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9 tháng 7 2018
A. \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+3x+2x+6\right)-\left(x^2+5x-2x-10\right)=0\)
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(\Leftrightarrow x^2+3x+2x-x^2-5x+2x=-6-10\)
\(\Leftrightarrow2x=-16\)
\(\Leftrightarrow x=-8\) .Vậy \(S=\left\{-8\right\}\)
B. \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x+5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x+5x-20\)
\(\Leftrightarrow2x^2-8x+3x+x^2-2x-5x-3x^2+12x-5x=12-10-20\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\) . Vậy \(S=\left\{\dfrac{18}{5}\right\}\)
C. \(\left(8-4x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)
\(\Leftrightarrow8x-4x^2-8x+4x^2+4x-8x=-16+8\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\) . Vậy \(S=\left\{2\right\}\)
D. \(\left(2x-3\right)\left(8x+2\right)=\left(4x+1\right)\left(4x-1\right)-3\)
\(\Leftrightarrow16x^2+4x-24x-6=16x^2+1^2-3\)
\(\Leftrightarrow16x^2+4x-24x-16x^2=6+1-3\)
\(\Leftrightarrow-20x=4\)
\(\Leftrightarrow x=-\dfrac{1}{5}\) . Vậy \(S=\left\{-\dfrac{1}{5}\right\}\)
9 tháng 7 2018
a)(x+2)(x+3)-(x-2)(x+5)=0
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
<=>2x=-16
<=>x=-8
b)(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
\(\Leftrightarrow5x=22\Leftrightarrow x=\dfrac{22}{5}\)
c)(8-4x)(x+2)+4(x-2)(x+1)=0
\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)
\(\Leftrightarrow-4x=-8\Leftrightarrow x=2\)
d)(2x-3)(8x+2)=(4x+1)(4x-1)-3
\(\Leftrightarrow16x^2+4x-24x-6=16x^2-4x+4x-1-3\)
\(\Leftrightarrow-20x=-2\Leftrightarrow x=\dfrac{-1}{10}\)
14 tháng 2 2022
bạn viết rõ đề câu a;b nhé
c, \(2x\left(x-5\right)-\left(x-5\right)=0\Leftrightarrow\left(2x-1\right)\left(x-5\right)=0\Leftrightarrow x=\dfrac{1}{2};x=5\)
d, \(\left(x+3\right)\left(x+3-5+x\right)=0\Leftrightarrow\left(x+3\right)\left(2x-2\right)=0\Leftrightarrow x=-3;x=1\)
e, \(\left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\Leftrightarrow x=-2;x=\dfrac{1}{5}\)
D
21 tháng 10 2016
8x2+30x+7=0
8x2+16x+14x+7=0
8x(x+2) +7(x+2)=0
(8x+7)(x+2)=0
=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)
16 tháng 9 2019
a) x3 - 16x = 0
x(x2 - 16) = 0
=> x = 0 hoặc x2 - 16 = 0
x = 4
Vậy x = 0 hoặc x = 4
b) x4 -2x3 + 10x2 - 20x = 0
x3 (x - 2) + 10x(x - 2) = 0
(x - 2)(x3 + 10x) = 0
=> x - 2 = 0 hoặc x3 + 10x = 0
x = 2 x(x2 + 10) = 0
+ TH1: x = 0
+ TH2: x2 + 10 = 0
x2 = -10 (vô lí)
Vậy x = 2 hoặc x = 0
c) (2x - 3)2 = (x + 5)2
(2x)2 + 2 . 2x . 3 + 32 = x2 + 2.x.5 + 52
4x2 + 12x + 9 = x2 + 10x + 25
4x2 + 12x - x2 - 10x = 25 - 9
3x2 + 2x = 16
x(3x + 2) = 16
Đến đây bạn làm nốt câu c nhé!
PA
30 tháng 11 2016
\(2x^2-7x+5=0\)
\(2x^2-2x-5x+5=0\)
\(2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)
\(x\left(2x-5\right)-4x+10=0\)
\(x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(x-2\right)=0\)
\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)
\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)
\(x^2-25-x^2+2x=15\)
\(2x=15+25\)
\(2x=40\)
\(x=\frac{40}{2}\)
\(x=20\)
\(x^2\left(2x-3\right)-12+8x=0\)
\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2+4\right)=0\)
\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
\(4x\left(5-x\right)-8x\left(5-x\right)=0\)
\(\Rightarrow x\left(5-x\right).\left(4-8\right)=0\)
\(\Rightarrow x\left(5-x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy \(x\in\left\{0;5\right\}\)
Chúc bạn học tốt!!!
Ta có:4x(5-x)-8x(5-x)=0<=>4x(5-x)-2.4x(5-x)<=>4x(5-x)(1-2)=0
<=>-4x(5-x)=0
=>4x=0 hoặc 5-x=0=>x=0 hoặc x=5
Vậy S={0;5}
<=> 20x-4x2-40x+8x2=0
<=> 4x2-20x=0
<=> x(4x-20)=0
<=> \(\left\{{}\begin{matrix}x=0\\4x-20=0< =>4x=20< =>x=5\end{matrix}\right.\)
\(4x\left(5-x\right)-8x\left(5-x\right)=0\\ \Leftrightarrow20x-4x^2-40x+8x^2=0\\ \Leftrightarrow4x^2-20x=0\\ \Leftrightarrow4x\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={0;5}
\(4x\left(5-x\right)-8x\left(5-x\right)=0\Leftrightarrow\left(4x-8x\right)\left(5-x\right)=0\Leftrightarrow-4x\left(5-x\right)=0\Leftrightarrow\left[{}\begin{matrix}-4x=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)