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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)⇔ 2x-1/2 -1 = x2+x-3/x-1 - 5x-2/2.(x-1)
⇔ ( 2x-1 ).(x-1) - 2.(x-1)=2.(x2+x-3) - (5x-2)
⇔2x2-3x+1-2x+2=2x2+2x-6-5x+2
⇔2x2-3x+1-2x+2-2x2-2x+6+5x-2=0
⇔-2x+7=0
⇔x=7/2
Vậy ....
b) ⇔3.(x-1)2-(x-1).(x+1)=0
⇔ (x-1).(3x-3-x-1)=0
⇔ (x-1).(2x-4)=0
⇔x=1 hoặc x=2
Vậy....
c) ⇔ 4x2-4x+x-1=0
⇔4x(x-1)+(x-1)=0
⇔(x-1)(4x+1)=0
⇔x=1 hoặc x=-1/4
Vậy....
d) ⇔4x2-4x-3=0
⇔ 4x2-6x+2x-3 = 0
⇔ 2x( 2x-3)+(2x-3)=0
⇔ (2x+3)(2x+1)=0
⇔ x=-3/2 hoặc x=-1/2
vậy ....
\(a,\frac{2x-1}{2}-1=\frac{x^2+x-3}{x-1}-\frac{5x-2}{2-2x}ĐKXĐ:x\ne1\)
\(\left(2x-1\right)\left(x-1\right)\left(1-x\right)-2\left(x-1\right)\left(1-x\right)=2\left(x^2+x-3\right)\left(1-x\right)-\left(5x-2\right)\left(x-1\right)\)
\(7x^2-8x+3=-5x^2+15x-8\)
\(7x^2-8x+3+5x^2-15x+8=0\)
\(12x^2-23x+11=0\)
\(\left(12x-11\right)\left(x-1\right)=0\)
\(\left[{}\begin{matrix}12x=11\\x=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\frac{11}{12}\\x=1\end{matrix}\right.\)Theo ĐKXĐ => x= \(\frac{11}{12}\)
Bài 2 :
a, Ta có : \(\left(x+4\right)\left(x-1\right)=0\)
=> \(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
b, Ta có : \(\left(3x-2\right)\left(4x-7\right)=0\)
=> \(\left[{}\begin{matrix}3x-2=0\\4x-7=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}3x=2\\4x=7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{7}{4}\end{matrix}\right.\)
c, Ta có : \(\left(x+5\right)\left(x^2+1\right)=0\)
=> \(\left[{}\begin{matrix}x+5=0\\x^2+1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-5\\x^2+1=0\left(VL\right)\end{matrix}\right.\)
d, Ta có : \(x\left(x-1\right)\left(x^2+4\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\x-1=0\\x^2+4=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=1\\x^2+4=0\left(VL\right)\end{matrix}\right.\)
e, Ta có : \(\left(3x+2\right)\left(x+\frac{1}{2}\right)=0\)
=> \(\left[{}\begin{matrix}3x+2=0\\x+\frac{1}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{1}{2}\end{matrix}\right.\)
f, Ta có : \(\left(x+2\right)\left(x+3\right)\left(x^2+7\right)=0\)
=> \(\left[{}\begin{matrix}x+2=0\\x-3=0\\x^2+7=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-2\\x=3\\x^2+7=0\left(VL\right)\end{matrix}\right.\)
Bài 1 :
a, Ta có : \(1-\frac{x+3}{4}-\frac{x-2}{6}=0\)
=> \(\frac{12}{12}-\frac{3\left(x+3\right)}{12}-\frac{2\left(x-2\right)}{12}=0\)
=> \(12-3\left(x+3\right)-2\left(x-2\right)=0\)
=> \(12-3x-9-2x+4=0\)
=> \(-5x=-7\)
=> \(x=\frac{7}{5}\)
a đkxđ khi x khác 2 và -2 \(\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)^2-\left(x-2\right)^2}{x^2-4}=\frac{4}{x^2-4}\)
\(\Rightarrow\left(x+2\right)^2-\left(x-2\right)^2=4\)\(\Rightarrow\left(x+2-x+2\right)\left(x+2+x-2\right)=4\Rightarrow4\cdot2x=4\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)(thảo mãn)
b đkxđ khi x+3 khác 0 suy ra x khác -3
\(\frac{x^2-9}{x+3}=\frac{\left(x-3\right)\left(x+3\right)}{x+3}=x-3=0\Rightarrow x=3\)(thảo mãn)
Câu c : \(x^4-3x^3+2x^2-9x+9=0\)
<=>\(x^4-x^3-2x^3+2x^2-9x+9=0\)
<=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)-9\left(x-1\right)=0\)
<=>\(\left(x-1\right)\left(x^3-2x^2-9\right)=0\)
<=> \(x-1=0\) hoặc \(x^3-2x^2-9=0\)
Nếu x-1=0 <=> x=1
Nếu \(x^3-2x^2-9=0\)
<=> \(x^3-3x^2+x^2-9=0\)
<=>\(x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)=0\)
<=>\(\left(x-3\right)\left(x^2+x+3\right)=0\)
Vì \(x^2+x+3=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\) >0 nên x-3=0 <=> x=3
Vậy \(S=\left\{1;3\right\}\)
Câu b : \(x^2+\left(\frac{x}{x+1}\right)^2=\frac{5}{4}\)
<=> \(4x^2\left(x^2+2x+2\right)=5\left(x^2+2x+1\right)\)
<=> \(4x^4+8x^3+8x^2=5x^2+10x+5\)
<=>\(4x^4+8x^3+3x^2-10x-5=0\)
<=>\(4x^4-4x^3+12x^3-12x^2+15x^2-15x+5x-5=0\)
<=>\(\left(x-1\right)\left(4x^3+12x^2+15x+5\right)=0\)
<=>\(\left(x-1\right)\left(2x+1\right)\left(2x^2+5x+5\right)=0\)
<=>x=1 hoặc \(x=\frac{-1}{2}\)
Phương trình \(2x^2+5x+5=0\) Vô nghiệm
@Nguyễn Lê Phước Thịnh bạn có thể chỉ chỗ mình sai sót được không ạ? Mình mò không ra ._.
a) ĐKXĐ: x≠0
Ta có: \(\frac{9}{x}+2=-6\)
⇔\(\frac{9}{x}+2+6=0\)
⇔\(\frac{9}{x}+8=0\)
⇔\(\frac{9}{x}+\frac{8x}{x}=0\)
⇔9+8x=0
⇔8x=-9
hay \(x=-\frac{9}{8}\)
Vậy: \(x=-\frac{9}{8}\)
b) ĐKXĐ: x≠0;x≠-1;x≠-3
Ta có: \(\frac{7}{x+1}+\frac{-18x}{x\left(x^2+4x+3\right)}=\frac{-4}{x+3}\)
⇔\(\frac{7}{x+1}+\frac{-18x}{x\left(x+1\right)\left(x+3\right)}-\frac{-4}{x+3}=0\)
⇔\(\frac{7x\left(x+3\right)}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}+\frac{-18x}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}-\frac{-4x\left(x+1\right)}{\left(x+3\right)\cdot x\cdot\left(x+1\right)}=0\)
⇔\(7x^2+21x-18x+4x\left(x+1\right)=0\)
\(\Leftrightarrow7x^2+21x-18x+4x^2+4x=0\)
⇔\(11x^2+7x=0\)
\(\Leftrightarrow x\left(11x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\11x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\11x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\frac{-7}{11}\end{matrix}\right.\)
Vậy: \(x=\frac{-7}{11}\)
c) ĐKXĐ: x≠1; x≠-3
Ta có: \(\frac{3x-1}{x-1}-1=\frac{2x+5}{x+3}+\frac{4}{x^2-2x+3}\)
⇔\(\frac{3x-1}{x-1}-1-\frac{2x+5}{x+3}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
⇔\(\frac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
⇔\(\left(3x-1\right)\left(x+3\right)-\left(x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)-4=0\)
\(\Leftrightarrow3x^2+9x-x-3-\left(x^2+3x-x-3\right)-\left(2x^2-2x+5x-5\right)-4=0\)
\(\Leftrightarrow3x^2+8x-3-\left(x^2+2x-3\right)-\left(2x^2+3x-5\right)-4=0\)
\(\Leftrightarrow3x^2+8x-3-x^2-2x+3-2x^2-3x+5-4=0\)
\(\Leftrightarrow3x+1=0\)
\(\Leftrightarrow3x=-1\)
hay \(x=\frac{-1}{3}\)
Vậy: \(x=\frac{-1}{3}\)
\(a,x^2-10x-39=0\)
\(\Leftrightarrow x^2-10x-39+64=64\)
\(\Leftrightarrow x^2-10x+25=64\)
\(\Leftrightarrow\left(x-5\right)^2=64\)
làm nốt
\(x^2-10x-39=0\Leftrightarrow x^2-13x+3x-39=0\Leftrightarrow x\left(x-13\right)+3\left(x-13\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=13\\x=-3\end{cases}}\)
b) \(\frac{x-5}{2017}+\frac{x-2}{2020}=\frac{x-6}{2016}+\frac{x-68}{1954}\)
\(\Leftrightarrow\)\(\frac{x-5}{2017}-1+\frac{x-2}{2020}-1=\frac{x-6}{2016}-1+\frac{x-68}{1954}-1\)
\(\Leftrightarrow\)\(\frac{x-2022}{2017}+\frac{x-2022}{2020}=\frac{x-2022}{2016}+\frac{x-2022}{1954}\)
\(\Leftrightarrow\)\(\left(x-2022\right)\left(\frac{1}{2017}+\frac{1}{2020}-\frac{1}{2016}-\frac{1}{1954}\right)=0\)
\(\Leftrightarrow\)\(x-2022=0\) (vì 1/2017 + 1/2020 - 1/2016 - 1/1954 \(\ne0\))
\(\Leftrightarrow\)\(x=2022\)
Vậy...
b) \(\frac{x-5}{2017}+\frac{x-2}{2020}=\frac{x-6}{2016}+\frac{x-68}{1954}\)
\(\Leftrightarrow\)\(\frac{x-5}{2017}-1+\frac{x-2}{2020}-1=\frac{x-6}{2016}-1+\frac{x-68}{1954}-1\)
\(\Leftrightarrow\)\(\frac{x-2022}{2017}+\frac{x-2022}{2020}=\frac{x-2022}{2016}+\frac{x-2022}{1954}\)
\(\Leftrightarrow\)\(\left(x-2022\right)\left(\frac{1}{2017}+\frac{1}{2020}-\frac{1}{2016}-\frac{1}{1954}\right)=0\)
\(\Leftrightarrow\)\(x-2022=0\) (vì 1/2017 + 1/2020 - 1/2016 - 1/1954 \(\ne0\))
\(\Leftrightarrow\)\(x=2022\)
Vậy,....
ĐK: \(x\ne\pm2\)
\(pt\Rightarrow x+2+x^2-4x=0\Leftrightarrow x^2-3x+2=0\Leftrightarrow x^2-x-2x+2=0\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=1\left(N\right)\end{matrix}\right.\)
Vậy \(S=\left\{1\right\}\)
phần đầu khử mẫu sao bạn?