\(a.\)

\(\Leftrightarrow x-5+x-7=-12.\)

\(\Leftrightarrow x=-12+-12\)

\(\Leftrightarrow x=-24\)

\(b.\)

\(\Leftrightarrow3x+2x=-109+44\)

\(\Leftrightarrow5x=-65\)

\(\Rightarrow x=-13\)

\(c.\)

\(\Leftrightarrow2x-10-15+5x=-109\)

\(\Leftrightarrow2x-5x=-109+15+10\)

\(\Leftrightarrow3x=-84\)

\(\Rightarrow x=-28\)

\(d.\)

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d) \(\left(x-3\right)\left(x+5\right)< 0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3< 0\\x+5< 0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x< 3\\x< -5\end{cases}}\)

\(\Leftrightarrow\frac{x+100+9}{3}+\frac{x+100+25}{5}+\frac{x+100+49}{7}+\frac{100+81}{9}=2\)

<=> \(\frac{x+100}{3}+3+\frac{x+100}{5}+5+\frac{x+100}{7}+7+\frac{x+100}{9}+9=24\)

<=> \(\left(x+100\right)\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+24=24\)

<=> \(\left(x+100\right)\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)=0\Leftrightarrow x+100=0\Leftrightarrow x=-100\)

Chào Trang! Em kiểm tra lại đề nhé!

\(\frac{x+109}{3}+\frac{x+125}{5}+\frac{x+149}{7}+\frac{x+181}{9}=24?\)

D

D

D. 109 x (7 + 3)

\(x+\frac{109}{3}+x+\frac{125}{5}+x+\frac{149}{7}+x+\frac{181}{9}=2\)

\(\Rightarrow4x+\left(\frac{109}{3}+\frac{125}{5}+\frac{149}{7}+\frac{181}{9}\right)=2\)

\(\Rightarrow4x+\frac{6472}{63}=2\)

\(\Rightarrow4x=2-\frac{6472}{63}=-\frac{6346}{63}\)

\(\Rightarrow x=-\frac{6346}{63}:4\)

\(\Rightarrow x=-\frac{3173}{126}\)

`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`

`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`

Lời giải:

a.

$|2x-5|=12-3x$

Nếu $x\geq \frac{5}{2}$ thì $2x-5=12-3x$

$\Leftrightarrow x=3,4$ (thỏa mãn)

Nếu $x< \frac{5}{2}$ thì: $5-2x=12-3x$

$\Leftrightarrow x=7$ (loại)

Vậy......

b.

$4x=|x+1|+|x+2|+|x+3|\geq 0$

$\Rightarrow x\geq 0$

Do đó: $|x+1|+|x+2|+|x+3|=(x+1)+(x+2)+(x+3)=3x+6$

Vậy: $3x+6=4x$

$\Leftrightarrow x=6$ (thỏa mãn)

c.

$|x^2+|x+2||=x^2+3$

$\Leftrightarrow x^2+|x+2|=x^2+3$
$\Leftrightarrow |x+2|=3$

$\Leftrightarrow x+2=3$ hoặc $x+2=-3$

$\Leftrightarrow x=1$ hoặc $x=-5$

d.

$|x^2-3|=6$

$\Leftrightarrow x^2-3=6$ hoặc $x^2-3=-6$

$\Leftrightarrow x^2=9$ (chọn) hoặc $x^2=-3< 0$ (loại)

$\Leftrightarrow x=\pm 3$

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

1) \(x^2-9x=0\Rightarrow x\left(x-9\right)=0\Rightarrow x=0;9\)

2) \(x\left(x-4\right)-x^2=7\Rightarrow-4x=7\Rightarrow x=-\dfrac{7}{4}\)

3) \(3x+2\left(x-5\right)=5\Rightarrow5x-10=5\Rightarrow5x=15\Rightarrow x=3\)

4) \(25x^2-1=0\Rightarrow x^2=\dfrac{1}{25}\Rightarrow x=\pm\dfrac{1}{5}\)

5) \(3x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-5\right)=0\Rightarrow x=2;\dfrac{5}{3}\)

6) \(3x\left(x-7\right)+4\left(x-7\right)\Rightarrow\left(3x+4\right)\left(x-7\right)=0\Rightarrow x=-\dfrac{4}{3};7\)

7) \(4x^2-9=0\Rightarrow x^2=\dfrac{9}{4}\Rightarrow x=\pm\dfrac{3}{2}\)

8) \(10x\left(x-4\right)+2x-8=0\Rightarrow2\left(x-4\right)\left(5x+1\right)=0\Rightarrow x=4;-\dfrac{1}{5}\)

9) \(x\left(2x-5\right)-2x^2=0\Rightarrow x\left(2x-5-2x=0\right)\Rightarrow x=0\)

10) \(2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\Rightarrow x=0;2\)

11) \(2x\left(3-4x\right)+3\left(4x-3\right)=0\Rightarrow2x\left(4x-3\right)-3\left(4x-3\right)=0\Rightarrow\left(4x-3\right)\left(2x-3\right)=0\Rightarrow x=\dfrac{3}{4};\dfrac{3}{2}\)

12) \(2x\left(x-5\right)-2x^2=3\Rightarrow-10x=3\Rightarrow x=-\dfrac{3}{10}\)

a) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=\left(x^2+3x+1\right)^2+x\)

\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=\left(x^2+3x+1\right)^2+x\)

\(\Leftrightarrow\left(t-1\right)\left(t+1\right)=t^2+x\) (với \(t=x^2+3x+1\))

\(\Leftrightarrow t^2-1=t^2+x\)

\(\Leftrightarrow x=-1\).

b) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)=\left(x^2+8x+11\right)^2+2x\)

\(\Leftrightarrow\left(x^2+8x+7\right)\left(x^2+8x+15\right)=\left(x^2+8x+11\right)^2+2x\)

\(\Leftrightarrow\left(t-4\right)\left(t+4\right)=t^2+2x\) (với \(t=x^2+8x+11\))

\(\Leftrightarrow t^2-16=t^2+2x\)

\(\Leftrightarrow x=-8\)

c) \(\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)=63\)

\(\Leftrightarrow\left(x^3-1\right)\left(x^3+1\right)=63\)

\(\Leftrightarrow x^6-1=63\)

\(\Leftrightarrow x^6=64\)

\(\Leftrightarrow x=\pm2\)

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