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d, 2x2-5x-3 = 0
\(\Leftrightarrow\)2x2-6x+x-3= 0
\(\Leftrightarrow\)(2x2-6x) +(x-3) = 0
\(\Leftrightarrow\)2x(x-3) + (x-3) = 0
\(\Leftrightarrow\)(x-3) (2x+1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S =\(\left\{3;\frac{-1}{2}\right\}\)
a)
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\\ \Leftrightarrow\frac{201-x}{99}+\frac{99}{99}+\frac{203-x}{97}+\frac{97}{97}+\frac{205-x}{95}+\frac{95}{95}+4=4\\ \Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\) (*)
Do \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)\ne0\)
nên (*) \(\Leftrightarrow300-x=0\\ \Leftrightarrow x=300\)
b)
\(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\\ \Leftrightarrow\frac{2-x}{2002}+\frac{2002}{2002}-1+1=\frac{1-x}{2003}+\frac{2003}{2003}-\frac{x}{2004}+\frac{2004}{2004}\\ \Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\\ \Leftrightarrow\frac{2004-x}{2002}-\frac{2004-x}{2003}+\frac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\) (*)
Do \(\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)\ne0\)
nên (*) \(\Leftrightarrow2004-x=0\)
\(\Leftrightarrow x=2004\)
c) \(\left|2x-3\right|=2x-3\) (1)
ĐKXĐ: \(\\ 2x-3\ge0\)
\(\Leftrightarrow x\ge\frac{3}{2}\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}2x-3=2x-3\\2x-3=-2x+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}0x=0\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\forall x\in R\\x=\frac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\frac{3}{2}\right\}\)
\(\frac{-5}{9}x+1=\frac{2}{3}x-10\)
\(\frac{-5}{9}x+\frac{9}{9}=\frac{6}{9}x-\frac{90}{9}\)
\(-5x+9=6x-90\)
\(-5x-6x=-90-9\)
\(-11x=-99\)
\(x=\frac{-99}{-11}=9\)
b. \(\frac{x-22}{8}+\frac{x-21}{9}+\frac{x-20}{10}+\frac{x-19}{11}=4\)
\(\frac{x-22}{8}-1+\frac{x-21}{9}-1+\frac{x-20}{10}-1+\frac{x-19}{11}-1=0\)
\(\frac{x-30}{8}+\frac{x-30}{9}+\frac{x-30}{10}+\frac{x-30}{11}=0\)
\(\left(x-30\right)\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)=0\)
x=30
Chúc bạn học tốt!!
Giải các pt sau:
a) (x+4)(2x-3)=0
TH1: x+4=0 => x=-4
TH2 : 2x-3=0 => 2x=3 =>x=3/2
Bài 1:
a) \(\frac{2x+1}{3}-\frac{x}{4}=2\)
\(\Leftrightarrow\frac{4\left(2x+1\right)}{12}-\frac{3x}{12}-\frac{24}{12}=0\)
\(\Leftrightarrow8x+4-3x-24=0\)
\(\Leftrightarrow5x-20=0\)
\(\Leftrightarrow5x=20\)
\(\Leftrightarrow x=4\)
Vậy \(S=\left\{4\right\}\)
b) \(\frac{2x+5}{2x}-\frac{x}{x+5}=0\)
ĐKXĐ: \(x\ne0;x\ne-5\)
\(\Leftrightarrow\frac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\frac{x\left(2x+5\right)}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow2x^2+10x+5x+25-2x^2-5x=0\)
\(\Leftrightarrow10x+25=0\)
\(\Leftrightarrow10x=-25\)
\(\Leftrightarrow x=-\frac{5}{2}\left(TM\right)\)
Vậy \(S=\left\{-\frac{5}{2}\right\}\)
#Học tốt!
a) \(\left(2x+1\right)\left(3x-2\right)=\left(2x+1\right)\left(5x-8\right)\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2\right)-\left(2x+1\right)\left(5x-8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=0\\6-2x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-0,5\\x=3\end{cases}}\)
Vậy...
b) \(ĐKXĐ:\) \(x\ne-2;\) \(x\ne4\)
\(\frac{3}{x+2}+\frac{2}{x-4}=0\)
\(\Leftrightarrow\)\(\frac{3\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{3x-12+2x+4}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{5x-8}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Rightarrow\)\(5x-8=0\)
\(\Leftrightarrow\)\(x=\frac{8}{5}\) (T/m đkxđ)
Vậy...
c) \(x^3+4x^2+4x+3=0\)
\(\Leftrightarrow\)\(x^3+3x^2+x^2+3x+x+3=0\)
\(\Leftrightarrow\)\(x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\)\(x+3=0\) (do \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) \(\forall x\))
\(\Leftrightarrow\)\(x=-3\)
Vậy...
a) \(\frac{x}{x+1}-\frac{2x-3}{x-1}=\frac{2x+3}{x^2-1}\) \(\left(ĐKXĐ:x\ne\pm1\right)\)
\(\Leftrightarrow x\left(x-1\right)-\left(2x-3\right)\left(x+1\right)=2x+3\)
\(\Leftrightarrow x^2-x-2x^2-2x+3x+3=2x+3\)
\(\Leftrightarrow-x^2-2x=0\Leftrightarrow-x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
b) \(\frac{x-1}{x}-\frac{x-2}{x+1}=2\) \(\left(ĐKXĐ:x\ne0;x\ne-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-x\left(x-2\right)=2x\left(x+1\right)\)
\(\Leftrightarrow x^2-1-x^2+2x=2x^2+2x\)
\(\Leftrightarrow2x^2=-1\left(\text{vô lí}\right)\)
Vậy phương trình vô nghiệm.
câu a
x/3 +20 =x/2
x/2 - x/3 = 20
(3x-2x)/6 = 20
x/6 = 20
x = 20*6
x=120
câu b
x/(x-1) + 2x/x*x = 0 (x khác 0 ,1)
(x*x*x + 2x *(x-1)) / (x-1) * x*x = 0
x*x*x + 2*x*x - 2*x = 0
x*(x*x + 2*x -2 ) =0
x=0 hoặc x*x+2*x-2=0
x=0 hoặc (x*x + 2x + 1)-3 =0
x=0 hoặc (x + 1)*(x+1)=3
x=0 hoặc x+1 = căn 3 hoặc x=âm căn3
x=0 hoặc x =căn 3 trừ 1 hoặc x = âm căn 3 trừ một