8(3x-2)-10x=2(4-7x)+15

\(\Leftrightarrow24x-16-10x=8-14x+15\)

\(\Leftrightarrow14x-16=-14x+23\)

\(\Leftrightarrow14x+14x=23+16\)

\(\Leftrightarrow28x=39\)

\(\Leftrightarrow x=\frac{39}{28}\)

8(3x-2)-10x = 2(4-7x)+15

24x - 16 - 10x = 8 - 14x + 15

14x - 16 = 23 - 14x

14x + 14x = 23 + 16

28x = 39

x = 39:28

x = 39/28

a) 6x(5x + 3) + 3x(1 – 10x) = 7  

⇒ 30x²+18x+3x-30x²=7

⇒21x=7

⇒x=\(\dfrac{7}{21}\)

⇒x= \(\dfrac{1}{3}\)

b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44

⇒15x-63x²-15+63x + 63x²-35x+36x-20=44

⇒79x-35=44

⇒79x=44+35

⇒79x=79

⇒x=1

a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)

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

b. 

PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$

$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$

$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$

$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$

$\Leftrightarrow (x-3)^2(2x+1)^2=0$

$\Leftrightarrow (x-3)(2x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $2x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$

d.

$x^2-2x=24$

$\Leftrightarrow x^2-2x-24=0$

$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$

$\Leftrightarrow x=-4$ hoặc $x=6$

dap an:

x=1

\(a,=\left(x-2\right)^2\\ b,=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\\ c,=\left(1-2x\right)\left(1+2x+4x^2\right)\\ d,=\left(x+1\right)^3\\ e,Sửa:\left(x+y\right)^2-9x^2=\left(x+y-3x\right)\left(x+y+3x\right)\\ =\left(y-2x\right)\left(4x+y\right)\\ f,=\left(x+3\right)^2\\ g,=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\\ h,=8\left(x^3-\dfrac{1}{64}\right)=8\left(x-\dfrac{1}{4}\right)\left(x^2+\dfrac{1}{4}x+\dfrac{1}{16}\right)\)

a) \(\left(x-2\right)^2\)

b) \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)

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

d) \(\left(x+1\right)^3\)

e) \(\left(x+y-3\sqrt{x}\right)\left(x+y+3\sqrt{x}\right)\)

f) \(\left(x+3\right)^2\)

g) \(-\left(x-5\right)^2\)

h) \(\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

het thoirui pan oi

câu 1/ 5x(\(4x^2\)-2x+1) - 2x(\(10x^2\)-5x-2)

= 5x.\(4x^2\)-5x.2x+ 5x.1 - ( 2x.\(10x^2\)-2x.5x-2x.2)

= 9\(x^3\)-10\(x^2\)+5x - 20\(x^3\)+10\(x^2\)+4x

= (9\(x^3\)-\(20x^3\)) + (-10\(x^2\)+10\(x^2\)) + (5x+4x)

= \(-11x^3\) + 9x

à cj ơi, e 2k6, đọc phần lí thuyết r lm, nên có lỗi sai j mong cj thông cảm

`(3x^2+10x-8)=(5x^2-2x+10)^2`

`<=>(3x^2+10x-8+5x^2-2x+10)(3x^2+10x-8-5x^2+2x-10)=0`

`<=> (8x^2+8x+2)(-2x^2+12x-18)=0`

\(\Leftrightarrow\left[{}\begin{matrix}8x^2+8x+2=0\\-2x^2+12x-18=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy `S={-1/2 ; 3}`.