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28 tháng 5 2022
Bài 3:
a: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=21\)
\(\Leftrightarrow x^3-27-x\left(x^2-16\right)=21\)
\(\Leftrightarrow x^3-27-x^3+16x=21\)
=>16x=48
hay x=3
b: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=4\)
\(\Leftrightarrow x^3+8-x^3-2x=4\)
=>-2x=4-8=-4
hay x=2
TN
13 tháng 3 2020
\(a.\frac{x-6}{x-4}=\frac{x}{x-2}\\\Leftrightarrow \frac{\left(x-6\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=\frac{x\left(x-4\right)}{\left(x-4\right)\left(x-2\right)}\\\Leftrightarrow \left(x-6\right)\left(x-2\right)=x\left(x-4\right)\\\Leftrightarrow \left(x-6\right)\left(x-2\right)-x\left(x-4\right)=0\\ \Leftrightarrow x^2-2x-6x+12-x^2+4x=0\\\Leftrightarrow -4x+12=0\\\Leftrightarrow -4x=-12\\ \Leftrightarrow x=3\)
\(b.1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\\ \Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}+\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)=0\\ \Leftrightarrow x^2-x-2x+3+2x^2-2x-5x+5-3x^2+6x+5x-10=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \)
TH
21 tháng 7 2016
a)\(\frac{2x-5}{x+5}\)=3 ĐKXĐ: x khác -5
=> 2x-5=3(x+5)
<=>2x-5=3x+15
<=>-x=20
<=>x =-20
TH
21 tháng 7 2016
b)\(\frac{x2-6}{x}\)=x+\(\frac{3}{2}\)ĐKXĐ\(x\ne0\)
=>2(x2-6)=2x2+3x
<=>2x2-12=2x2+3x
<=>-3x=12
<=>x=-4
12 tháng 4 2017
a) x = 1
b) x = 6; x = -3
c) x = 5,5; x = 1,5
d) x = 1; x = -1
e) x = -2; x = -1,000000371....
AO
21 tháng 1 2018
\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\)
\(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{16}{x^2-1}\)
\(\Rightarrow\left(x+1\right)^2-\left(x-1\right)^2=16\)
\(\Rightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=16\)
\(\Rightarrow2\left(2x\right)=16\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
vậy \(x=4\)
\(\frac{6x+1}{x^2-7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)
\(\frac{6x+1}{\left(x-2\right)\left(x-5\right)}+\frac{5\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}=\frac{3\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}\)
\(\Rightarrow6x+1+5x-5=3x-6\)
\(\Rightarrow11x-3x=-6+4\)
\(\Rightarrow8x=-2\)
\(\Rightarrow x=\frac{-1}{4}\)
3) \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\frac{x^2+x+1}{x^3-1}+\frac{\left(2x^2-5\right)}{x^3-1}=\frac{4\left(x-1\right)}{x^3-1}\)
\(\Rightarrow x^2+x+1+2x^2-5=4x-4\)
\(\Rightarrow3x^2-3x=-4+4\)
\(\Rightarrow3x\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
16 tháng 4 2020
a, \(\frac{6x+1}{x^2+7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)
\(11x^3-31x^2-72x-240=3\left(x+2\right)\left(x+5\right)\left(x-2\right)\)
\(11x^3-31x^2-72x-240-3\left(x+2\right)\left(x+5\right)\left(x-2\right)=0\)
\(8x^3-46x^2-60x-180=0\)
=> vô nghiệm
16 tháng 4 2020
b) \(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\left(x\ne0;x\ne\pm2\right)\)
\(\Leftrightarrow\frac{2x}{\left(x-2\right)\left(x+2\right)x}-\frac{\left(x+2\right)\left(x-1\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x+4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x}{x\left(x-2\right)\left(x+2\right)}-\frac{x^2+x-2}{x\left(x-2\right)\left(x+2\right)}+\frac{x^2+2x-8}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x-x^2-x+2+x^2+2x-8}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{3x-6}{x\left(x-2\right)\left(x+2\right)}=0\)
=> 3x-6=0
<=> x=2 (ktm)
Vậy pt vô nghiệm
\(\left(x^2+x-1\right)\left(x^2+x+3\right)=5\\ \Leftrightarrow\left(x^2+x-1\right)\left(x^2+x-1+4\right)-5=0\\ \Leftrightarrow\left(x^2+x-1\right)^2+4\left(x^2+x-1\right)-5=0\\ \Leftrightarrow\left[\left(x^2+x-1\right)^2+5\left(x^2+x-1\right)^2\right]-\left[\left(x^2+x-1\right)+5\right]=0\\ \Leftrightarrow\left(x^2+x-1\right)\left(x^2+x-1+5\right)-\left(x^2+x-1+5\right)=0\\ \Leftrightarrow\left(x^2+x-1+5\right)\left(x^2+x-1-1\right)=0\\ \Leftrightarrow\left(x^2+x+4\right)\left(x^2+x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{15}{4}=0\\\left(x^2+2x\right)-\left(x+2\right)=0\end{matrix}\right. \)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\left(vô.lí\right)\\x\left(x+2\right)-\left(x+2\right)=0\end{matrix}\right.\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Đặt x\(^2\) +x+1=a
=>(a-2)(a+2)=5
=>a^2=9
=>a=3
và a=-3
thay ngược vào ta được
1,x^2+x+1=3
<=>x^2+x-2=0
<=>(x-1)(x+2)=0
<=>x=1 hoặc x=-2
2,x^2+x+1=-3
<=>x^2+x+4=0
<=>(x+\(\dfrac{1}{2}\) )^2+\(\dfrac{15}{4}\) =0 (vô nghiệm)
Vậy tập nghiệm S=(1;-2)