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D
1
18 tháng 1 2022
\(a.x^2-11x+15=-15.\Leftrightarrow x^2-11x+30=0.\)
\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=6.\\x=5.\end{matrix}\right.\)
\(b.2x-3x+10=x.\Leftrightarrow-2x+10=0.\Leftrightarrow x=5.\)
\(c.x^3-4=4.\Leftrightarrow x^3=8.\Leftrightarrow x^3=2^3.\Rightarrow x=2.\)
\(d.x^4+x^3-x^2-x=0.\Leftrightarrow x^2\left(x^2+x\right)-\left(x^2+x\right)=0.\Leftrightarrow\left(x^2-1\right)\left(x^2+x\right)=0.\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)x\left(x+1\right)=0.\Leftrightarrow\left(x-1\right)\left(x+1\right)^2x=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\x+1=0.\\x=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-1.\\x=0.\end{matrix}\right.\)
NV
0
CT
1
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2023
ĐKXĐ: \(x\ne\left\{-4;-5;-6;-7\right\}\)
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow x\left(x-2\right)+13\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
NN
1
12 tháng 2 2023
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+...+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
=>\(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+...+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
=>1/x+2-1/x+6=1/8
=>\(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)
=>x^2+8x+12=32
=>x^2+8x-20=0
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
TN
19 tháng 5 2016
x2(x+2)2+4x2=12(x+2)2
=>x2(x+2)2+4x2-12(x+2)2=0
VT=(x2-2x-4)(x2+6x+12)
pt trở thành (x2-2x-4)(x2+6x+12)=0
=>x2-2x-4=0 hoặc x2+6x+12=0
Th1:x2-2x-4=0
denta:(-2)2-(-4(1.4))=20
x1:(2+\(\sqrt{20}\)):2=1+\(\sqrt{5}\)
x2:(2-\(\sqrt{20}\)):2=\(\sqrt{5}\)+1
Th2:x2+6x+12=0
denta:62-4(1.12)=-12
=>\(\Delta< 0\)
=>vô nghiệm
vậy pt có nghiệm là 1-\(\sqrt{5}\)và \(\sqrt{5}\)+1
16 tháng 8 2018
\(\left(x^2+3x+2\right)\left(x^2+11x+30\right)-60=0\)
\(\Rightarrow\left[\left(x+1\right)\left(x+2\right)\right].\left[\left(x+5\right)\left(x+6\right)\right]-60=0\)
\(\Rightarrow\left[\left(x+1\right)\left(x+6\right)\right].\left[\left(x+2\right)\left(x+5\right)\right]-60=0\)
\(\Rightarrow\left(x^2+7x+6\right)\left(x^2+7x+10\right)-60=0\left(1\right)\)
Đặt \(x^2+7x+6=a\Rightarrow x^2+7x+10=a+4\)
Thay vào (1), ta có:
\(a\left(a+4\right)-60=0\)
\(\Rightarrow a^2+4a-60=0\)
\(\Rightarrow a^2+10a-6a-60=0\)
\(\Rightarrow a\left(a+10\right)-6\left(a+10\right)=0\)
\(\Rightarrow\left(a-6\right)\left(a+10\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a=6\\a=-10\end{cases}}\)
- Nếu \(x^2+7x+6=6\)
\(\Rightarrow x^2+7x=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}\)
- Nếu \(x^2+7x+6=-10\)
\(\Rightarrow x^2+7x+16=0\)
Mà \(x^2+7x+16=x^2+2.x.\frac{7}{2}+\frac{49}{4}+\frac{15}{4}=\left(x+\frac{7}{2}\right)^2+\frac{15}{4}>0\forall x\)
Vậy \(x=0,x=-7\)
Học tốt.
PL
1
MV
8 tháng 1 2018
\(x^4+2x^3-2x^2+2x-3=0\\ \Leftrightarrow x^4+3x^3-x^3-3x^2+x^2+3x-x-3=0\\ \Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)+x\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^3-x^2+x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left[x^2\left(x-1\right)+\left(x-1\right)\right]=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\\x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\left(\text{vì }x^2+1\ge1>0\right)\)
Vậy ...
\(\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[\left(x^2+5x-2\right)-\left(x^2+x+1\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy ...
\(x^2+\left(x+2\right)\left(11x-7\right)=4\\ \Leftrightarrow x^2-4+\left(x+2\right)\left(11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2\right)+\left(11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2+11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\\ \Leftrightarrow3\left(x+2\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\4x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy ...
27 tháng 2 2018
nghiệm đâu bạn ưi...nó là phương trình vô nghiệm hay vô số nghiệm vậy m :))
VD
10 tháng 3 2022
\(\left(x^2+5x+6\right)\left(x^2-11x+3x\right)=180\\ \Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x-5\right)\left(x-6\right)=180\\ \Leftrightarrow\left[\left(x+2\right)\left(x-5\right)\right]\left[\left(x+3\right)\left(x-6\right)\right]\\ \Leftrightarrow\left(x^2-3x-10\right)\left(x^2-3x-18\right)=0\left(1\right)\)
Đặt \(x^2-3x-14=a\)
\(\left(1\right)\Leftrightarrow\left(a+4\right)\left(a-4\right)=180\\ \Leftrightarrow a^2-16=180\\ \Leftrightarrow a^2=196\\ \Leftrightarrow\left[{}\begin{matrix}a=14\\a=-14\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-3x-14=14\\x^2-3x-14=-14\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-3x=0\left(vô.lí\right)\\x^2-3x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
CC
2
10 tháng 3 2019
\(a.\left(x^2+3x+2\right)\left(x^2+11x+30\right)-60=0\)
\(\Leftrightarrow\left(x^2+7x-4x+16-14\right)\left(x^2+7x+4x+16+14\right)-60=0\)
\(\Leftrightarrow\left(x^2+7x+16-4x-14\right)\left(x^2+7x+16+4x+14\right)=0\)
\(\Leftrightarrow\left(x^2+7x+16\right)^2-\left(4x+14\right)^2-60=0\)
Vì \(\left(x^2+7x+16\right)^2>0;\left(4x+14\right)^2>0\)
Nên \(\left(x^2+7x+16\right)^2-\left(4x+14\right)^2-60\ge-60\)
V...\(S=\varnothing\)
\(b.4^x-12.2^x+32=0\)
\(\Leftrightarrow\left(2^x\right)^2-2.2^x.6+36-4=0\)
\(\Leftrightarrow\left(2^x-6\right)^2-4=0\)
\(\Leftrightarrow\left(2^x-4\right)\left(2^x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2^x-4=0\\2^x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2^x=4\\2^x=8\end{cases}\Leftrightarrow}\orbr{\begin{cases}2^x=2^2\\2^x=2^3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)
V...\(S=\left\{2;3\right\}\)
^^ đúng ko ta
29 tháng 12 2019
a) (x+1)(x+2)(x+5)(x+6)-60=0
[(x+1)(x+6)][(x+2)(x+5)]-60=0
(x^2 + 7x + 6)(x^2 + 7x + 10) - 60 = 0
đặt t = x^2 + 7x + 8
pt trở thành
(t-2)(t+2)-60=0
t^2 - 64=0 .....
t=8 hoặc t=-8.
tìm x ....
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