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1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~
\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\ne\left\{-4;-3;-2;-1\right\}\)
\(\Leftrightarrow\frac{x^2+x+1}{x+1}-1+\frac{x^2+2x+2}{x+2}-1=\frac{x^2+3x+3}{x+3}-1+\frac{x^2+4x+4}{x+4}-1\)
\(\Leftrightarrow\frac{x^2}{x+1}+\frac{x^2+x}{x+2}-\frac{x^2+2x}{x+3}-\frac{x^2+3x}{x+4}=0\)
\(\Leftrightarrow x\left(\frac{x}{x+1}+\frac{x+1}{x+2}-\frac{x+2}{x+3}-\frac{x+3}{x+4}\right)=0\)
\(\Leftrightarrow x\left(1-\frac{1}{x+1}+1-\frac{1}{x+2}+\frac{1}{x+3}-1+\frac{1}{x+4}-1\right)=0\)
\(\Leftrightarrow x\left(\frac{1}{x+3}+\frac{1}{x+4}-\frac{1}{x+1}-\frac{1}{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x+3}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x+4}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{-2}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+2\right)\left(x+4\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+4\right)+\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow2x^2+10x+11=0\Rightarrow x=\frac{-5\pm\sqrt{3}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có : \(3\left(x-1\right)-2\left(x+3\right)=-15\)
=> \(3x-3-2x-6=-15\)
=> \(3x-3-2x-6+15=0\)
=> \(x=-6\)
Vậy phương trình có nghiệm là x = -6 .
b, Ta có : \(3\left(x-1\right)+2=3x-1\)
=> \(3x-3+2=3x-1\)
=> \(3x-3+2-3x+1=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
c, Ta có : \(7\left(2-5x\right)-5=4\left(4-6x\right)\)
=> \(14-35x-5=16-24x\)
=> \(14-35x-5-16+24x=0\)
=> \(-35x+24x=7\)
=> \(x=\frac{-7}{11}\)
Vậy phương trình có nghiệm là \(x=\frac{-7}{11}\) .
Bài 2 :
a, Ta có : \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)
=> \(\frac{x}{30}+\frac{3\left(5x-1\right)}{30}=\frac{2\left(x-8\right)}{30}-\frac{5\left(2x+3\right)}{30}\)
=> \(x+3\left(5x-1\right)=2\left(x-8\right)-5\left(2x+3\right)\)
=> \(x+15x-3=2x-16-10x-15\)
=> \(x+15x-3-2x+16+10x+15=0\)
=> \(24x+28=0\)
=> \(x=\frac{-28}{24}=\frac{-7}{6}\)
Vậy phương trình có nghiệm là \(x=\frac{-7}{6}\) .
b, Ta có : \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
=> \(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)
=> \(6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)
=> \(6x+24-30x+120=10x-15x+30\)
=> \(6x+24-30x+120-10x+15x-30=0\)
=> \(-19x+114=0\)
=> \(x=\frac{-114}{-19}=6\)
Vậy phương trình có nghiệm là x = 6 .
![](https://rs.olm.vn/images/avt/0.png?1311)
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pt đầu \(\Leftrightarrow x+1+\frac{1}{x+1}+x+7+\frac{7}{x+7}=x+3+\frac{3}{x+3}+x+5+\frac{5}{x+5}\)
\(\Rightarrow\frac{1}{x+1}+\frac{7}{x+7}=\frac{3}{x+3}+\frac{5}{x+5}\\ \Rightarrow\frac{8x+14}{x^2+8x+7}=\frac{8x+30}{x^2+8x+15}\)
\(\Leftrightarrow\left(4x+7\right)\left(x^2+8x+15\right)=\left(4x+15\right)\left(x^2+8x+7\right)\)
Đặt a=4x+7
b=x2 +8x+7
như vậy ta được pt mới có dạng \(a\left(b+8\right)=b\left(a+8\right)\Leftrightarrow ab+8a=ab+8b\Rightarrow a=b\)
hay\(4x+7=x^2+8x+7\Rightarrow x^2+4x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
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\(\frac{5}{x^2+x-6}-\frac{2}{x^2+4x+3}=-\frac{3}{2x-1}\)
<=> \(\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> 5(x + 1)(2x - 1) - 2(x - 2)(2x - 1) = -3(x - 2)(x + 3)(x + 1)
<=> 6x2 + 15x - 9 = -3x3 - 6x2 + 15x + 18
<=> 6x2 - 9 = -3x3 - 6x2 + 18
<=> 6x2 - 9 + 3x3 + 6x2 - 18 = 0
<=> 12x2 - 27 + 3x3 = 0
<=> 3(4x2 - 9 + x3) = 0
<=> 3(x2 + x - 3)(x + 3) = 0
<=> \(\orbr{\begin{cases}x=-3\\x=\frac{-1\pm\sqrt{13}}{2}\end{cases}}\)
DKXD \(x\ne\frac{1}{2};2;-1;3,;-3\)
<=> \(\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
<=> \(\frac{1}{x+3}\left(\frac{5}{x-2}-\frac{2}{x+1}\right)=\frac{-3}{2x-1}\)
<=> \(\frac{1}{x+3}\left(\frac{5x+5-2x+4}{\left(x-2\right)\left(x+1\right)}\right)=\frac{-3}{2x-1}\)
<=> \(\frac{1}{x+3}\left(\frac{3\left(x+3\right)}{\left(x-2\right)\left(x+1\right)}\right)=\frac{3}{1-2x}\)
<=> \(\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{3}{1-2x}\)
<=> \(x^2-x-2=1-2x\)
<=> \(x^2+x-3=0\)
<=> \(\orbr{\begin{cases}x=\frac{-1+\sqrt{13}}{2}\\x=\frac{-1-\sqrt{13}}{2}\end{cases}}\)
chuc ban hoc tot
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a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x^2+2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)
(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)
\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)
\(-x^3-x^2+9x+9=0\)
\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)
\(\left(x+1\right)\left(9-x^2\right)\)=0
(x+1)(3-x)(3+x)=0
*x+1=0 =>x=-1
*3-x=0=>x=3
*3+x=0=>x=-3
\(\frac{x-1}{2}\left(x-2\right)=\frac{\left(x-1\right)}{2}\left(x+2\right)\)
<=> \(\frac{\left(x-1\right)\left(x-2\right)}{2}=\frac{\left(x-1\right)\left(x+3\right)}{2}\)
<=> (x - 1)(x - 2) = (x - 1)(x + 3)
<=> x2 - 2x - x + 2 = x2 + 3x - x - 3
<=> -3x + 2 = 2x - 3
<=> 2 = 2x - 3 + 3x
<=> 2 = 3 + 5x
<=> 2 + 3 = 5x
<=> 5 = 5x
<=> x = 1