\(a)5-\left(x-6\right)=4\left(3-2x\right)\)

\(\Leftrightarrow5-x+6=12-8x\)

\(\Leftrightarrow-x+8x=12-5-6\)

\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)

a) 5-(x-6)=4(3-2x)

<=>5-x-6=12-8x

<=>-x+8x=2-5-6

<=>7x=1

<=>x=1/7

\(b)3-4x\left(25-2x\right)=8x^2+x-300\)

\(\Leftrightarrow3-100x+8x^2=8x^2+x-300\)

\(\Leftrightarrow101x-303=0\)

\(\Leftrightarrow101\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

\(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)

\(\Leftrightarrow\left(\dfrac{x-3}{113}+1\right)+\left(\dfrac{x-5}{115}+1\right)=\left(\dfrac{x-7}{117}+1\right)+\left(\dfrac{x-9}{119}+1\right)\)\(\Leftrightarrow\dfrac{x+110}{113}+\dfrac{x+110}{115}=\dfrac{x+110}{117}+\dfrac{x+110}{119}\)

\(\Leftrightarrow\dfrac{x+110}{113}+\dfrac{x+110}{115}-\dfrac{x+110}{117}-\dfrac{x+110}{119}=0\)

\(\Leftrightarrow\left(x+110\right)\left(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\right)=0\)

Mà \(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\ne0\)

\(\Rightarrow x+110=0\)

\(\Rightarrow x=-110\)

\(\dfrac{x-3}{133}+\dfrac{x-5}{155}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)

\(\Leftrightarrow\left(\dfrac{x-3}{113}+1\right)+\left(\dfrac{x-5}{115}+1\right)=\left(\dfrac{x-7}{117}+1\right)+\left(\dfrac{x-9}{119}+1\right)\)

\(\Leftrightarrow\dfrac{x+130}{113}+\dfrac{x+130}{115}=\dfrac{x+130}{117}+\dfrac{x+130}{119}\)

\(\Leftrightarrow\dfrac{x+130}{113}+\dfrac{x+130}{115}-\dfrac{x+130}{117}-\dfrac{x+130}{119}=0\)

\(\Leftrightarrow\left(x+130\right)\left(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\right)=0\)

Mà \(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\ne0\)

\(\Leftrightarrow x+130=0\)

\(\Leftrightarrow x=-130\)

Vậy..

Giải:

Ta có:

\(\frac{x+1}{15}+\frac{x+2}{7}+\frac{x+4}{4}+6=0\)

\(\Leftrightarrow\frac{x}{15}+\frac{1}{15}+\frac{x}{7}+\frac{2}{7}+\frac{x}{4}+\frac{4}{4}+6=0\)

\(\Leftrightarrow\frac{x}{15}+\frac{x}{7}+\frac{x}{4}=-\frac{772}{105}\)

\(\Leftrightarrow x\left(\frac{1}{15}+\frac{1}{7}+\frac{1}{4}\right)=-\frac{772}{105}\)

\(\Leftrightarrow x=-16\)

Vậy phương trình trên có nghiệm là x = -16.

b. Cách làm tương tự.

Chúc bạn học tốt@@

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)

<=>\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x^2-49\right)=0\)

<=>\(4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

<=>2x+255=0 

<=>2x=-255

<=>x=-255/2

\(a,\left(3x-7\right)\left(x+5\right)=\left(5+x\right)\left(3-2x\right)\)

\(\Leftrightarrow\left(3x-7\right)\left(x+5\right)-\left(x+5\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-7-3+2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\5x-10=0\end{matrix}\right.\)

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

\(b,\dfrac{-x+3}{2}=\dfrac{x-2}{3}\left(MSC=6\right)\)

Suy ra :

\(3\left(-x+3\right)=2\left(x-2\right)\)

\(\Leftrightarrow-3x+9-2x+4=0\)

\(\Leftrightarrow-5x+13=0\)

\(\Leftrightarrow x=\dfrac{13}{5}\)

\(c,\dfrac{x-1}{x-2}+\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)\(\left(dkxd:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+2\right)+5\left(x-2\right)-12-x^2+4}{x^2-4}=0\)

\(\Leftrightarrow x^2+2x-x-2+5x-10-12-x^2+4=0\)

\(\Leftrightarrow6x-20=0\)

\(\Leftrightarrow x=\dfrac{10}{3}\)\(\left(n\right)\)

Vậy \(S=\left\{\dfrac{10}{3}\right\}\)

a
b

phải ko bn

\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{97}+\frac{x+100}{96}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

Dễ thấy \(\left(\frac{1}{99}< \frac{1}{98}< \frac{1}{97}< \frac{1}{96}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)\ne0\)

\(\Rightarrow x+100=0\Rightarrow x=-100\)

Vậy x = -100

\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}+4=0\)

\(\Rightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)

\(\Rightarrow\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)

\(\Rightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)=0\)

Dễ thấy \(\left(\frac{1}{91}>\frac{1}{93}>\frac{1}{95}>\frac{1}{97}\right)\)nên \(\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)\ne0\)

\(\Rightarrow200-x=0\Rightarrow x=200\)

Vậy x = 200

e: =>x=0 hoặc x=1

a:565-13.x=30

13.x=565-370

13.x=195

x=195:13

x=15

Nếu sai sót gì thì e xin lỗi ạ ^^

Mình khuyên bạn thế này : 

Bạn nên tách những câu hỏi ra 

Như vậy các bạn sẽ dễ giúp

Và cũng có nhiều bạn giúp hơn !

Bài 1.

a) ( x - 3 )( x + 7 ) = 0

<=> x - 3 = 0 hoặc x + 7 = 0

<=> x = 3 hoặc x = -7

Vậy S = { 3 ; -7 }

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

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

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

<=> x - 2 = 0 hoặc 2x - 5 = 0

<=> x = 2 hoặc x = 5/2

Vậy S = { 2 ; 5/2 }

c) x² - 5x + 6 = 0

<=> x² - 2x - 3x + 6 = 0

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

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

<=> x - 2 = 0 hoặc x - 3 = 0

<=> x = 2 hoặc x = 3

Bài 2.

a) \(\frac{x}{x+1}-1=\frac{3}{2}x\)

ĐKXĐ : x khác -1

<=> \(\frac{x}{x+1}-\frac{x+1}{x+1}=\frac{3}{2}x\)

<=> \(\frac{-1}{x+1}=\frac{3x}{2}\)

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

<=> 3x² + 3x + 2 = 0

Vi 3x² + 3x + 2 = 3( x² + x + 1/4 ) + 5/4 = 3( x + 1/2 )² + 5/4 ≥ 5/4 > 0 ∀ x

=> phương trình vô nghiệm

b) \(\frac{4x}{x-2}-\frac{7}{x}=4\)

ĐKXĐ : x khác 0 ; x khác 2

<=> \(\frac{4x^2}{x\left(x-2\right)}-\frac{7x-14}{x\left(x-2\right)}=\frac{4x^2-8x}{x\left(x-2\right)}\)

=> 4x² - 7x + 14 = 4x² - 8x

<=> 4x² - 7x - 4x² + 8x = -14

<=> x = -14 ( tm )

Vậy phương trình có nghiệm x = -14

a) Ta thấy:
\(\left(x+4\right)\left(x-4\right)=x\left(x-\frac{2}{3}\right)\)
\(\Rightarrow\left(x^2-4x\right)+\left(4x-16\right)=x^2-\frac{2}{3}x\)
\(\Rightarrow\left(x^2-16\right)-\left(4x-4x\right)=x^2-\frac{2}{3}x\)
\(\Rightarrow x^2-16-0=x^2-\frac{2}{3}x\)
\(\Rightarrow x^2-16=x^2-\frac{2}{3}x\)
\(\Rightarrow16=\frac{2}{3}x\)    ( do có cùng hiệu và cùng số bị trừ )
\(\Rightarrow x=16:\frac{2}{3}\)
\(\Rightarrow x=24\)
Vậy x = 24

b.) x^3-x^2-2x=0

    x(x^2-x-2)=0

   x(x^2-2x+x-2)=0

   x(x(x-2)+x-2)=0

  x(x-2)(x+1)=0

suy ra x=0 hoặc x-2=0 hoặc x+1=0 

    vậy x=0 hoặc x=2 hoặc x=-1 

hình như câu c đề phải là (x+4)/120 thì phải đó bạn 

c.)(x+4)/120+(x+8)/116=(x+5)/119+(x+7)/117

   (x+4)/120+(x+8)/116-(x+5)/119-(x+7)/117=0

   (x+4)/120+1+(x+8)/116+1-(x+5)/119-1-(x+7)/117-1=0

   (x+4)/120+1+(x+8)/116+1-((x+5)/119+1)-((x+7)/117+1)=0

   (x+124)/120+(x+124)/116-(x+124)/119-(x+124)/117=0

(x+124)(1/120+1/116-1/119-1/117)=0

suy ra x+124=0

 x=-124