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Bài 1.
a) ( x - 3 )( x + 7 ) = 0
<=> x - 3 = 0 hoặc x + 7 = 0
<=> x = 3 hoặc x = -7
Vậy S = { 3 ; -7 }
b) ( x - 2 )2 + ( x - 2 )( x - 3 ) = 0
<=> ( x - 2 )( x - 2 + x - 3 ) = 0
<=> ( x - 2 )( 2x - 5 ) = 0
<=> x - 2 = 0 hoặc 2x - 5 = 0
<=> x = 2 hoặc x = 5/2
Vậy S = { 2 ; 5/2 }
c) x2 - 5x + 6 = 0
<=> x2 - 2x - 3x + 6 = 0
<=> x( x - 2 ) - 3( x - 2 ) = 0
<=> ( x - 2 )( x - 3 ) = 0
<=> x - 2 = 0 hoặc x - 3 = 0
<=> x = 2 hoặc x = 3
Đề ntn hả bạn: \(\frac{1}{x^2-x}\)+\(\frac{1}{x^2+x}\)+\(\frac{1}{x^2+3x}\)+ 2 = \(\frac{3}{4}\)?
\(x-5=\frac{1}{3\left(x+2\right)}\left(đkxđ:x\ne-2\right)\)
\(< =>3\left(x-5\right)\left(x+2\right)=1\)
\(< =>3\left(x^2-3x-10\right)=1\)
\(< =>x^2-3x-10=\frac{1}{3}\)
\(< =>x^2-3x-\frac{31}{3}=0\)
giải pt bậc 2 dễ r
\(\frac{x}{3}+\frac{x}{4}=\frac{x}{5}-\frac{x}{6}\)
\(< =>\frac{4x+3x}{12}=\frac{6x-5x}{30}\)
\(< =>\frac{7x}{12}=\frac{x}{30}< =>12x=210x\)
\(< =>x\left(210-12\right)=0< =>x=0\)
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
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a) Lập bảng xét dấu
x 0 1 2
x - 0 + | + | +
x - 1 - | - 0 + | +
x - 2 - | - | - | +
Xét các TH xảy ra
TH1: x \(\le\)0 => pt trở thành: -x - 2(1 - x) + 3(2 - x) = 4
<=> - x - 2 + 2x + 6 - 3x = 4 <=> -2x = 4 - 4 <=> -2x = 0 <=> x = 0 (tm)
TH2: 0 < x \(\le\)1 => pt trở thành: x - 2(1 - x) + 3(2 - x) = 4
<=> x - 2 + 2x + 6 - 3x = 4 <=> 4 = 4 (luôn đúng)
TH3: 1 < x \(\le\)2 => pt trở thành: x - 2(x - 1) + 3(2 - x) = 4
<=> x - 2x + 2 + 6 - 3x = 4 <=> -4x = 4 - 8 <=> -4x = -4 <=> x = 1 (ktm)
TH4: x > 2 => pt trở thành: x - 2(x - 1) + 3(x - 2) = 4
<=> x - 2x + 2 + 3x - 6 = 4 <=> 2x = 4 + 4 <=> 2x = 8 <=> x = 4 (tm)
Vậy ....
Bài 1 :
\(\frac{4x-5}{x-1}=\frac{2+x}{x-1}\)ĐK : x \(\ne\)1
\(\Leftrightarrow\frac{4x-5}{x-1}-\frac{2-x}{x-1}=0\Leftrightarrow\frac{4x-5-2+x}{x-1}=0\)
\(\Rightarrow5x-7=0\Leftrightarrow x=\frac{7}{5}\)( tmđk )
Vậy tập nghiệm của phuwong trình là S= { 7/5 }
b, \(\frac{x-1}{x-2}-3+x=\frac{1}{x-2}\)ĐK : x \(\ne\)2
\(\Leftrightarrow\frac{x-1}{x-2}-\left(3-x\right)=\frac{1}{x-2}\)
\(\Leftrightarrow\frac{x-1}{x-2}-\frac{\left(3-x\right)\left(x-2\right)}{x-2}=\frac{1}{x-2}\)
\(\Leftrightarrow\frac{x-1-3x+6+x^2-2x-1}{x-2}=0\)
\(\Rightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)( ktmđkxđ )
Vậy phương trình vô nghiệm
c, \(1+\frac{1}{2+x}=\frac{12}{x^3+8}\)ĐK : x \(\ne\)-2
\(\Leftrightarrow\frac{\left(x+2\right)\left(x^2-2x+4\right)+x^2-2x+4-12}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)
\(\Rightarrow x^3+8+x^2-2x+4-12=0\)
\(\Leftrightarrow x^3+x^2-2x=0\Leftrightarrow x\left(x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\Leftrightarrow x=0;x=1;x=-2\left(ktm\right)\)
Vậy tập nghiệm của phương trình là S = { 0 ; 1 }
d, đưa về dạng hđt
Bài 2 : làm tương tự, chỉ khác ở chỗ mẫu số phức tạp hơn tí thôi
ta có :
\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)
\(\Leftrightarrow\left|x-1\right|\left|x+1\right|-\left|x-1\right|-\left|x+1\right|+1=0\)
\(\Leftrightarrow\left(\left|x-1\right|-1\right)\left(\left|x+1\right|-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=1\\\left|x+1\right|=1\end{cases}}\)
\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)