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a: \(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-4\right)=5\)
\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-20=5\)
\(\Leftrightarrow30x-15=5\)
hay x=2/3
b: \(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=42\)
\(\Leftrightarrow26x+28=42\)
=>26x=14
hay x=7/13
a: \(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15x^2-60-5=0\)
\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-65=0\)
\(\Leftrightarrow30x-60=0\)
hay x=2
b: \(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=42\)
\(\Leftrightarrow9x^3+6x^2+27x+28-9x^3-6x^2-x=42\)
=>26x=14
hay x=7/13
\(a,5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)\(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-4\right)-5=0\)\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60-5=0\)\(\Leftrightarrow30x=60\Leftrightarrow x=2\)
\(b,\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\)\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2-42=0\)\(\Leftrightarrow x^3-9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2-42=0\)\(\Leftrightarrow26x=14\Rightarrow x=\dfrac{7}{13}\)
Học tốt nha<3
a: \(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15x^2-60-5=0\)
\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-65=0\)
\(\Leftrightarrow30x-60=0\)
hay x=2
b: \(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=42\)
\(\Leftrightarrow9x^3+6x^2+27x+28-9x^3-6x^2-x=42\)
=>26x=14
hay x=7/13
Bài 1:
a) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\) (1)
\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=42\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=42\)
\(\Leftrightarrow26x+28=42\)
\(\Leftrightarrow26x=42-28\)
\(\Leftrightarrow26x=14\)
\(\Leftrightarrow x=\dfrac{7}{13}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{7}{13}\right\}\)
1b) \(5x\left(x+3\right)^2-5\left(x+1\right)^3+15\left(x+2\right)\left(x-2\right)=5\Leftrightarrow5x\left(x^2+6x+9\right)-5\left(x^3+3x^2+3x+1\right)+15\left(x^2-4\right)=5\Leftrightarrow30x-65=5\Leftrightarrow30x=70\Leftrightarrow x=\dfrac{7}{3}\)
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)