2 + 5x2 - x5 - 4x4 + 1 - 4x5
P(x) = (5x5 - x5 - 4x5) + (5x4 - 4x4) - (2x2 - 5x2) + 1
P(x) = x4 + 3x2 + 1
Ta có: x4 \(\ge\)0 \(\forall\)x; 3x2 \(\ge\)0 \(\forall\)x
=> x4 + 3x2 + 1 \(\ge\)1 \(\forall\)x
=> P(x) \(\ne\)0
=> P(x) vô nghiệm
a) \(A\left(x\right)=3x^3-4x^4-2x^3+4x^4-5x+3\)
\(\Rightarrow A\left(x\right)=-4x^4+4x^4+3x^3-2x^3-5x+3\)
\(\Rightarrow A\left(x\right)=x^3-5x+3\)
\(B\left(x\right)=5x^3-4x^2-5x^3-4x^2-5x-3\)
\(\Rightarrow B\left(x\right)=5x^3-5x^3-4x^2-4x^2-5x-3\)
\(\Rightarrow B\left(x\right)=-8x^2-5x-3\)
b) \(A\left(x\right)+B\left(x\right)=x^3-5x+3+\left(-8x^2-5x-3\right)\)
\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-5x+3-8x^2-5x-3\)
\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-8x^2-5x-5x+3-3\)
\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-8x^2-10x\)
\(A\left(x\right)-B\left(x\right)=x^3-5x+3-\left(-8x^2-5x-3\right)\)
\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3-5x+3+8x^2+5x+3\)
\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3+8x^2-5x+5x+3+3\)
\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3+8x^2+6\)
\(\left(5x-1\right)^2+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
\(=\left(5x-1\right)^2-2\left(5x-1\right)\left(5x+4\right)+\left(5x+4\right)^2\)
\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2\)
\(=\left(5x-1-5x-4\right)^2\)
\(=\left(-5\right)^2\)
\(=25\)