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`@` `\text {Ans}`
`\downarrow`
`a)`
`3x ( 12x - 4 ) - 9x( 4x - 3 ) = 30`
`=> 3x (12x-4) - 3*3x (4x - 3) = 30`
`=> 3x [12x - 4 - 3(4x-3)] = 30`
`=> 3x (12x - 4 - 12x + 9) = 30`
`=> 3x (-4+9)=30`
`=> 3x*5=30`
`=> 3x=6`
`=> x=2`
Vậy, `x=2`
`b)`
`x( 5 - 2x) + 2x( x - 1)`
`=> x(5-2x) + 2x^2 - 2x=15`
`=> 5x - 2x^2 + 2x^2 - 2x =15`
`=> 3x = 15`
`=> x=5`
Vậy, `x=5.`
a: =>36x^2-12x-36x^2+27x=30
=>15x=30
=>x=2
b: =>5x-2x^2+2x^2-2x=15
=>3x=15
=>x=5
b) ta có \(\left(4x-5\right)^3-\left(2x+5\right)\left(16x^2-25\right)=0\)
\(\left(4x-5\right)^3-\left(2x+5\right)\left(4x+5\right)\left(4x-5\right)=0\)
\(\left(4x-5\right)\left[\left(4x-5\right)^2-\left(2x+5\right)\left(4x+5\right)\right]=0\)
\(\left(4x-5\right)\left(16x^2-40x+5^2-8x^2-10x-20x-5^2\right)=0\)
\(\left(4x-5\right)\left(8x^2-70x\right)=0\)
=> \(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}=>\orbr{\begin{cases}4x=5\\x\left(8x-70\right)=0\end{cases}< =>}\orbr{\begin{cases}x=\frac{5}{4}\\\orbr{\begin{cases}x=0\\8x-70=0=>x=\frac{35}{4}\end{cases}}\end{cases}}}\) \(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}< =>\orbr{\begin{cases}x=\frac{5}{4}\\x\left(8x-70\right)=0\end{cases}\orbr{\begin{cases}x=\frac{5}{4}\\\orbr{\begin{cases}x=0\\8x-70=0=>x=\frac{35}{4}\end{cases}}\end{cases}}}}\)
\(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}=>\orbr{\begin{cases}x=\frac{5}{4}\\x\left(8x-70\right)=0\end{cases}}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{35}{4}\end{cases}}\) Vậy \(\orbr{\begin{cases}x=\frac{5}{4}\\x=0\end{cases}}\) hoặc \(x=\frac{35}{4}\)
a) \(\left(3x-4\right)^3=\left(9x-8\right)\left(3x^2-8\right)\)
\(\Leftrightarrow27x^3+108x^2+144x+64=27x^3-72x-24x^2+64\)
\(\Leftrightarrow27x^3+108x^2+144x+64-27x^3+72x+24x^2-64=0\)
\(\Leftrightarrow132x^2+216x=0\)
\(\Leftrightarrow12x\left(11x+18\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\11x+18=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-\frac{18}{11}\end{array}\right.\)
a.
$x^4-25x^3=0$
$\Leftrightarrow x^3(x-25)=0$
\(\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.\)
b.
$(x-5)^2-(3x-2)^2=0$
$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$
$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix}
-2x-3=0\\
4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}
x=\frac{-3}{2}\\
x=\frac{7}{4}\end{matrix}\right.\)
c.
$x^3-4x^2-9x+36=0$
$\Leftrightarrow x^2(x-4)-9(x-4)=0$
$\Leftrightarrow (x-4)(x^2-9)=0$
$\Leftrightarrow (x-4)(x-3)(x+3)=0$
\(\Leftrightarrow \left[\begin{matrix} x-4=0\\ x-3=0\\ x+3=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=4\\ x=3\\ x=-3\end{matrix}\right.\)
d. ĐK: $x\neq 0$
$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$
$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$
$\Leftrightarrow -2(-x^2+3x-4)=0$
$\Leftrightarrow x^2-3x+4=0$
$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)
Vậy pt vô nghiệm.
\(a,\) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+6\)
\(< =>20x-5x^2+5x^2-12-x-6=0\)
\(< =>19x-18=0\)
\(< =>x=\dfrac{18}{19}\)
\(b,\left(2x-7\right)\left(5+4x\right)-8\left(x^2-4x+5\right)=-30\)
\(< =>10x+8x^2-35-28x-8x^2+24x-40+30=0\)
\(< =>6x-45=0< =>x=\dfrac{45}{6}=7,5\)
a) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+\Rightarrow6\\ \Leftrightarrow20x-5x^2+5x^2-12=x+6\\ \Leftrightarrow20x-12=x+6\\\Rightarrow20x-x=6+12\\ \Rightarrow19x=18\\ \Rightarrow x=\dfrac{18}{19}\)
b) \(\left(2x-7\right)\left(5+4x\right)-8\left(x^2-3x+5\right)=-30\\ \Rightarrow10x+8x^2-35-28x-8x^2+24x-40=-30\\ \Rightarrow6x-75=-30\\ \Rightarrow6x=45\\ \Rightarrow x=\dfrac{15}{2}\)