1: \(\Leftrightarrow\left(x-4\right)^2+14=-9\left(x-4\right)\)

\(\Leftrightarrow x^2-8x+16+14+9x-36=0\)

\(\Leftrightarrow x^2+x-6=0\)

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

=>x=-3(nhận) hoặc x=2(nhận)

2: \(\Leftrightarrow\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9=0\)

\(\Leftrightarrow16x^2-8x+2x-1-4x^2-2x-12x^2+9=0\)

=>-8x+8=0

hay x=1(nhận)

c: \(\dfrac{1}{2\left(x-3\right)}-\dfrac{3x-5}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{2}\)

\(\Leftrightarrow x-1-2\left(3x-5\right)=\left(x-3\right)\left(x-1\right)\)

\(\Leftrightarrow x^2-4x+3=x-1-6x+10=-5x+9\)

\(\Leftrightarrow x^2+x-6=0\)

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

=>x=-3(nhận) hoặc x=2(nhận)

het thoirui pan oi

Chia nhỏ ra ik ạ

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

\(5x\left(x-3\right)+\left(x-2\right)^2=5x^2-15x+x^2-4x+4=6x^2-19x+4\)

\(x\left(x+2\right)-\left(x-3\right)\left(x+3\right)=x^2+2x-\left(x^2-9\right)=x^2+2x-x^2+9=2x+9\)

`Answer:`

1) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(=[x\left(x+3\right)][\left(x+1\right)\left(x+2\right)]+1\)

\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)

\(=\left(x^2+3x\right)^2+2.\left(x^2+3x\right)+1\)

\(=\left(x^2+3x+1\right)^2\)

2) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)

\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)

\(=[\left(12x^2+11x+0,5\right)+1,5][\left(12x^2+11x+0,5\right)-1,5]-4\)

\(=\left(12x^2+11x+0,5\right)^2-\left(1,5\right)^2-4\)

\(=\left(12x^2+11x+0,5\right)^2-\left(2,5\right)^2\)

\(=\left(12x^2+11x+0,5-2,5\right)\left(12x^2+11x+0,5+2,5\right)\)

\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)

3) \(\left(x^2+6x+5\right)\left(x^2+10x+21\right)+15\)

\(=\left(x^2+x+5x+5\right)\left(x^2+3x+7x+21\right)+15\)

\(=\left(x+1\right)\left(x+5\right)\left(x+3\right)\left(x+7\right)+15\)

\(=[\left(x+1\right)\left(x+7\right)][\left(x+5\right)\left(x+3\right)]+15\)

\(=\left(x^2+x+7x+7\right)\left(x^2+3x+5x+15\right)+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(v=x^2+=8x+11\)

Đa thức có dạng sau: \(\left(v-4\right)\left(v+4\right)+15\)

\(=v^2-4^2+15\)

\(=v^2-1\)

\(=\left(v+1\right)\left(v-1\right)\)

\(=\left(x^2+8x+11+1\right)\left(x^2+8x+11-1\right)\)

\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

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

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

\(=\left(x^2-a\right).x^2-a\left(x^2-a\right)-6x^2+4x+2a\)

\(=x^4-ax^2-a.\left(x^2-a\right)-6x^2+4x+2a\)

\(=x^4-ax^2-\left(ax^2-aa\right)-6x^2+4x+2a\)

\(=x^4-2ax^2+a^2-6x^2+2a+4x\)

6) \(a^2-b^2-c^2+2bc-2a+1\)

\(=\left(a^2-2a+1\right)-\left(b^2-2bc+c^2\right)\)

\(=\left(a-1\right)^2-\left(b-c\right)^2\)

\(=\left(a-b+c-1\right)\left(a+b-c-1\right)\)

7) \(4a^2-4b^2+16bc-16c^2\)

\(=4a^2-\left(4b^2-16bc+16c^2\right)\)

\(=\left(2a\right)^2-\left(2b-4c\right)^2\)

\(=\left(2a-2b+4c\right)\left(2a+2b-4c\right)\)

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

\(=4\left(a-b-2c\right)\left(a+b-2c\right)\)

đề là gì

a)\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}3x-2=0\\x+6=0\\x^2+5=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x=2\\x=-6\\x^2=-5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{2}{3}\\x=-6\\x\in\varnothing\end{cases}}}\)

vậy x=2/3 hoặc x=-6

a, (3x-2) (x+6) (x^2 +5) = 0 

<=> 3x - 2 = 0 hoặc x + 6 = 0 hoặc x² + 5 = 0 (loại vì x² \(\ge\)0 => x² + 5 > 0)

<=> x = 2/3 hoặc x = -6 

b, (2x+5)^2 = (3x-1)^2 

<=> (2x + 5)² - (3x - 1)² = 0

<=> (2x + 5 - 3x + 1)(2x + 5 + 3x - 1) = 0

\(\Leftrightarrow\orbr{\begin{cases}2x-3x+6=0\\2x+3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}-x=-6\\5x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=6\\x=\frac{4}{5}\end{cases}}}\)

c, 4x² (x-1) - x+1 = 0

<=> 4x²(x - 1) - (x - 1) = 0

<=> (x - 1)(4x² - 1) = 0

<=> (x - 1)(2x - 1)(2x + 1) = 0

vậy x - 1 = 0 hoặc 2x - 1 = 0 hoặc 2x + 1 = 0

hay x = 1 hoặc x = 1/2 hoặc x = -1/2

1,=\(x^2-3x-2x^2+6x=-x^2+3x\)

2,=\(3x^2-x-5+15x=3x^2+14x-5\)

3,=\(5x+15-6x^2-6x=-6x^2-x+15\)

4,=\(4x^2+12x-x-3=4x^2+11x-3\)

5: =>(x+5)^3=0

=>x+5=0

=>x=-5

6: =>(2x-3)^2=0

=>2x-3=0

=>x=3/2

7: =>(x-6)(x-10)=0

=>x=10 hoặc x=6

8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)

=>(x-4)^3=0

=>x-4=0

=>x=4

a. ĐKXĐ: $x\geq 2$ hoặc $x=1$

PT $\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}$

$\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0$

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)

b.

PT $\Leftrightarrow \sqrt{(x-2)^2}=\sqrt{(2x-3)^2}$

$\Leftrightarrow |x-2|=|2x-3|$

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

c. ĐKXĐ: $x=2$ hoặc $x\geq 3$

PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$

$\Leftrightarrow \sqrt{x-2}(\sqrt{x-3}-1)=0$

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x-3}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=4\end{matrix}\right.\) (đều tm)

d.

PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$

$\Leftrightarrow |2x-1|=|x-3|$

\(\Leftrightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)

a: Ta có: \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b: Ta có: \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)

\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)

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

c: Ta có: \(\sqrt{x^2-5x+6}=\sqrt{x-2}\)

\(\Leftrightarrow x^2-5x+6=x-2\)

\(\Leftrightarrow x^2-6x+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

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