x^2 - 5x - 14
3x^2 + 9x - 30
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MH
6 tháng 1 2022
⇔ \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
⇔ \(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)
⇔ \(\dfrac{4}{x^2+8x+12}=\dfrac{1}{8}\)
⇔ \(x^2+8x+12=32\)
⇔ \(x^2+8x-20=0\)
⇔ \(\left(x-2\right)\left(x+10\right)=0\)
⇔ \(\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)
MH
18 tháng 10 2023
a) \(x^2-5x+6\)
\(=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
b) \(x^2-9x+18=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)
c) \(x^2-6x+5=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
d) \(3x^2+5x-30=3\left(x^2+\dfrac{5x}{3}-10\right)=3\left(x^2+2.x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{5347}{500}\right)\)
Câu này bạn xem lại đề nha
e) \(3x^2-5x-2=3x^2-6x+x-2\)
\(3x\left(x-2\right)+x-2=\left(x-2\right)\left(3x+1\right)\)
T
21 tháng 10 2023
\(a,x^2-5x+6\\=x^2-3x-2x+6\\=x(x-3)-2(x-3)\\=(x-3)(x-2)\\---\\b,3x^2+9x-30\\=3x^2-6x+15x-30\\=3x(x-2)+15(x-2)\\=(x-2)(3x+15)\\=3(x-2)(x+5)\\---\)
\(c,x^2-3x+2\\=x^2-x-2x+2\\=x(x-1)-2(x-1)\\=(x-1)(x-2)\\---\\d,3x^2-5x-2\\=3x^2-6x+x-2\\=3x(x-2)+(x-2)\\=(x-2)(3x+1)\\Toru\)
15 tháng 10 2017
a. x3 + 5x2 +9x + 45 = 0
<=> x2(x + 5) + 9(x + 5) = 0
<=> (x + 5)(x2 +9)=0
(x+5)= 0 hoặc (x2 + 9)=0 (vô lý)
<=> x = -5
2 tháng 9 2021
Đặt f(x)=0
nên 3x-6=0
hay x=2
Đặt h(x)=0
nên 30-5x=0
hay x=6
Đặt g(x)=0
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
2 tháng 9 2021
Ta có:
\(f\left(x\right)=0\Leftrightarrow3x-6=0\Leftrightarrow x=2\)
Vậy nghiệm của đa thức f(x) là 2
\(h\left(x\right)=0\Leftrightarrow-5x+30=0\Leftrightarrow x=6\)
Vậy nghiệm của đa thức h(x) là 6
N
5
TN
25 tháng 5 2017
Đk:\(x\ne2;x\ne3;x\ne4;x\ne5;x\ne6\)
\(pt\Leftrightarrow\frac{1}{\left(x-6\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-4\right)}+...+\frac{1}{\left(x-3\right)\left(x-2\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-5}+\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-4}+...+\frac{1}{x-3}-\frac{1}{x-2}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-2}=\frac{1}{8}\)\(\Leftrightarrow\frac{x-2}{\left(x-6\right)\left(x-2\right)}-\frac{x-6}{\left(x-2\right)\left(x-6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4}{\left(x-6\right)\left(x-2\right)}=\frac{1}{8}\Leftrightarrow\left(x-2\right)\left(x-6\right)=32\)
\(\Leftrightarrow x^2-8x+12=32\Leftrightarrow x^2-8x-20=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)
13 tháng 4 2021
ĐK : \(x\ne-2.-3;-4;-5;-6\)
\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\Leftrightarrow\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\Leftrightarrow x^2+8x-20=0\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\Leftrightarrow x=2;x=-10\)( tmđkxđ )
Vậy tập nghiệm phương trình là S = { -10 ; 2 }
H
13 tháng 4 2021
ĐKXĐ \(x\notin\left\{-2;-3;...;-6\right\}\)
Phương trình tương đương với:
\(\dfrac{1}{\left(x^2+2x\right)+\left(3x+6\right)}+\dfrac{1}{\left(x^2+3x\right)+\left(4x+12\right)}+\dfrac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\dfrac{1}{\left(x^2+5x\right)+\left(6x+30\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{\left(x+3\right)-\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}+\dfrac{\left(x+4\right)-\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}+\dfrac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\dfrac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x+2\right)\left(x+6\right)}=\dfrac{4}{32}\\ \Rightarrow\left(x+2\right)\left(x+6\right)=32\\\Leftrightarrow x^2+8x-20=0\\ \Leftrightarrow\left(x+10\right)\left(x-2\right)=0\\ \Leftrightarrow\begin{matrix}x=2\\x=-10\end{matrix}\left(t.m\right)\)
TN
1
11 tháng 1 2017
\(\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{x+2}-\frac{1}{\left(x+6\right)}\)
\(\frac{1}{t}-\frac{1}{t+4}=\frac{4}{t\left(t+4\right)}=\frac{1}{8}=\frac{4}{32}\Rightarrow t=4\Rightarrow x=2\)
\(x^2-5x-14\)
\(=x^2+2x-7x-14\)
\(=x\left(x+2\right)-7\left(x+2\right)\)
\(=\left(x-7\right)\left(x+2\right)\)
\(3x^2+9x-30\)
\(=3\left(x^2+3x-10\right)\)
\(=3\left(x^2-2x+5x-10\right)\)
\(=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)
\(=3\left(x-2\right)\left(x+5\right)\)