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a) \(\left(x+1\right)^2=3\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)^2-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
b) \(\left(2x-7\right)^3=8\left(7-2x\right)^2\)
\(\Leftrightarrow\left(2x-7\right)^3-8\left(2x-7\right)^2=0\)
\(\Leftrightarrow\left(2x-7\right)^2\left(2x-15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-7\right)^2=0\\2x-15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=15\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{15}{2}\end{cases}}\)
a, \(\left(x+1\right)^2=3\left(x+1\right)\Leftrightarrow x^2+2x+1=3x+3\)
\(\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)
b, \(\left(2x-7\right)^3=8\left(7-2x\right)^2\)
\(\Leftrightarrow8x^3-116x^2+518x-735=0\Leftrightarrow\orbr{\begin{cases}x=3,5\\x=7,5\end{cases}}\)
\(a.4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\\ 4x+8-14x+7+27x-36=30\\ 17x+15=66\\ 17x=51\Rightarrow x=3\)
\(b.2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ =10x-16-12x+15=12x-16+11\\ -2x-1=12x-5\\ \Leftrightarrow-2x-12x=1-5\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{7}{2}\)
\(c.4x^2+3\left(2x^2+1\right)=2x\left(5x-7\right)\\ 4x^2+6x^2+3=10x^2-14x\\ 10x^2+3=10x^2-14x\\ \Leftrightarrow3=14x\\\Rightarrow x=\dfrac{3}{14}\)
\(d.x\left(x^2-7\right)=2x\left(\dfrac{1}{2}x^2+6\right)+8\\ x^3-7x=x^3+12x+8\\ \Leftrightarrow-7x=12x+8\\ \Leftrightarrow-7x-12x=8\\ \Leftrightarrow-19x=8\Rightarrow x=-\dfrac{8}{19}\)
f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)
(2x_1)2+(x+3)2_5(x+7)(x_7)=0
=>4x2-4x+1+x2+6x+9-5x2+245=0
=>(4x2+x2-5x2)+(-4x+6x)+(9+245)=0
=>2x+255=0
=>2x=-255
=>x=-255/2
P/S : Câu 2,3 kết quả bằng bao nhiêu mới tìm được x ?
1.\(\left(2x-7\right)^2-4\left(x-3\right)=5\)
=> \(\left(2x\right)^2-2\cdot2x\cdot7+7^2-4x+12=5\)
=> \(4x^2-28x+49-4x+12=5\)
=> \(4x^2-32x+61=5\)
=> \(4x^2-32x+61-5=0\)
=> \(4x^2-32x+56=0\)
=> \(4\left(x^2-8x+14\right)=0\)
=> \(x^2-8x+14=0\)
=> \(\orbr{\begin{cases}x=4-\sqrt{2}\\x=\sqrt{2}+4\end{cases}}\)
4.\(\left(3x-1\right)^2-6\left(x-1\right)\left(x+1\right)-3x\left(x-2\right)=7\)
=> \(\left(3x\right)^2-2\cdot3x\cdot1+1^2-6\left(x^2-1\right)-3x^2+6x=7\)
=> \(9x^2-6x+1-6x^2+6-3x^2+6x=7\)
=> \(\left(9x^2-6x^2-3x^2\right)+\left(-6x+6x\right)+\left(1+6\right)=7\)
=> 7 = 7(đúng)
5. \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
=> \(x^2+2\cdot x\cdot3+3^2-x\left(x+8\right)+4\left(x+8\right)=1\)
=> x2 + 6x + 9 - x2 - 8x + 4x + 32 = 1
=> (x2 - x2) + (6x - 8x + 4x) + (9 + 32) = 1
=> 2x + 41 = 1
=> 2x = -40
=> x = -20
\((2x-7)^3=8(7-2x)^2\)
⇔ \((2x-7)^3=8(2x-7)^2\) (*)
\(TH1: (2x-7)^2=0\)
Khi đó: \(2x-7=0\) ⇔ \(x=\dfrac{7}{2} \)
\(TH2:\left(2x-7\right)^2\ne0\)
Khi đó: (*) ⇔ \(2x-7=8\) (chia 2 vế cho \((2x-7)^2\))
⇔ \(x=\dfrac{15}{2} \)
Vậy \(x=\dfrac{15}{2}\); \(x=\dfrac{7}{2}\)
chuyển vế sang là thành \(-8(2x-7)^2 \) chứ ạ