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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 \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
1 tháng 9 2020
( 2x - 3 )( x + 1 ) - 2x2 + 6x = 0
<=> 2x2 - x - 3 - 2x2 + 6x = 0
<=> 5x - 3 = 0
<=> 5x = 3
<=> x = 3/5
( x2 - x + 1 )( x - 3 ) - x3 + 4x2 = 0
<=> x3 - 4x2 + 4x - 3 - x3 + 4x2 = 0
<=> 4x - 3 = 0
<=> 4x = 3
<=> x = 3/4
( x2 - 2 )( x2 + 2 ) - x4 - 2x + 5 = 0
<=> ( x2 )2 - 4 - x4 - 2x + 5 = 0
<=> x4 + 1 - x4 - 2x = 0
<=> 1 - 2x = 0
<=> 2x = 1
<=> x = 1/2
( x - 3 )( x2 - 3x + 2 ) - ( x2 - 2x - 7 )( x - 2 ) + 2x2 - 2x = 0
<=> x3 - 6x2 + 11x - 6 - ( x3 - 4x2 - 3x + 14 ) + 2x2 - 2x = 0
<=> x3 - 6x2 + 11x - 6 - x3 + 4x2 + 3x - 14 + 2x2 - 2x = 0
<=> 12x - 20 = 0
<=> 12x = 20
<=> x = 20/12 = 5/3
A
1 tháng 9 2020
a, \(\left(2x-3\right)\left(x+1\right)-2x^2+6x=0\)
\(\Leftrightarrow2x^2+2x-3x-3-2x^2+6x=0\Leftrightarrow5x-3=0\Leftrightarrow x=\frac{3}{5}\)
b, \(\left(x^2-x+1\right)\left(x-3\right)-x^3+4x^2=0\)
\(\Leftrightarrow x^3-3x^2-x^2+3x+x-3-x^3+4x^2=0\Leftrightarrow4x-3=0\Leftrightarrow x=\frac{3}{4}\)
c ; d tương tự nhé !
5 tháng 11 2017
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 !
30 tháng 9 2018
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
21 tháng 12 2021
a: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
2 tháng 7 2018
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
15 tháng 8 2016
a) \(x\left(x-1\right)-x^2+2x=5\)
\(\Leftrightarrow x^2-x-x^2+2x=5\)
\(\Leftrightarrow x=5\)
b) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)
\(\Leftrightarrow-x=21\)
\(\Leftrightarrow x=-21\)
c) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)
\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)
\(\Leftrightarrow x=-4\)
15 tháng 8 2016
a) x ( x - 1 ) - x^2 + 2x = 5
=>x2-x-x2+2x=5
=>-x+2x=5
=>x=5
b) 4x ( 3x + 2 ) - 6x ( 2x + 5 ) + 21 ( x - 1 ) = 0
=>12x2+8x-12x2-30x-21+21x=0
=>-x-21=0
=>x=-21
c) 2x( x + 1) - x^2 ( x + 2 ) + x^3 - x + 4 = 0
=>2x2+2x-x3-2x2+x3-x+4=0
=>x+4=0
=>x=-4
KN
28 tháng 7 2015
1) (2x-1)(x+3)(2-x)=0
=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0
=>x=1/2 hoặc x=-3 hoặc x=2
2)x^3 + x^2 + x + 1 = 0
=>.x^2(x+1)+(x+1)=0
=>(x^2+1)(x+1)=0
=>x^2+1=0 hoặc x+1=0
=> x =-1
3) 2x(x-3)+5(x-3) =0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>x=-5/2 hoặc x=3
4)x(2x-7)-(4x-14)=0
=> (x-2)(2x-7)=0
=> x-2 =0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
5)2x^3+3x^2+2x+3=0
=>x^2(2x+3)+2x+3=0
=>(x^2+1)(2x+3)=0
=>x^2+1=0 hoặc 2x+3=0
=> x =-3/2
24 tháng 8 2017
\(A\))\(\left(x-1\right)^2+\left(x-3\right)^2-2x^2+1=0\)
\(x^2-2x+1+x^2-6x+9-2x^2+1=0\)
\(11-8x=0\)
\(\Rightarrow x=\frac{11}{8}\)
\(B\))\(\left(x-1\right)\left(x^2+x+1\right)-\left(x+1\right)\left(x^2-x+1\right)+2x=0\)
\(x^3-1-x^3-1+2x=0\)
\(2x-2=0\)
\(\Rightarrow x=1\)
TP
24 tháng 8 2017
\(A=\left(x-1\right)^2+\left(x-3\right)^2-2x^2+1=0\)
\(\Rightarrow x^2-2x+1+x^2-6x+9-2x^2+1=0\)
\(\Rightarrow\left(x^2+x^2-2x^2\right)+\left(-2x-6x\right)+\left(1+9+1\right)\)
\(\Rightarrow-8x+12=0\Leftrightarrow x=\frac{-11}{-8}=\frac{11}{8}\)
\(B=\left(x-1\right).\left(x^2+x-1\right)-\left(x+1\right).\left(x^2-x+1\right)+2x=0\)
\(\Rightarrow x.\left(x^2+x-1\right)-x^2-x+1-x.\left(x^2-x+1\right)-x^2+x-1+2x=0\)
\(\Rightarrow x^3+x^2-1-x^2-x+1-x^3+x^2-x-x^2+x-1+2x=0\)
\(\Rightarrow\left(x^3-x^3\right)+\left(x^2-x^2+x^2-x^2\right)+\left(-1+1-1\right)+\left(-x-x+x\right)+2x=0\)
\(\Rightarrow-1+x=0\Leftrightarrow x=1\)
\(C=\left(x-5\right).\left(x-5\right)+\left(2x+1\right)^2-3x^2=0\)
\(\Rightarrow x.\left(x-5\right)-5.\left(x-5\right)+\left(2x\right)^2+2.2x.1+1^2-3x^2=0\)
\(\Rightarrow x^2-5x-5x+25+4x^2+4x+1-3x^2=0\)
\(\Rightarrow\left(x^2-3x^2+4x^2\right)+\left(-5x-5x+4x\right)+26=0\)
\(\Rightarrow2x^2-6x+26=0\Leftrightarrow x=\)
\(D=\left(x-1\right)-9=0\Leftrightarrow x-1=9\Leftrightarrow x=10\)
\((x-2)(x^2+2x+5+2x+4-5)=0<=> (x-2)(x^2+4x+4)=0 <=> (x-2)(x+2)^2 = 0 <=> x = 2 ; x = -2\)
\(\left(x-2\right)\left(x^2+2x+5\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+5+2x+4-5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\left(x+2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)