Tìm x,biết:

2x(3x-5)=10+6x

=>6x²-10x=10+6x

=>6x²-10x-10-6x=0

>6x²-16x-10=0

=>6(x²-8/3x-5/3)=0

=>x²-8/3x-5/3=0....

Cảm ơn bạn Lê Phúc Huy nhưng bạn có thể làm 2x(3x-5)-10-6x=0 đc k ạ

\(\left(x+2\right)^2-9=0\)

\(\Rightarrow\left(x+2\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

Vậy...

b: =>(x-3)(2x+5)+(2x+5)(2x-5)=0

=>(2x+5)(x-3-2x+5)=0

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

=>x=2 hoặc x=-5/2

c: =>3x^2-6x+15-3x^2+30x=0

=>24x+15=0

=>x=-5/8

A=(-3/4+2/3).11/9+(-1/4+1/3):|-9/11|

\(=\left(-\frac{3}{4}+\frac{2}{3}\right).\frac{11}{9}+\left(-\frac{1}{4}+\frac{1}{3}\right).\frac{11}{9}\)

\(=\frac{11}{9}\left(-\frac{3}{4}+\frac{2}{3}-\frac{1}{4}+\frac{1}{3}\right)\)

\(=\frac{11}{9}\left(-1+1\right)\)

\(=\frac{11}{9}.0\)

=0

2x(8x-1)²(4x-1)= 9

<=> 2x(64x²-16x+1)(4x-1)=9

<=>(128x³  - 32x² + 2x)(4x-1)=9

<=>512x⁴ - 256x³ + 40x²  - 2x=9

<=>64x⁴ - 32x³ + 5x² - 0,25x - 1,125=0

<=>64x³(x-0,5) + 5x(x-0,5) + 2,5x  -0,25x - 1,125 = 0

<=> (x-0,5)(64x³ + 5x - 2,25) = 0

<=> (x-0,5)(64x³  + 16x² - 16x² - 4x + 9x - 2,25)=0

<=>(x-0,5)[64x² (x + 0,25 ) -16x(x + 0,25) + 9(x + 0,25) = 0

<=> (x-0,5)(x+0,25)(64x² -16x +9) = 0  (vì 64x² -16x +9 > 0)

<=>\(\orbr{\begin{cases}x-0,5=0\\x+0,25=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=0,5\\x=-0,25\end{cases}}\)

Vậy phương trình có hai nghiệm là S={\(\frac{1}{2}\) ; \(\frac{-1}{4}\)}

1: \(\left(2x+3\right)^2+4\left(3-x\right)\left(3+x\right)\)

\(=4x^2+12x+9+4\left(9-x^2\right)\)

\(=4x^2+12x+9+36-4x^2=12x+45\)

2: \(\left(1-\frac12x\right)^2-\frac14\left(x+2\right)^2\)

\(=\frac14x^2-x+1-\frac14\left(x^2+4x+4\right)\)

\(=\frac14x^2-x+1-\frac14x^2-x-1=-2x\)

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

\(=9x^2+30x+25+9x^2-30x+25-\left(9x^2-4\right)\)

\(=18x^2+50-9x^2+4=9x^2+54\)

4: \(\left(x^2+2\right)^2-\left(x+2\right)\left(x^2+4\right)\left(x-2\right)\)

\(=x^4+4x^2+4-\left(x^2-4\right)\left(x^2+4\right)\)

\(=x^4+4x^2+4-x^4+16=4x^2+20\)

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

\(=x^3-6x^2+12x-8-x\left(x^2-1\right)+6x^2-18x\)

\(=x^3-6x-8-x^3+x=-5x-8\)

6: \(\left(x+2\right)\left(x^2+2x+4\right)\left(x-2\right)\left(x^2-2x+4\right)\)

\(=\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x^2+2x+4\right)\)

\(=\left(x^3+8\right)\cdot\left(x^3-8\right)=x^6-64\)

3x.(x-2)-x²+2x=0

⇔3x²-6x-x²+2x=0

⇔2x²-4x=0

⇔2x(x-2)=0

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

vậy x=0 và x=2

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

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

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

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

<=>2x=0 hoặc x-2=0

<=>x=0 hoặc x=2