`3/7xx14/5+32/5xx3/7+3/7xx9/5`

`=3/7xx(14/5+32/5+9/5)`

`=3/7xx55/5`

`=3/7xx5`

`=15/7`

\(=\frac{3}{7}\times[\frac{14}{5}+\frac{32}{5}+\frac{9}{5}]=\frac{3}{7}\times\frac{55}{5}=\frac{3}{7}\times11=\frac{33}{7}\)

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

\(\dfrac{3}{x}-\left(-\dfrac{4}{x}\right)=2+5\)

\(\dfrac{3}{x}+\dfrac{4}{x}=7\)

\(\dfrac{1}{x}.\left(3+4\right)=7\)

\(\dfrac{1}{x}.7=7\)

\(\dfrac{1}{x}=7:7=1\)

\(\Rightarrow x=1\)

`3/x -5 = -4/x +2(đk : x ne 0)`

`3/x +4/x = 2+5`

`7/x = 7`

`x  =7/7=1(t//m)`

Vậy `x=1`

\(2-x:\dfrac{3}{5}=\dfrac{5}{6}\)

\(x:\dfrac{3}{5}=\dfrac{17}{6}\)

     \(x=\dfrac{17}{6}\times\dfrac{3}{5}\)

     \(x=\dfrac{51}{30}\)

\(3\left(x+2\right)=35-5\\ 3\left(x+2\right)=30\\ x+2=30:3\\ x+2=10\\ x=10-2\\ x=8\)

⇔3(x+2)=30

⇔x+2=10

⇔x=8

=>x^2+8x+15=x^2+6x+8

=>8x+15=6x+8

=>2x=-7

=>x=-7/2

Đề: 

`=> x^2 + 3x + 5x + 15 = 2x + 8 + x^2 + 4x`

`=> x^2 + 8x + 15 = x^2 + 6x + 8`

`=> x^2 - x^2 + 8x - 6x + 15 - 8 = 0`

`=> 2x + 7 = 0`

`=> 2x = -7`

`=> x = -7/2`

Vậy `x = -7/2`

\(\left(x+3\right)\left(x+5\right)=\left(x+4\right)\left(2+x\right)\)

\(\Leftrightarrow\left(x+3\right)\left(x+5\right)\left(4+x\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\\4+x=0\\x+2=0\end{matrix}\right.\)                 \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\\x=-4\\x=-2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{-3;-5;-4;-2\right\}\)

    3/2.(x-5/3)+4/5=x+1

3/2.x-3/2.5/3+4/5=x+1

          3/2.x-5/2-x=1+4/5

          3/2.x-5/2-x=9/5

                3/2.x-x=9/5+5/2

                   1/2.x=43/10

                         x=43/10:1/2

                         x=43/5

(x+1)+(x+3)+(x+5)+...+(x+59)=1710

=>(x+x+x+...+x)+(1+3+5+...+59)=1710

     (30 số x)         

=>30 x x+900=1710

=>30 x x = 810

=>x= 27

Vậy x =27

\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+59\right)=1710\)

\(\Rightarrow\left(x+x+x+...+x\right)+\left(1+3+5+...+59\right)=1710\)

\(\Rightarrow x\times30+\left[\left(59+1\right)\times30:2\right]=1710\)

\(\Rightarrow x\times30+900=1710\)

\(\Rightarrow x\times30=810\)

\(\Rightarrow x=27\).

Nhận thấy vế trái có 30 hạng tử

Ta có : (x + 1) + (x + 3) + (x + 5) + .... + (x + 59) = 1710 

<=> (x + x + x + ... + x) + (1 + 3 + 5 + .. + 59) = 1710 

<=> \(30x+\left[\left(59-1\right):2+1\right].\dfrac{59+1}{2}=1710\)

<=> 30x + 900 = 1710 

<=> 30x = 810 

<=> x = 27

Vậy x = 27 

\(2x-3-3x+3.5=4.3-4.x-18\)

<=> \(2x-3-3x+15=12-4x-18\)

<=> \(2x-3x+4x=12-18-15+3\)

<=>\(3x=-18\)

<=> x=-18/3=-6

\(2x-3-3\left(x-5\right)=4\left(3-x\right)-18\)

\(\Leftrightarrow2x-3-3x+15=12-4x-18\)

\(\Leftrightarrow\left(2x-3x\right)+\left(15-3\right)=\left(12-18\right)-4x\)

\(\Leftrightarrow-x+12=-6-4x\)

\(\Leftrightarrow12=-6-3x\)

\(\Leftrightarrow3x=-6-12\)

\(\Leftrightarrow3x=-18\)

\(\Leftrightarrow x=\frac{-18}{3}\)

\(\Leftrightarrow x=-6\)

\(\text{Vậy }x=-6\)

3x(x-3)+5(x-3)=0

\(\Leftrightarrow\)(x-3)(3x+5)=0

\(\Leftrightarrow\)x-3=0 hoặc 3x+5=0

\(\Leftrightarrow\)x=0+3 hoặc 3x=-5

\(\Leftrightarrow\)x=3 hoặc x=-5/3

Vậy x=3; x=-5/3

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-3=0\\3x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{3}\end{cases}}\)

Vậy \(S=\left\{3;\frac{-5}{3}\right\}\)

Bước 1) Nhóm nhân tử chung

Bước 2) Vì VP là 0 => 2TH bằng 0

Bước 3) Tìm x

Bước 4) KL

Cảm ơn! Mong là mình làm đúng.

3x (x - 3) + 5 (x - 3) = 0

(3x + 5) (x - 3) = 0

\(\Rightarrow\orbr{\begin{cases}3x+5=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=3\end{cases}}}\)

Vậy\(x=-\frac{5}{3}\)hoặc x = 3

<Giải thích: dựa vào công thức a (b + c) + d (b + c) = (a + d) (b + c) và công thức a.b = 0 => a = 0 hoặc b = 0>

#Học tốt!!!

~NTTH~

`7/8 - 3/5 =-x`

`=> x=-7/8 + 3/5`

`=> x= -35/40 + 23/40`

`=> x= -12/40`

`=> x= -3/10`

Vậy `x=-3/10`