Tìm x
2 . ( x - 7 )- 3 . ( 5 - x ) = -109
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DT
2
TR
10 tháng 8 2018
\(a.\)
\(\Leftrightarrow x-5+x-7=-12.\)
\(\Leftrightarrow x=-12+-12\)
\(\Leftrightarrow x=-24\)
\(b.\)
\(\Leftrightarrow3x+2x=-109+44\)
\(\Leftrightarrow5x=-65\)
\(\Rightarrow x=-13\)
\(c.\)
\(\Leftrightarrow2x-10-15+5x=-109\)
\(\Leftrightarrow2x-5x=-109+15+10\)
\(\Leftrightarrow3x=-84\)
\(\Rightarrow x=-28\)
\(d.\)
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LT
0
LT
2
17 tháng 4 2019
\(\Leftrightarrow\frac{x+100+9}{3}+\frac{x+100+25}{5}+\frac{x+100+49}{7}+\frac{100+81}{9}=2\)
<=> \(\frac{x+100}{3}+3+\frac{x+100}{5}+5+\frac{x+100}{7}+7+\frac{x+100}{9}+9=24\)
<=> \(\left(x+100\right)\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+24=24\)
<=> \(\left(x+100\right)\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)=0\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
17 tháng 4 2019
Chào Trang! Em kiểm tra lại đề nhé!
\(\frac{x+109}{3}+\frac{x+125}{5}+\frac{x+149}{7}+\frac{x+181}{9}=24?\)
LT
1
KN
16 tháng 4 2019
\(x+\frac{109}{3}+x+\frac{125}{5}+x+\frac{149}{7}+x+\frac{181}{9}=2\)
\(\Rightarrow4x+\left(\frac{109}{3}+\frac{125}{5}+\frac{149}{7}+\frac{181}{9}\right)=2\)
\(\Rightarrow4x+\frac{6472}{63}=2\)
\(\Rightarrow4x=2-\frac{6472}{63}=-\frac{6346}{63}\)
\(\Rightarrow x=-\frac{6346}{63}:4\)
\(\Rightarrow x=-\frac{3173}{126}\)
27 tháng 8 2021
`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`
27 tháng 8 2021
`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`
AH
Akai Haruma
Giáo viên
29 tháng 7 2021
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)
$\Leftrightarrow x=\pm 3$
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
7 tháng 7 2023
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) \(\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)=63\)
\(\Leftrightarrow\left(x^3-1\right)\left(x^3+1\right)=63\)
\(\Leftrightarrow x^6-1=63\)
\(\Leftrightarrow x^6=64\)
\(\Leftrightarrow x=\pm2\)
7 tháng 7 2023
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đề thiếu bn ơi
2.(x-7)-3.(5-x)=-109
=>2x-14-15+3x=-109
=>5x-29=-109
=>5x=-80
=>x=-16