e, 3x(2-x) =15(x-2)

\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)

\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Vậy..

f, (x+5)(x+4)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)

Vậy..

g, x(x+4)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

,h, (2x -4)(x-2)=0

\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

i, (x+1/5)(2x-3)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)

k, x²-4x=0

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

m, 4x²-1=0

\(\Leftrightarrow\left(2x\right)^2-1^2=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)

n, x²-6x+9=0

\(\Leftrightarrow x^2-2.x.3+3^2=0\)

\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)

<=> x=3

l, (3x-5)²-(x+4)²=0

\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)

\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy ..

o, 7x(x+2)-5(x+2)=0

\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)

Vậy....

p, 3x(2x-5)-4x+10=0

\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)

\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)

Vậy...

q, (2-2x)-x²+1=0

\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)

\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

Vậy ....

r, x(1-3x)=5(1-3x)

\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)

\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)

s, 2x-3/4+x+1/6=3

\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)

r, x(1-3x)=5(1-3x)

➜x(1-3x)-5(1-3x)=0

➜(x-5)(1-3x)=0

➜\(\left[{}\begin{matrix}x-5=0\\1-3x=0\end{matrix}\right.\text{➜}\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)

Mk lười lắm mai nha!!!~~~~~~~~~~~~

Làm dần:

a, 3x+6=0

➜3x=-6

➜x=2

b, 2x-10=0

➜2x=10

➜x=5

c, 3x-7=11

➜3x=11+7

➜3x=18

➜x=6

d, 3x-9=0

➜3x=9

➜x=3

Mọi người giúp mình nhanh nhé ! Mình đang cần gấp

Bài 1. a) 4x - 3 = 0

⇔ x = \(\dfrac{3}{4}\)

KL.....

b) - x + 2 = 6

⇔ x = - 4

KL...

c) -5 + 4x = 10

⇔ 4x = 15

⇔ x = \(\dfrac{15}{4}\)

KL....

d) 4x - 5 = 6

⇔ 4x = 11

⇔ x = \(\dfrac{11}{4}\)

KL....

h) 1 - 2x = 3

⇔ -2x = 2

⇔ x = -1

KL...

Bài 2. a) ( x - 2)( 4 + 3x ) = 0

⇔ x = 2 hoặc x = \(\dfrac{-4}{3}\)

KL......

b) ( 4x - 1)3x = 0

⇔ x = 0 hoặc x = \(\dfrac{1}{4}\)

KL.....

c) ( x - 5)( 1 + 2x) = 0

⇔ x = 5 hoặc x = \(\dfrac{-1}{2}\)

KL.....

d) 3x( x + 2) = 0

⇔ x = 0 hoặc x = -2

KL.....

Bài 3.a) 3( x - 4) - 2( x - 1) ≥ 0

⇔ x - 10 ≥ 0

⇔ x ≥ 10

b) 3 - 2( 2x + 3) ≤ 9x - 4

⇔ - 4x - 3 ≤ 9x - 4

⇔ 13x ≥1

⇔ x ≥ \(\dfrac{1}{13}\)

a: (2x-3)(3x+6)>0

=>(2x-3)(x+2)>0

=>x<-2 hoặc x>3/2

b: (3x+4)(2x-6)<0

=>(3x+4)(x-3)<0

=>-4/3<x<3

c: (3x+5)(2x+4)>4

\(\Leftrightarrow6x^2+12x+10x+20-4>0\)

\(\Leftrightarrow6x^2+22x+16>0\)

=>\(6x^2+6x+16x+16>0\)

=>(x+1)(3x+8)>0

=>x>-1 hoặc x<-8/3

f: (4x-8)(2x+5)<0

=>(x-2)(2x+5)<0

=>-5/2<x<2

h: (3x-7)(x+1)<=0

=>x+1>=0 và 3x-7<=0

=>-1<=x<=7/3

Bài 1:

a) (3x - 2)(4x + 5) = 0

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

c) (4x + 2)(x^2 + 1) = 0

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)

(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0

(3x+2)(x+1)(1-2x)=0

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

tự tìm x nha

Bài 1.

\( a)\dfrac{{4x - 8}}{{2{x^2} + 1}} = 0 (x \in \mathbb{R})\\ \Leftrightarrow 4x - 8 = 0\\ \Leftrightarrow 4x = 8\\ \Leftrightarrow x = 2\left( {tm} \right)\\ b)\dfrac{{{x^2} - x - 6}}{{x - 3}} = 0\left( {x \ne 3} \right)\\ \Leftrightarrow \dfrac{{{x^2} + 2x - 3x - 6}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{x\left( {x + 2} \right) - 3\left( {x + 2} \right)}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x - 3} \right)}}{{x - 3}} = 0\\ \Leftrightarrow x - 2 = 0\\ \Leftrightarrow x = 2\left( {tm} \right) \)

Bài 2.

\(c)\dfrac{{x + 5}}{{3x - 6}} - \dfrac{1}{2} = \dfrac{{2x - 3}}{{2x - 4}}\)

ĐK: \(x\ne2\)

\( Pt \Leftrightarrow \dfrac{{x + 5}}{{3x - 6}} - \dfrac{{2x - 3}}{{2x - 4}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{x + 5}}{{3\left( {x - 2} \right)}} - \dfrac{{2x - 3}}{{2\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{2\left( {x + 5} \right) - 3\left( {2x - 3} \right)}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{ - 4x + 19}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow 2\left( { - 4x + 19} \right) = 6\left( {x - 2} \right)\\ \Leftrightarrow - 8x + 38 = 6x - 12\\ \Leftrightarrow - 14x = - 50\\ \Leftrightarrow x = \dfrac{{27}}{5}\left( {tm} \right)\\ d)\dfrac{{12}}{{1 - 9{x^2}}} = \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} \)

ĐK: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)

\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)

Bài 2.

\(a)5 + \dfrac{{96}}{{{x^2} - 16}} = \dfrac{{2x - 1}}{{x + 4}} - \dfrac{{3x - 1}}{{4 - x}}\)

ĐK: \(x\ne\pm4\)

\( Pt \Leftrightarrow \dfrac{{96}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} - \dfrac{{2x - 1}}{{x + 4}} - \dfrac{{3x - 1}}{{x - 4}} = - 5\\ \Leftrightarrow \dfrac{{96 - \left( {2x - 1} \right)\left( {x - 4} \right) - \left( {3x - 1} \right)\left( {x + 4} \right)}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = - 5\\ \Leftrightarrow \dfrac{{ - 5{x^2} - 2x + 96}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = - 5\\ \Leftrightarrow - 5{x^2} - 2x + 96 = - 5\left( {{x^2} - 16} \right)\\ \Leftrightarrow 96 - 2x = 80\\ \Leftrightarrow - 2x = - 16\\ \Leftrightarrow x = 8\left( {tm} \right)\\ b)\dfrac{{3x + 2}}{{3x - 2}} - \dfrac{6}{{2 + 3x}} = \dfrac{{9{x^2}}}{{9{x^2} - 4}} \)

ĐK: \(x \ne \dfrac{2}{3};x \ne -\dfrac{2}{3}\)

\( Pt \Leftrightarrow \dfrac{{3x + 2}}{{3x - 2}} - \dfrac{6}{{2 + 3x}} - \dfrac{{9{x^2}}}{{9{x^2} - 4}} = 0\\ \Leftrightarrow \dfrac{{{{\left( {2 + 3x} \right)}^2} - 6\left( {3x - 2} \right) - 9{x^2}}}{{\left( {3x - 2} \right)\left( {2 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{16 - 6x}}{{\left( {3 - 2x} \right)\left( {2 + 3x} \right)}} = 0\\ \Leftrightarrow 16 - 6x = 0\\ \Leftrightarrow - 6x = - 16\\ \Leftrightarrow x = \dfrac{8}{3}\left( {tm} \right)\\ c)\dfrac{{x + 1}}{{{x^2} + x + 1}} - \dfrac{{x - 1}}{{{x^2} - x + 1}} = \dfrac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}} \)

Ta có: \(x(x^4+x^2+1)=x[(x^2+1)^2-x^2]=x(x^2+x+1)(x^2-x+1)\)

Do \(\left\{ \begin{array}{l} {x^2} + x + 1 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\forall x\\ {x^2} - x + 1 = \left( {x - \dfrac{1}{2}} \right) + \dfrac{3}{4} > 0\forall x \end{array} \right.\) nên phương trình xác định với mọi $x \ne 0$

Quy đồng, rồi biến đổi phương trình về dạng \(2x=3 \Leftrightarrow x =\dfrac{3}{2} (tm)\)

giúp mình với ;-;

ghi này chả hiểu j bn ak

ghi rõ ra coi

tui chịu,???

a) ĐKXĐ: \(x\ne0\)

Ta có: \(\dfrac{3x^2+7x-10}{x}=0\)

Suy ra: \(3x^2+7x-10=0\)

\(\Leftrightarrow3x^2-3x+10x-10=0\)

\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{10}{3}\right\}\)

a/ \(\dfrac{3x^2+7x-10}{x}=0\)

\(< =>3x^2+7x-10=0\)

\(< =>3x^2+10x-3x-10=0\)

\(< =>\left(3x^2+10x\right)-\left(3x+10\right)=0\)

\(< =>x\left(3x+10\right)-\left(3x+10\right)=0\)

\(< =>\left(3x+10\right)\left(x-1\right)=0\)

\(=>\left\{{}\begin{matrix}3x+10=0=>x=-\dfrac{10}{3}\\x-1=0=>x=1\end{matrix}\right.\)

Vậy tập nghiệm của .....

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

^^