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`#3107.101107`
\(x(x+5)(x-5) - (x+2)(x^2-2x+4)=5\)
`<=> x(x^2 - 25) - (x^3 + 2^3) = 5`
`<=> x^3 - 25x - x^3 - 8 = 5`
`<=> -25x - 8 = 5`
`<=> -25x = 13`
`<=> x = -13/25`
Vậy, `x = -13/25`
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\((x+1)^3 - (x-1)^3 -6(x-1)^2 = -19\)
`<=> x^3 + 3x^2 + 3x + 1 - (x^3 - 3x^2 + 3x - 1) - 6(x^2 - 2x + 1) = -19`
`<=> x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 - 6x^2 + 12x - 6 = -19`
`<=> (x^3 - x^3) + (3x^2 + 3x^2 - 6x^2) + (3x - 3x + 12x) + (1 + 1 - 6) = -19`
`<=> 12x - 4 = -19`
`<=> 12x = -15`
`<=> x = -15/12 = -5/4`
Vậy, `x = -5/4.`
________
`@` Sử dụng các hđt:
`1)` `A^2 + B^2 = (A - B)(A + B)`
`2)` `A^3 + B^3 = (A + B)(A^2 - AB + B^2)`
`3)` `(A - B)^3 = A^3 - 3A^2B + 3AB^2 - B^3`
`4)` `(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3`
`5)` `(A - B)^2 = A^2 - 2AB + B^2.`
a: \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=5\)
=>\(x\left(x^2-25\right)-x^3-8=5\)
=>\(x^3-25x-x^3-8=5\)
=>-25x=13
=>\(x=-\dfrac{13}{25}\)
b: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)
=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-19\)
=>\(6x^2+2-6x^2+12x-6=-19\)
=>12x-4=-19
=>12x=-15
=>x=-5/4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đặt x^2+2x+2=t
\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)
\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)
Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x+2)(x^2-2x+4)-x(x^2+2)=15
<=> x^3 + 8 - x^3 - 2x = 15
<=> -2x = 7
<=> x = -7/2
b) (x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28
<=> x^3 + 9x² + 27x + 27 - x(9x² + 6x + 1) + 8x^3 + 1 = 28
<=> x^3 + 9x² + 27x + 27 - 9x^3 - 6x² - x + 8x^3 + 1 - 28 = 0
<=> 3x² + 26x = 0
<=> x(3x + 26) = 0
Vậy x = 0 và x = -26/3
c) (x^2-1)^3-(x^4+x^2+1)(x^2-1)=0
<=> (x² - 1)[(x² -1)² - x^4 - x² - 1] = 0
<=> (x-1)(x+1)(x^4 - 2x² + 1 - x^4 - x² - 1 ) = 0
<=> -(x-1)(x+1)3x² = 0
Vậy nghiệm là x = 1 ; -1 ; 0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-5x-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
Vậy....
\(2x\left(x+6\right)=7x+42\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)
Vậy......
\(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\)
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
c: Ta có: \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b.
PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$
$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$
$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$
$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$
$\Leftrightarrow (x-3)^2(2x+1)^2=0$
$\Leftrightarrow (x-3)(2x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$
d.
$x^2-2x=24$
$\Leftrightarrow x^2-2x-24=0$
$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$
$\Leftrightarrow x=-4$ hoặc $x=6$