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a) \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6}{1-x}\)
\(=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(x-1\right)\left(2x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{4x^2-3x+17+2x^2-x-2x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=-\frac{12}{x^2+x+1}\)
b) \(\frac{1}{x^2-x+1}-\frac{x^2+2}{x^3+1}+1=\frac{x+1-x^2-2+x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x-x^2+x^3}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x}{x+1}\)
c) \(N=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac+abc^2+abc}\)
\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac\left(1+bc+b\right)}\)
\(N=\frac{1+b}{b+1+bc}+\frac{bc}{1+bc+b}\)
\(N=\frac{1+b+bc}{b+1+bc}\)
\(N=1.\)
\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
Bài làm
j) \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\) ĐKXĐ: \(x\ne\pm5\)
\(\Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}=\frac{20}{x^2-25}\)
\(\Rightarrow x^2+10x+25-x^2+10x-25=20\)
\(\Leftrightarrow20x=20\)
\(\Leftrightarrow x=1\)
Vậy x = 1 là nghiệm phương trình.
k) \(\frac{3}{x-4}+\frac{5x-2}{x^2-16}=\frac{4}{x+4}\)
\(\Leftrightarrow\frac{3\left(x+4\right)}{x^2-16}+\frac{5x-2}{x^2-16}=\frac{4\left(x-4\right)}{x^2-16}\)
\(\Rightarrow3x+12+5x-2=4x-16\)
\(\Leftrightarrow4x=-26\)
<=> \(x=-\frac{13}{2}\)
Vậy x = -13/2 là nghiệm phương trình.
l) \(\frac{2x-1}{3}-\frac{5x+2}{4}=2x\)
\(\Leftrightarrow4x-4-15x-6=24x\)
\(\Leftrightarrow-35x=10\)
\(\Leftrightarrow x=-\frac{2}{7}\)
Vậy x = -2/7 là nghiệm phương trình.
Bài làm
2 - x = 3x + 1
<=> - x - 3x = -2 + 1
<=> -4x = -1
<=> x = 1/4
Vậy x = 1/4 là nghiệm phương trình.
4x + 7( x - 2 ) = -9x + 5
<=> 4x + 7x - 14 = -9x + 5
<=> 4x + 7x + 9x = 14 + 5
<=> 20x = 19
<=> x = 19/20
Vậy x = 19/20 là nghiệm phương trình.
5x - 2( 3x - 5 ) = 7x + 11
<=> 5x - 6x + 10 = 7x + 11
<=> 5x - 6x - 7x = 11 - 10
<=> -8x = -21
<=> x = 21/8
Vậy x = 21/8 là nghiệm phương trình.
( 5x + 2 )( x - 7 ) = 0
<=> \(\left[{}\begin{matrix}5x+2=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{5}\\x=7\end{matrix}\right.\)
Vậy tập nghiệm phương trình S = { -2/5; 7 }
2x( x - 5 ) + 3( x - 5 ) = 0
<=> ( 2x + 3 )( x - 5 ) = 0
<=> \(\left[{}\begin{matrix}2x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=5\end{matrix}\right.\)
Vậy tập nghiệm phương trìh S = { -3/2; 5 }
\(\frac{5x-3}{6}=\frac{-2x+5}{9}\)
\(\Rightarrow6\left(-2x+5\right)=9\left(5x-3\right)\)
\(\Leftrightarrow-12x+30=45x-27\)
\(\Leftrightarrow-57x=-57\)
\(\Leftrightarrow x=1\)
Vậy x = 1 là nghiệm phương trình.
\(\frac{x}{3}-\frac{2x+1}{2}=\frac{5x}{6}\)
\(\Leftrightarrow2x-3\left(2x+1\right)=5x\)
\(\Leftrightarrow2x-6x-3=5x\)
\(\Leftrightarrow-9x=3\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy x = -1/3 là nghiệm phương trình.
\(\frac{x}{3}-\frac{2x+1}{2}=\frac{x}{6}-x\)
\(\Leftrightarrow2x-3\left(2x+1\right)=x-6x\)
\(\Leftrightarrow2x-6x-3=x-6x\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy x = 3/2 là nghiệm phương trình.
\(\frac{3}{x+1}=\frac{5}{2x+2}\) ĐKXĐ: x khác 1
<=> \(\frac{6}{2x+2}=\frac{5}{2x+2}\)( vô lí )
Vậy phương trình trên vô nghiệm.
# Học tốt #
Lời giải:
ĐKXĐ: $x\neq \pm 2; x\neq -3$
Ta có:
$M=1+\frac{x+3}{(x+2)(x+3)}:\left(\frac{8x}{4x^2(x-2)}-\frac{3x}{3(x-2)(x+2)}-\frac{1}{x+2}\right)$
$=1+\frac{1}{x+2}:\left(\frac{2}{x(x-2)}-\frac{x}{(x-2)(x+2)}-\frac{1}{x+2}\right)$
$=1+\frac{1}{x+2}:\frac{2(x+2)-x^2-x(x-2)}{x(x-2)(x+2)}$
$=1+\frac{1}{x+2}:\frac{-2(x^2-2x-2)}{x(x-2)(x+2)}$
$=1-\frac{x(x-2)}{2x^2-4x-4}=\frac{x^2-2x-4}{2x^2-4x-4}$
a/ ĐKXĐ ....
A=\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
=\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+...+\frac{1}{x-5}-\frac{1}{x-4}\)
=\(\frac{1}{x}-\frac{1}{x-5}\)
=\(-\frac{5}{x^2-5x}\)
b/ \(x^3-x+2=0\Leftrightarrow\left(x+1\right)\left(\left(x-1\right)^2+1\right)=0\)
<=> x=-1, thay vào tính nốt
\(\text{ĐKXĐ:}\hept{\begin{cases}x\ne1\\x\ne2\\x\ne3\end{cases}}\)
\(\frac{x+2}{\left(x-2\right)\left(x-3\right)}+\frac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}:\frac{2x-2-x}{x-1}\)
\(=\frac{x+2+x^2-9+x^2-4}{\left(x-2\right)\left(x-3\right)}.\frac{x-1}{x-2}=\frac{2x^2+x-11}{\left(x-2\right)\left(x-3\right)}\cdot\frac{x-1}{x-2}=\frac{\left(x-1\right)\left(2x^2+x-11\right)}{\left(x-2\right)^2\cdot\left(x-3\right)}\)