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a) (x-1)(5x+3)=(3x-8)(x-1)
= (x-1)(5x+3)-(3x-8)(x-1)=0
=(x-1)[(5x+3)-(3x-8)]=0
=(x-1)(5x+3-3x+8)=0
=(x-1)(2x+11)=0
\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0
\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)
Vậy S={1;\(\dfrac{-11}{2}\)}
b) 3x(25x+15)-35(5x+3)=0
=3x.5(5x+3)-35(5x+3)=0
=15x(5x+3)-35(5x+3)=0
=(5x+3)(15x-35)=0
\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0
\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)
Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}
c) (2-3x)(x+11)=(3x-2)(2-5x)
=(2-3x)(x+11)-(3x-2)(2-5x)=0
=(3x-2)[(x+11)-(2-5x)]=0
=(3x-2)(x+11-2+5x)=0
=(3x-2)(6x+9)=0
\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0
\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)
Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}
d) (2x2+1)(4x-3)=(2x2+1)(x-12)
=(2x2+1)(4x-3)-(2x2+1)(x-12)=0
=(2x2+1)[(4x-3)-(x-12)=0
=(2x2+1)(4x-3-x+12)=0
=(2x2+1)(3x+9)=0
\(\Leftrightarrow\)2x2+1=0 hoặc 3x+9=0
\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3
Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}
e) (2x-1)2+(2-x)(2x-1)=0
=(2x-1)[(2x-1)+(2-x)=0
=(2x-1)(2x-1+2-x)=0
=(2x-1)(x+1)=0
\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0
\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1
Vậy S={\(\dfrac{-1}{2}\);-1}
f)(x+2)(3-4x)=x2+4x+4
=(x+2)(3-4x)=(x+2)2
=(x+2)(3-4x)-(x+2)2=0
=(x+2)[(3-4x)-(x+2)]=0
=(x+2)(3-4x-x-2)=0
=(x+2)(-5x+1)=0
\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0
\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)
Vậy S={-2;\(\dfrac{1}{5}\)}
a)(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)(-2x+6)=0
⇔2x+1=0 hoặc -2x+6=0
1.2x+1=0⇔2x=-1⇔x=-1/2
2.-2x+6=0⇔-2x=-6⇔x=3
phương trình có 2 nghiệm x=-1/2 và x=3
TA CÓ:
\(a,\left(4x-1\right)\left(x-3\right)=\left(x-3\right)\left(5x+2\right)\Leftrightarrow\left(4x-1\right)\left(x-3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\left(x-3\right)\left(4x-1-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-x-3\right)=0\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
\(b,\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\Leftrightarrow\left(x+3\right)\left(x-5+3x-4\right)=0\)
\(\left(x-3\right)\left(4x-9\right)=0\orbr{\begin{cases}x=3\\x=\frac{9}{4}\end{cases}}\)
\(c,\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\Leftrightarrow\left(1-x\right)\left(5x+3\right)=\left(7-3x\right)\left(1-x\right)\)
\(\left(1-x\right)\left(5x+3-7+3x\right)=0\Leftrightarrow\left(1-x\right)\left(8x-4\right)=0\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
a) 0,75x(x + 5) = (x + 5)(3 - 1,25x)
<=> 0,75x(x + 5) - (x + 5)(3 - 1,25x) = (x + 5)(3 - 1,25x) - (x + 5)(3 - 1,25x)
<=> 0,75x(x + 5) - (x + 5)(3 - 1,25x) = 0
<=> (x + 5)(0,75 + 1,25x - 3) = 0
<=> (x + 5)(2x - 3) = 0
<=> x + 5 = 0 hoặc 2x - 3 = 0
<=> x = -5 hoặc x = 3/2
b) 4/5 - 3 = 1/5x(4x - 15)
<=> -11/5 = x(4x - 15)/5
<=> -11 = x(4x - 15)
<=> -11 = 4x2 - 15x
<=> 11 + 4x2 - 15x = 0
<=> 4x2 - 4x - 11x + 11 = 0
<=> 4x(x - 1) - 11(x - 1) = 0
<=> (4x - 11)(x - 1) = 0
<=> 4x - 11 = 0 hoặc x - 1 = 0
<=> x = 11/4 hoặc x = 1
c) \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
<=> 12x - 36 - 2(x - 3)(2x - 5) = 3(x - 3)(3 - x)
<=> 12x - 36 - 4x2 + 10x + 12x - 30 = 9x - 3x2 - 27 + 9x
<=> 34x - 66 - 4x2 = 18x - 3x2 - 27
<=> 34x - 66 - 4x2 - 18x + 3x2 + 27 = 0
<=> 16x - 39x - x2 = 0
<=> x2 - 16x + 39x = 0
<=> (x - 3)(x - 13) = 0
<=> x - 3 = 0 hoặc x - 13 = 0
<=> x = 3 hoặc x = 13
d) \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
<=> (3x + 1)(3x - 2) + 15(3x + 1) = 2(2x + 1)(3x + 1) + 6x(3x + 1)
<=> 9x2 - 6x + 3x - 2 + 45x + 15 = 12x3 + 4x + 6x + 2 + 18x2 + 6x
<=> 9x2 + 42x + 13 = 30x2 + 16x + 2
<=> 9x2 + 42x + 13 - 30x2 - 16x - 2 = 0
<=> -21x2 + 26x + 11 = 0
<=> 21x2 - 26x - 11 = 0
<=> 21x2 + 7x - 33x - 11 = 0
<=> 7x(3x + 1) - 11(3x + 1) = 0
<=> (7x - 11)(3x + 1) = 0
<=> 7x - 11 = 0 hoặc 3x + 1 = 0
<=> x = 11/7 hoặc x = -1/3
a) \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+8\)
\(\Rightarrow\left(3x+2+3x-2\right)\left(3x+2-3x+2\right)=5x+8\)
\(\Rightarrow4.6x=5x+8\Rightarrow24x=5x+8\)
\(\Rightarrow19x=8\Rightarrow x=\frac{8}{19}\)
b) \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Rightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)
\(\Rightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)
\(\Rightarrow-12x+12+9x-9=3x-9\)
\(\Rightarrow-3x+3=3x-9\)
\(\Rightarrow6x=12\Rightarrow x=2\)
a) Ta có: \(\left(x-3\right)\left(x-4\right)-2\left(3x-2\right)=\left(4-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)-2\left(3x-2\right)-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)-\left(x-4\right)\right]-2\left(3x-2\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-3-x+4\right)-6x+4=0\)
\(\Leftrightarrow x-4-6x+4=0\)
\(\Leftrightarrow-5x=0\)
mà -5<0
nên x=0
Vậy: x=0
\(\left(x-5\right)\left(x-1\right)=2x\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-5-2x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy............
\(5\left(x+3\right)\left(x-2\right)-3\left(x+5\right)\left(x+2\right)=0\)
\(\Leftrightarrow5\left(x^2+x-6\right)-3\left(x^2+7x+10\right)=0\)
\(\Leftrightarrow2x^2-16x-60=0\)
\(\Leftrightarrow x^2-8x-30=0\)
làm tiếp nhé!!!!!
\(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)
\(\Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(S=\left\{3;-2\right\}\)
Chúc bạn học tốt !!!
\(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)
\(\Leftrightarrow\frac{x^2+2x-3x-6}{x-3}=0\)
\(\Leftrightarrow\frac{x\left(x+2\right)-3\left(x+2\right)}{x-3}=0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)
<=> x + 2 = 0
=> x = -2
$pt⇔(x-2)^3-(x+1)^3+9x^2-1=0$
$⇔(x-2-x-1)^3+3.(x-2)(x+1)(x-2-x-1)+9x^2-1=0$
$⇔-27-9x^2+9x+18+9x^2-1=0$
$⇔9x=10$
$⇔x=\dfrac{10}{9}$
vậy hệ phương trình cho có tập nghiệm $S=\dfrac{10}{9}$