\(\dfrac{2x+3}{3}=\dfrac{3y-2}{6}=\dfrac{2x+6y-1}{5x}\left(1\right)\)

Từ `2` tỉ số đầu , ta áp dụng t/c của DTSBN , ta đc :

\(\dfrac{2x+3}{3}=\dfrac{3y-2}{6}=\dfrac{2x+3+3y-2}{3+6}=\dfrac{2x+3y+1}{9}\left(2\right)\)

Từ `(1);(2)=>`\(\dfrac{2x+6y-1}{5x}=\dfrac{2x+3y+1}{9}\left(3\right)\)

Từ `(3)` ta xét `2` trường hợp :

+, Nếu `2x+3y+1 \ne  0` thì :

`(3)=>5x=9=>x=9/5`

Thay `x=9/5` vào \(\dfrac{2x+3}{3}=\dfrac{3y-2}{6}\), ta đc :

\(\dfrac{2\cdot\dfrac{9}{5}+3}{3}=\dfrac{3y-2}{6}\\ \Rightarrow\dfrac{\dfrac{18}{5}+3}{3}=\dfrac{3y-2}{6}\\ \Rightarrow\dfrac{11}{5}=\dfrac{3y-2}{6}\\ 3y-2=6\cdot\dfrac{11}{5}\\ 3y-2=\dfrac{66}{5}\\ 3y=\dfrac{76}{5}\\ y=\dfrac{76}{16}\)

+, Nếu `2x+3y+1=0` thì :

`(1)=>` \(\dfrac{2x+3}{3}=\dfrac{3y-2}{6}=0\\ \Rightarrow\left\{{}\begin{matrix}2x+3=0\\3y-2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\)

3\(x^2\).(5\(x\) + 1) + 6\(x^3\).(5\(x\) + 2) = 9\(x^3\) .(5\(x\) + 3)

15\(x^3\) + 3\(x^2\) + 30\(x^4\) + 12\(x^3\) = 45\(x^4\) + 27\(x^3\)

(15\(x^3\) + 12\(x^3\)) + 3\(x^2\) + 30\(x^4\) - 45\(x^4\) - 27\(x^3\) = 0

       27\(x^3\) + 3\(x^2\)  - 15\(x^4\) - 27\(x^3\) = 0

                     3\(x^2\) - 15\(x^4\)      = 0

                     3\(x^2\).(1 - 5\(x^2\)) = 0

                         \(\left[{}\begin{matrix}x^2=0\\1-5x^2=0\end{matrix}\right.\)

                         \(\left[{}\begin{matrix}x=0\\5x^2=1\end{matrix}\right.\)

                          \(\left[{}\begin{matrix}x=0\\x=\mp\dfrac{\sqrt{5}}{5}\end{matrix}\right.\)  

Ta có: \(5x^3+4x^2-3x\left(2x^2+7x-1\right)\)

\(=5x^3+4x^2-6x^3-21x^2+3x\)

\(=-x^3-17x^2+3x\)

6x³ - 7x² + 5x - 2
= 6x³ - 4x² - 3x² + 2x + 3x - 2
= 6x²(x - 2/3) - 3x(x - 2/3) + 3(x - 2/3)
= (x - 2/3)(6x² - 3x + 3)
= 3(x - 2/3)(2x² - x + 1)

4x³ + 5x² + 10x - 12
= 4x³ - 3x² + 8x² - 6x + 16x - 12
= 4x²(x - 3/4) + 8x(x - 3/4) + 16(x - 3/4)
= (x - 3/4)(4x² + 8x + 16)
= 4(x - 3/4)(x² + 2x + 4)

4x³ - 7x² - x + 3
= 4x³ - 3x² - 4x² + 3x - 4x + 3
= 4x²(x - 3/4) - 4x(x - 3/4) - 4(x - 3/4)
= (x - 3/4)(4x² - 4x - 4)
= 4(x - 3/4)(x² - x - 1)

4x³ - 5x² + 6x + 9
= 4x³ + 3x² - 8x² - 6x + 12x + 9
= 4x²(x + 3/4) - 8x(x + 3/4) + 12(x + 3/4)
= (x + 3/4)(4x² - 8x + 12)
= 4(x + 3/4)(x² - 2x + 3)

3x³ - 5x² + 5x - 2
= 3x³ - 2x² - 3x² + 2x + 3x - 2
= 3x²(x - 2/3) - 3x(x - 2/3) + 3(x - 2/3)
= (x - 2/3)(3x² - 3x + 3)
= 3(x - 2/3)(x² - x + 1)

1) 2x – (3 – 5x) = 4( x +3)

<=>2x-3+5x=4x+12

<=>2x-3+5x-4x-12=0

<=>3x-15=0

<=>x=5

2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)

<=>10x-15-20x+28=19-2x-22

<=>10x-15-20x+28-19+2x+22=0

<=>-8x+16=0

<=>x=2

tham khảo

1) 2x – (3 – 5x) = 4( x +3)

<=>2x-3+5x=4x+12

<=>2x-3+5x-4x-12=0

<=>3x-15=0

<=>x=5

2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)

<=>10x-15-20x+28=19-2x-22

<=>10x-15-20x+28-19+2x+22=0

<=>-8x+16=0

<=>x=2

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

\(=\left(x+2\right)\left(5x-1-2x+6\right)-3\left(x^2+4x+4\right)\)

\(=\left(x+2\right)\left(3x+5\right)-3x^2-12x-12=-x-2\)

\(\left(3x-4\right)\left(2x+1\right)\left(5x-2\right)=0\)

\(\Rightarrow\hept{\begin{cases}3x-4=0\\2x+1=0\\5x-2=0\end{cases}\Rightarrow}\hept{\begin{cases}3x=4\\2x=-1\\5x=2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{4}{3}\\x=-\frac{1}{2}\\x=\frac{2}{5}\end{cases}}}\)

Vậy ...

Ối ối nhầm rồi :(

\(\left(3x-4\right)\left(2x+1\right)\left(5x-2\right)=0\)

\(\Rightarrow\hept{\begin{cases}3x-4=0\\2x+1=0\\5x-2=0\end{cases}\Rightarrow\hept{\begin{cases}3x=4\Leftrightarrow x=\frac{4}{3}\\2x=-1\Leftrightarrow x=-\frac{1}{2}\\5x=2\Leftrightarrow x=\frac{2}{5}\end{cases}}}\)

Vậy ... là nghiệm của pt

(3x - 4)(2x + 1)(5x - 2) = 0

<=> 3x - 4 = 0 hoặc 2x + 1 = 0 hoặc 5x - 2 = 0

<=> x = 4/3 hoặc x = -1/2 hoặc x = 2/5

(x²-5x+7)²-(2x-5)²=0

⇔(x²-5x+7+2x-5)(x²​-5x+7-2x+5)=0

⇔(x²-3x+2)(x²-7x+12)=0

⇔(x²-2x-x+2)(x²-3x-4x+12)=0

⇔[x(x-2)-(x-2)][x(x-3)-4(x-3)]=0

⇔(x-1)(x-2)(x-3)(x-4)=0

⇔x-1=0 hoặc x-2=0 hoặc x-3=0 hoặc x-4=0

⇔x=1 hoặc x=2 hoặc x=3 hoặc x=4.

Vậy tập nghiệm của pt trên là : S={1;2;3;4}

(x^2-5x+7)^2 - (2x-5)^2 = 0

<=> x^4 + 25^2 + 49 - 10x^3 - 70x + 14x^2 - (4x^2 - 20x + 25) = 0

<=> x^4 - 10x^3 + 39x^2 - 70x + 49 - 4x^2 + 20x - 25 = 0

<=> x^4 - 10x^3 + 35x^2 - 50x + 24 = 0

<=> x^4 - 4x^3 - 6x^3 + 24x^2 + 11x^2 - 44x - 6x + 24 = 0

<=> (x - 4)(x^3 - 6x^2 + 11x - 6) = 0

<=> (x - 4)(x^3 - 3x^2 - 3x^2 + 9x + 2x - 6) = 0

<=> (x - 4)(x - 3)(x^2 - 3x + 2) = 0

<=> (x - 4)(x - 3)(x - 2)(x - 1) = 0

<=> x ∈ {4,3,2,1}

\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)

\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)

\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)

\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)

a: \(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\left(3x-y\right)\cdot\left(3x+y\right)=9x^2-y^2\)

b: \(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\cdot\dfrac{1}{2}\)

\(=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\)

\(=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

c: \(\left(5y-3x\right)\cdot\dfrac{1}{4}\cdot\left(12x+20y\right)\)

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

\(=25y^2-9x^2\)

d: \(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(\dfrac{3}{2}y+x\right)\cdot2\)

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

\(=\dfrac{9}{4}y^2-x^2\)

e: \(\left(a+b+c\right)\left(a+b-c\right)\)

\(=\left(a+b\right)^2-c^2\)

\(=a^2+2ab+b^2-c^2\)

Lời giải:

$5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2$

$5^x(1+5+5^2+5^3)=88.89:2-16$

$5^x.156=3900$

$5^x=3900:156=25=5^2$

$\Rightarrow x=2$