\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\) giúp tớ với pls ❤
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a) Ta có: \(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\left(6x-2\right)^2-2\cdot\left(6x-2\right)\left(5x-2\right)+\left(5x-2\right)^2=0\)
\(\Leftrightarrow\left(6x-2-5x+2\right)^2=0\)
\(\Leftrightarrow x^2=0\)
hay x=0
Vậy: x=0
b) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-4\right)-5=0\)
\(\Leftrightarrow x^3-6-x^2+4x=0\)
\(\Leftrightarrow4x-6=0\)
\(\Leftrightarrow4x=6\)
hay \(x=\frac{3}{2}\)
Vậy: \(x=\frac{3}{2}\)
c) Ta có: \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)
\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+27\right)+3x^2-12-2=0\)
\(\Leftrightarrow x^3+3x-15-x^3-27=0\)
\(\Leftrightarrow3x-42=0\)
\(\Leftrightarrow3x=42\)
hay x=14
Vậy: x=14

a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1

a ) \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=5\)
\(\Leftrightarrow x^3-27-x\left(x^2-16\right)=5\)
\(\Leftrightarrow x^3-27-x^3+16x=5\)
\(\Leftrightarrow16x-27=5\)
\(\Leftrightarrow16x=32\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
b ) \(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
\(\Leftrightarrow x^3-6x^2+12x-8-\left(x^3+125\right)+6x^2=11\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3-125+6x^2=11\)
\(\Leftrightarrow12x-133=11\)
\(\Leftrightarrow12x=144\)
\(\Leftrightarrow x=12\)
Vậy \(x=12\)

a. \(9\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow9x+18-3x-6=0\)
\(\Leftrightarrow6x+12=0\)
\(\Leftrightarrow x=-2\)
e. \(\left(2x-1\right)^2-45=0\)
\(\Leftrightarrow4x^2-2x+1-45=0\)
\(\Leftrightarrow4x^2-2x-44=0\)
Đến đó tự giải tiếp nha!
c. \(2\left(2x-5\right)-3x=0\)
\(\Leftrightarrow4x-10-3x=0\)
\(\Leftrightarrow x-10=0\)
\(\Leftrightarrow x=10\)
g. \(2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
delll
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