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https://www.mathway.com/vi/popular-problems/Algebra/242673
https://www.mathway.com/vi/Algebra
a) \(x^2+2x+7=0\)
\(\Leftrightarrow\left(x+1\right)^2=-6\) ( vô lý )
Vậy pt vô nghiệm
b) \(x^3-x^2-21x+45=0\)
\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x+1\right)^2-4\right]=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
Vậy pt có tập nghiệm \(S=\left\{3,-5\right\}\)
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Bài làm
\(36^2+\frac{1}{x^2}+21x+\frac{7}{2x}-18=0\)
\(\Leftrightarrow\frac{36^2.2.x^2}{2x^2}+\frac{2}{2x^2}+\frac{2.x^2.21x}{2x^2}+\frac{7x}{2x^2}-\frac{2.x^2.18}{2x^2}=0\)
\(\Rightarrow2592x^2+2+42x^3+7x-36x^2=0\)
\(\Leftrightarrow2556x^2+42x^3+7x+2=0\)
tự giải nốt.
Không có cách khác à bạn? Mình làm cách đấy rồi mà thấy nó dài vl luôn nên đăng nên hỏi coi có cách khác không
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pt trên \(< =>1296+\frac{2}{2x^2}+\frac{7x}{2x^2}+21x-18=0\)
\(< =>1278+\frac{7x+2}{2x}+21x=0\)
\(< =>1278+\frac{9}{2}=-21x\)
\(< =>\frac{2565}{2}=-21x\)
\(< =>x=\frac{2565}{-42}=-\frac{855}{14}\)
Ko chắc lắm :P
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a) \(21x+7=15x+35\)
\(\Leftrightarrow21x-15x=35-7\)
\(\Leftrightarrow6x=28\)
\(\Leftrightarrow x=\frac{28}{6}\)
b)\(|5x+3|-2x=x-17\)
\(\Leftrightarrow|5x+3|=3x-17\)
\(\Leftrightarrow\orbr{\begin{cases}5x+3=3x-17\\5x+3=17-3x\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-20\\8x=14\end{cases}}}\Leftrightarrow\orbr{\begin{cases}x=-10\\x=\frac{4}{7}\end{cases}}\)
c) \(\left(5x+2\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x+2=0\\3x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=-2\\3x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{4}{3}\end{cases}}}\)
a) \(21x+7=15x+35\)
\(\Leftrightarrow\)\(6x=28\)
\(\Leftrightarrow\)\(x=\frac{14}{3}\)
Vậy...
b) \(\left|5x+3\right|-2x=x-17\)
\(\Leftrightarrow\)\(\left|5x+3\right|=3x-17\)
Nếu \(x\ge-\frac{3}{5}\)thì pt trở thành:
\(5x+3=3x-17\)
\(\Leftrightarrow\)\(2x=-20\)
\(\Leftrightarrow\)\(x=-10\)(loại)
Nếu \(x< -\frac{3}{5}\)thì pt trở thành:
\(-5x-3=3x-17\)
\(\Leftrightarrow\)\(8x=14\)
\(\Leftrightarrow\) \(x=\frac{7}{4}\) (loại)
Vậy pt vô nghiệm
c) \(\left(5x+2\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}5x+2=0\\3x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{4}{3}\end{cases}}\)
Vậy...
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\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
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Câu 7:
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
6, (x-5).(x-4)=10-2x
(x-5).(x-4)+2(x-5)=0
(x-5)(x-2)=0
=>x=5, x=2
7, (x^2+1)(x-2)+2x=4
x^3-2x^2+x-2+2x=4
x^3-2x^2+3x-2-4=0
x^3-2x^2+3x-6=0
x^2(x-2)+3(x-2)=0
(x-2)(x^2+3)=0
th1: x=2
th2: x^2+3>0 với mọi x thuộc Z
8, ( đề cs sai hông , giải hong ra:>)
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a)\(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow2\cdot\left(x^3+1\right)+7x\cdot\left(x+1\right)=0\)
\(\Leftrightarrow2\cdot\left(x+1\right)\cdot\left(x^2+x+1\right)+7x\cdot\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left[2\cdot\left(x^2+x+1\right)+7x\right]=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(2x^2-2x+2+7x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(2x^2+5x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(2x+1\right)\cdot\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{-1}{2}\\x=-2\end{matrix}\right.\)
b)\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+\frac{x+3}{63}-\frac{x+5}{61}-\frac{x+7}{59}=0\)
\(\Leftrightarrow\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)-\left(\frac{x+5}{61}+1\right)-\left(\frac{x+7}{59}+1\right)=0\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right)\cdot\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
\(\Rightarrow x+66=0\)
\(\Rightarrow x=-66\)
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\(\left(\dfrac{7}{2}x-7\right)\left(21x-63\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{2}x-7=0\\21x-63=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{2}x=7\\21x=63\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
TH1: 2
TH2: 3