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\(\frac{3x}{x-1}-\frac{2x}{x-3}+\frac{4x}{\left(x-1\right)\left(x-3\right)}=0\)đkxd \(x\ne1;3\)
\(\Leftrightarrow3x^2-9x-2x^2-2x+4x=0\)
\(\Leftrightarrow x^2-7x=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-7=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=7\end{cases}\left(tm\right)}\)
\(\frac{3x}{x-1}-\frac{2x}{x-3}+\frac{4x}{\left(x-1\right)\left(x-3\right)}=0\)\(ĐKXĐ:x\ne1;3\)
\(3x\left(x-3\right)\left(x+3\right)-2x\left(x-1\right)\left(x+3\right)+4x\left(x-3\right)=0\)
\(x^3-33x=0\)
\(x\left(x^2-33\right)=0\)
\(x=0;\pm\sqrt{33}\)
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)=1/3
<=>1/x-1/x+1+1/x+1-1/x+2+1/x+2-1/x+3+1/x+3-1/x+4=1/3
<=>1/x-1/x+4=1/3
<=>x+4/x(x+4)-x/x(x+4) ( quy dong mau ) =1/3
<=>4/x(x+4)=1/3
<=> 4.3=x(x+4) ( nhan cheo )
<=> x(x+4)=12
<=> x^2+4x-12=0
<=>x^2-2x+6x-12=0
<=>x(x-2) + 6(x-2) =0
<=> (x-2)(x+6)=0
<=> x-2 =0 hoac x +6=0
<=>x=2 hoac x= -6
Vay x thuoc ( 2,-6 )
K mk nha !!
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x\text{+}2\right)}\text{+}\frac{1}{\left(x\text{+}2\right)\left(x\text{+}3\right)}+\frac{1}{\left(x\text{+}3\right)\left(x\text{+}4\right)}=\frac{1}{3}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x\text{+}1}\text{+}\frac{1}{x\text{+}1}-\frac{1}{x\text{+}2}\text{+}.....\text{+}\frac{1}{x\text{+}3}-\frac{1}{x\text{+}4}=\frac{1}{3}\)
\(\Rightarrow\)\(\frac{1}{x}-\frac{1}{x\text{+}4}=\frac{1}{3}\)
\(\Rightarrow\frac{x\text{+}4}{x\left(x\text{+}4\right)}-\frac{x}{x\left(x\text{+}4\right)}=\frac{1}{3}\)
\(\Rightarrow\frac{4}{x\left(x\text{+}4\right)}=\frac{1}{3}\)
\(\Rightarrow\frac{4}{x\left(x\text{+}4\right)}=\frac{4}{12}\)
\(\Rightarrow x\left(x\text{+}4\right)=12\)
mà x và x+4 cách nhau 4 đơn vị \(\Rightarrow x=2\)và x+4\(=\)6
Vậy \(x=2\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
\(=\frac{1}{x}\)
ta có: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
=\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
= \(\frac{1}{x}\)
d) \(\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+3\right)}=\frac{1}{\left(x+2\right)\left(x+3\right)}\)
ĐKXĐ : \(x\ne-2;x\ne-3\)
\(\Leftrightarrow x+3+x+2=1\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-2\) (không nhận)
Vậy : \(S=\varnothing\)
Giai phương trình sau :
a) \(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\)
ĐKXĐ : \(x\ne1;x\ne-5\)
Với điều kiện trên ta có :
\(\Leftrightarrow\)\(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{-3}{x-1}=\frac{5}{x+5}\)
\(\Leftrightarrow10-3\left(x+5\right)=5\left(x-1\right)\)
\(\Leftrightarrow10-3x-15=5x-5\)
\(\Leftrightarrow-8x=0\)
\(\Leftrightarrow x=0\) (nhận)
Vậy : \(S=\left\{0\right\}\)
\(\frac{x+1}{x-3}-\frac{1}{x-1}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow x^2-1-x+3-2=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(\frac{x+1}{x-3}-\frac{1}{x-1}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
điều kiện: \(x\ne3;1\)
quy đồng mẫu hai phân số
\(\frac{\left(x+1\right)\left(x-1\right)-\left(x-3\right)}{\left(x-3\right)\left(x-1\right)}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
\(\frac{x^2-1-x+3}{\left(x-3\right)\left(x-1\right)}-\frac{2}{\left(x-1\right)\left(x-3\right)}=0\)
\(\frac{x^2-1-x+3-2}{\left(x-1\right)\left(x-3\right)}=0\)
\(\frac{x^2-x}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Rightarrow x^2-x=0\)
\(x\left(x-1\right)=0\)
vây x = 0
hoặc x = 1 (không thỏa điều kiện)
vậy x = 0