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d) \(\dfrac{x+1}{x-1}-\dfrac{x+2}{x+3}+\dfrac{4}{x^2+2x-3}=0\) (ĐKXĐ: \(x\ne1;-3\))
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)-\left(x+2\right)\left(x-1\right)+4}{\left(x+3\right)\left(x-1\right)}=0\)
\(\Rightarrow\left(x+1\right)\left(x+3\right)-\left(x+2\right)\left(x-1\right)+4=0\)
\(\Leftrightarrow x^2+4x+3-x^2-x+2+4=0\)
\(\Leftrightarrow3x+9=0\Leftrightarrow x=-3\left(loại\right)\)
vậy phương trình đã cho vô nghiệm
c)\(\dfrac{2}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{x}{x^2+x+1}\) (ĐKXĐ: \(x\ne1\))
\(\Leftrightarrow\dfrac{2\left(x^2+x+1\right)-3x^2}{x^3-1}=\dfrac{x\left(x-1\right)}{x^3-1}\)
\(\Rightarrow2x^2+2x+2-3x^2=x^2-x\)
\(-2x^2+3x+2=0\)
\(\left(x-2\right)\left(-2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\\-2x-1=0\Leftrightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
vậy tập nghiệm của phương trình là S={2;-0,5)
b: \(\Leftrightarrow\dfrac{2}{\left(x+7\right)\left(x-3\right)}=\dfrac{3x+21}{\left(x-3\right)\left(x+7\right)}\)
=>3x+21=2
=>x=-19/3
d: \(\Leftrightarrow\left(2x+1\right)^2-\left(2x-1\right)^2=8\)
\(\Leftrightarrow4x^2+4x+1-4x^2+4x-1=8\)
=>8x=8
hay x=1
a) Đk : \(x\ne0;\ne1\)
\(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}\)
\(\Rightarrow\dfrac{x^2+3x}{x\left(x+1\right)}+\dfrac{x^2-x-2}{x\left(x+1\right)}-\dfrac{2x^2+2x-2}{x\left(x+1\right)}=0\)
\(\Rightarrow\dfrac{x^2+3x+x^2-x-2-2x^2-2x+2}{x\left(x-1\right)}=0\)
\(\Rightarrow\dfrac{0}{x-1}=0\)
=> Phương trình có vô số nghiệm x
b) Đk : \(x\ne2;x\ne3\)
\(\dfrac{2}{x-2}-\dfrac{x}{x+3}=\dfrac{5x}{\left(x-2\right)\left(x+3\right)}-1\)
\(\Rightarrow\dfrac{2x+6}{\left(x-2\right)\left(x+3\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+3\right)}-\dfrac{5x}{\left(x-2\right)\left(x+3\right)}+\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}\)
=0
\(\Rightarrow\dfrac{2x+6-x^2+2x-5x+x^2+x+6}{\left(x-2\right)\left(x+3\right)}=0\)
\(\Rightarrow\dfrac{12}{\left(x-2\right)\left(x+3\right)}=0\)
=> Phương trình vô nghiệm
c)
\(\Leftrightarrow\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+x+1}{x^4+x^2+1}-\dfrac{1-2x}{x^4+x^2+1}=0\)
\(\Rightarrow\dfrac{x^2-x+1-x^2-x-1-1+2x}{x^4+x^2+1}=0\)
\(\Rightarrow\dfrac{-1}{x^4+x^2+1}=0\)
=> PTVN
d) Thôi tự làm đi, câu này dễ :Vvv
e)
\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)\)=40
\(\Rightarrow\left[\left(x+1\right)\left(x+5\right)\right]\cdot\left[\left(x+2\right)\left(x+4\right)\right]=40\)
\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt
\(x^2+6x+7=t\)
Phương trình tương đương
\(\left(t-1\right)\left(t+1\right)=40\)
\(t^2=41\)
\(\)\(t=\pm\sqrt{41}\)
Thay vào tìm x.
a) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\dfrac{x}{x+2}=\dfrac{x^2+4}{x^2-4}\)
\(\Leftrightarrow\dfrac{x}{x+2}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow x\left(x-2\right)=x^2+4\)
\(\Leftrightarrow x^2-2x=x^2+4\)
\(\Leftrightarrow-2x=4\Leftrightarrow x=-2\)(KTMĐK)
Vậy phương trình vô nghiệm
b) ĐKXĐ: \(x\ne3;x\ne-1\)
Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{2.2x}{2\left(x+1\right)\left(x-3\right)}=0\)
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)-2.2x=0\)
\(\Leftrightarrow x^2+x+x^2-3x-4x=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=3\left(KTMĐK\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=0\)
1)
\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
\(\Leftrightarrow\dfrac{x-5}{100}+1+\dfrac{x-4}{101}+1+\dfrac{x-3}{102}+1=\dfrac{x-100}{5}+1+\dfrac{x-101}{4}+1+\dfrac{x-102}{3}+1\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)
\(\Leftrightarrow x=105\)
b)
\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)
\(\Leftrightarrow50-x=0\)
\(\Leftrightarrow x=50\)
2)
\(\left(5x+1\right)^2=\left(3x-2\right)^2\)
\(\Leftrightarrow\left|5x+1\right|=\left|3x-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=3x-2\\5x+1=-3x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)
b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)
\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)
\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)
\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)
\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
a) \(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
\(=\dfrac{5x-1-5x+7}{3x+2-3x+1}\)
\(=\dfrac{-1+7}{2+1}\)
\(=\dfrac{6}{3}\)
\(=2\)
Với \(\dfrac{5x-1}{3x+2}=2\)
\(\Rightarrow5x-1=2\left(3x+2\right)\)
\(\Rightarrow5x-1-2\left(3x+2\right)=0\)
\(\Rightarrow5x-1-6x-4=0\)
\(\Rightarrow-x-5=0\)
\(\Rightarrow x=-5\)
a) \(\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{16}{x^2-1}\)
=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)
=>\(x^2+2x+1-x^2+2x-1=16\)
=>4x=16=>x=4
b)\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)
=>\(\dfrac{12}{x^2-4}-\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\)
=>\(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)
=>\(12-x^2-3x-2+x^2+5x-14=0\)
=>2x-4=0=>2x=4=>x=2
c)\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)
=>\(\dfrac{12}{8+x^3}=\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^3+8}\)
=>\(12=x^3+8+x^2-2x+4\)
=>\(x^3+x^2-2x=0\)
=>\(x^3-x+x^2-x=0\)
Ta có: 8\(\left(x+\dfrac{1}{x}\right)^2\)+4\(\left(x^2+\dfrac{1}{x^2}\right)^2\)\(\left(x+\dfrac{1}{x}\right)^2\)=(x+4)2
ĐKXĐ: x khác 0
<=>8\(\left(x+\dfrac{1}{x}\right)^2\)+4\(\left(x^2+\dfrac{1}{x^2}\right)\)\(\left(x^2+\dfrac{1}{x^2}-x^2-2-\dfrac{1}{x^2}\right)\)=(x+4)2
<=>8\(\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)
<=>8\(\left(x^2+2+\dfrac{1}{x^2}-x^2-\dfrac{1}{x^2}\right)\)=(x+4)2
=>(x+4)2=16
Vậy có 2 TH:
+) x+4=4 => x=0(KTMĐKXĐ)
+)x+4=-4 => x=-8(TMĐKXĐ)
Vậy tập nghiệm của phương trình S={-8}
ĐKXĐ: \(x\ne\pm1\)
\(\dfrac{x+1}{X-1}-\dfrac{x-1}{X+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)(loại)
Vậy phương trình vô nghiệm