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1) (2x + 1)(3x – 2) = (5x – 8)(2x + 1)
⇔ (2x + 1)(3x – 2) – (5x – 8)(2x + 1) = 0
⇔ (2x + 1).[(3x – 2) – (5x – 8)] = 0
⇔ (2x + 1).(3x – 2 – 5x + 8) = 0
⇔ (2x + 1)(6 – 2x) = 0
⇔\(\left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=3\end{matrix}\right.\)
Vậy.....
2) 4x2 -1 = (2x + 1)(3x - 5)
⇔ (2x-1)(2x+1)-(2x+1)(3x-5)=0
⇔ (2x+1)(2x-1-3x+5)=0
⇔ (2x+1)(4-x)=0
⇔ \(\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=4\end{matrix}\right.\)
Vậy...
3)
(x + 1)2 = 4(x2 – 2x + 1)
⇔ (x + 1)2 - 4(x2 – 2x + 1) = 0
⇔ x2 + 2x +1- 4x2 + 8x – 4 = 0
⇔ - 3x2 + 10x – 3 = 0
⇔ (- 3x2 + 9x) + (x – 3) = 0
⇔ -3x (x – 3)+ ( x- 3) = 0
⇔ ( x- 3) ( - 3x + 1) = 0
⇔\(\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy......
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a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)
\(\Rightarrow3x-6x^2+6x+14=29\)
\(\Rightarrow-6x^2+9x-15=0\)
\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)
\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)
Vậy \(S=\varnothing\)
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a, 3x.(x+2) - 5.(x+2) = 0
=> * (x+2) = 0 => x = 2
Và * (3x -5) = 0 => x = \(\frac{5}{3}\)
Vậy x = 2 ; x = \(\frac{5}{3}\)
b, 2x2 + 3x +10x -2x2 =26
13x = 26
=> x = 26 : 13
=> x = 2
Vậy: x=2
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b) Ta có: 2x(x – 5) – x(3 + 2x) = 26
⇔ 2 – 10x – 3x – 2 =26
⇔ - 13x = 26
⇔ x = - 2
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a)\(5\left(2x-1\right)-4\left(8-3x\right)=7\)
\(\Leftrightarrow10x-5+12x-32=7\)
\(\Leftrightarrow22x-37=7\)
\(\Leftrightarrow22x=44\Rightarrow x=2\)
b)\(5x\left(x-5\right)-x\left(2x+3\right)=26\)
\(\Leftrightarrow5x^2-25x-2x^2-3x=26\)
\(\Leftrightarrow3x^2-28x-26=0\)
\(\Leftrightarrow3\left(x-\dfrac{14}{3}\right)^2-\dfrac{274}{3}=0\)
\(\Rightarrow x=\dfrac{14}{3}\pm\dfrac{\sqrt{274}}{3}\)
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a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
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a, 2x(x-5) - x ( 3 + 2x ) = 26
=> 2x^2 - 10x - 3x - 2x ^ 2 = 26
=> - 13 x = 26
=> x = -2
a, \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=-2\)
Vậy x = -2
b, \(3x^3-48x=0\)
\(\Leftrightarrow3x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4;x=-4\end{cases}}\)
Vậy x = 0 hoặc x = 4 hoặc x = -4
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Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
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