a)x/2=y/3

=>x.3=y.2

x.3/2=y

x.1,5=y

=>y gấp 1,5 lần x

x=24:(1+1,5)

x=9,6

y =24-9,6=14,4

a. Ta có: x²+y²-2x+4y+5=0

⇌(x-1)²+(y-2)²=0

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

b. Ta có: 4x²+y²-4x-6y+10=0

⇌ (2x-1)²+(y-3)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\end{matrix}\right.\)

c.Ta có: 5x²-4xy+y²-4x+4=0

⇌(2x-y)²+(x-2)²=0

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

d.Ta có: 2x²-4xy+4y²-10x+25=0

⇌ (x-2y)²+(x-5)²=0

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

1)6x-8=3x+1

6x-3x=1+8

3x=9

x=3

Vậy x=3

2: 12-10x=25-30x

=>20x=13

=>x=13/20

3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)

=>6x+9-8x+10=10x+21

=>10x+21=-2x+19

=>12x=-2

=>x=-1/6

4: \(\Leftrightarrow25x-15-6x+12=11-5x\)

=>19x-3=11-5x

=>24x=14

=>x=7/12

5: \(\Leftrightarrow8-12x-5+10x=4-6x\)

=>4-6x=-2x+3

=>-4x=-1

=>x=1/4

6: \(\Leftrightarrow32x-24-6+9x=13-40x\)

=>41x-30=13-40x

=>81x=43

=>x=43/81

7: \(\Leftrightarrow10x-5+20x=5x-11\)

=>30x-5=5x-11

=>25x=-6

=>x=-6/25

a: \(=8x^3-10x^5+10x^5-5x^3=3x^3\)

b: \(=x^3-2x^2+4x-2x^2+4x-8+x^2+2x-4x-8\)

\(=x^3-3x^2+4x-16\)

a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42

2x+3/5x+2 = 4x+5/10x+2

<=> (2x+3)(10x+2)=(5x+2)(4x+5)

<=>2x(10x+2)+3(10x+2)= 5x(4x+5)+2(4x+5)

<=> 20x^2+4x+20x+6 = 20x^2+25x+9x+10

<=> 20x^2+4x+20x+6 - (20x^2+25x+9x+10)=0 => 20x^2+24x+6-(20x^2+34x+10)=0

                                                                  <=>   -10x-4=0

                                                                 <=>-10x=4

                                                                  <=> x= -4/10

2x+3/5x+2=4x+5/10x+2

=> (2x+3)(10x+2)=(5x+2)(4x+5)

=> 20x^2+4x+30x+6=10x^2+25x+8x+10 ( Vì cả hai vế đều có 10x^2 nên ta xóa đi )

=> 34x+6=33x+10

=> 34x-33x=-6+10

=> x=4

x⁴+5x²-9x +3

Tìm x?

\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\Leftrightarrow\left(2x+3\right)\left(10x+2\right)=\left(4x+5\right)\left(5x+2\right)\Leftrightarrow20x^2+34x+6=20x^2+33x+10\Leftrightarrow x=4\)