a) Ta có: \(x^3-3x^2-16x+48=0\)

\(\Leftrightarrow x^2\left(x-3\right)-16\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm4\end{cases}}\)

b) Ta có: \(10x^2-33x-7=0\)

\(\Leftrightarrow\left(10x^2-35x\right)+\left(2x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\5x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{5}\end{cases}}\)

x³ - 3x² - 16x + 48 = 0

<=> ( x³ - 3x² ) - ( 16x - 48 ) = 0

<=> x²( x - 3 ) - 16( x - 3 ) = 0

<=> ( x - 3 )( x² - 16 ) = 0

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

<=> x = 3 hoặc x = 4 hoặc x = -4

10x² - 33x - 7 = 0

<=> 10x² + 2x - 35x - 7 = 0

<=> ( 10x² + 2x ) - ( 35x + 7 ) = 0

<=> 2x( 5x + 1 ) - 7( 5x + 1 ) = 0

<=> ( 5x + 1 )( 2x - 7 ) = 0

<=> \(\orbr{\begin{cases}5x+1=0\\2x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{5}\\x=\frac{7}{2}\end{cases}}\)

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

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

\(2x^2-7x+3=4x^2+4x-3\)

\(2x^2-7x+3-4x^2-4x+3=0\)

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

\(2x^2+12x-x-6=0\)

\(2x\left(x+6\right)-\left(x+6\right)=0\)

\(\left(x+6\right)\left(2x-1\right)=0\)

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

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

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

\(x\left(2x-3\right)=0\)

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

chiu lop 3 ma

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

1) \(x^2-9x=0\Rightarrow x\left(x-9\right)=0\Rightarrow x=0;9\)

2) \(x\left(x-4\right)-x^2=7\Rightarrow-4x=7\Rightarrow x=-\dfrac{7}{4}\)

3) \(3x+2\left(x-5\right)=5\Rightarrow5x-10=5\Rightarrow5x=15\Rightarrow x=3\)

4) \(25x^2-1=0\Rightarrow x^2=\dfrac{1}{25}\Rightarrow x=\pm\dfrac{1}{5}\)

5) \(3x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-5\right)=0\Rightarrow x=2;\dfrac{5}{3}\)

6) \(3x\left(x-7\right)+4\left(x-7\right)\Rightarrow\left(3x+4\right)\left(x-7\right)=0\Rightarrow x=-\dfrac{4}{3};7\)

7) \(4x^2-9=0\Rightarrow x^2=\dfrac{9}{4}\Rightarrow x=\pm\dfrac{3}{2}\)

8) \(10x\left(x-4\right)+2x-8=0\Rightarrow2\left(x-4\right)\left(5x+1\right)=0\Rightarrow x=4;-\dfrac{1}{5}\)

9) \(x\left(2x-5\right)-2x^2=0\Rightarrow x\left(2x-5-2x=0\right)\Rightarrow x=0\)

10) \(2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\Rightarrow x=0;2\)

11) \(2x\left(3-4x\right)+3\left(4x-3\right)=0\Rightarrow2x\left(4x-3\right)-3\left(4x-3\right)=0\Rightarrow\left(4x-3\right)\left(2x-3\right)=0\Rightarrow x=\dfrac{3}{4};\dfrac{3}{2}\)

12) \(2x\left(x-5\right)-2x^2=3\Rightarrow-10x=3\Rightarrow x=-\dfrac{3}{10}\)

a)

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

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

\(\Rightarrow x+2=\pm3\)

\(\Rightarrow x=-5;1\)

b)

\(25x^2-10x+1=0\)

\(\left(5x\right)^2-2\cdot5x+1^2=0\)

\(\Rightarrow\left(5x+1\right)^2=0\)

\(\Rightarrow5x+1=0\)

\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)

c)

\(x^2+14x+49=0\)

\(\Rightarrow x^2+2\cdot7x+7^2=0\)

\(\Rightarrow\left(x+7\right)^2=0;x+7=0\)

\(\Rightarrow x=-7\)

d)

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)

\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)

\(\Rightarrow2x+255=0\)

\(\Rightarrow2x=-255\)

\(\Rightarrow x=\dfrac{-255}{2}\)

7)(16-8x)(2-6x)=0  

=> 16 - 8x = 0 hoặc 2 - 6x = 0

=> 16 = 8x hoặc 2 = 6x

=> x = 2 hoặc x = 1/3
8) (x+4)(6x-12)=0  

=> x + 4 = 0 hoặc 6x - 12 = 0

=> x = -4 hoặc x = 2
9) (11-33x)(x+11)=0 

=> 11 - 33x = 0 hoặc x + 11 = 0

=> x = 1/3 hoặc x = -11
10) (x-1/4)(x+5/6)=0 

=> x - 1/4 = 0 hoặc x + 5/6 = 0

=> x = 1/4 hoặc x = -5/6
11) (7/8-2x)(3x+1/3)=0  

=> 7/8 - 2x = 0 hoặc 3x + 1/3 = 0

=> 2x = 7/8 hoặc 3x = -1/3

=> x = 7/16 hoặc x = -1/9
12)3x-2x^2=0  

=> x(3 - 2x) = 0

=> x = 0 hoặc 3 - 2x = 0

=> x = 0 hoặc x = 3/2

\(a,\left(16-8x\right)\left(2-6x\right)=0\)

\(\hept{\begin{cases}16-8x=0\\2-6x=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=\frac{1}{3}\end{cases}}}\)

\(b,\left(x+4\right)\left(6x-12\right)=0\)

\(\hept{\begin{cases}x+4=0\\6x-12=0\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\x=2\end{cases}}}\)

\(c,\left(11-33x\right)\left(x+11\right)=0\)

\(\hept{\begin{cases}11-33x=0\\x+11=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\\x=-11\end{cases}}}\)

\(d,\left(x-\frac{1}{4}\right)\left(x+\frac{5}{6}\right)=0\)

\(\hept{\begin{cases}x-\frac{1}{4}=0\\x+\frac{5}{6}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{5}{6}\end{cases}}}\)

\(e,\left(\frac{7}{8}-2x\right)\left(3x+\frac{1}{3}\right)=0\)

\(\hept{\begin{cases}\frac{7}{x}-2x=0\\3x+\frac{1}{3}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{7}{4}\\x=-\frac{1}{9}\end{cases}}}\)

\(f,3x-2x^2=0\)

\(x\left(3-2x\right)=0\)

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

Để phương trình có nghiệm \(\Delta'\ge0\)

 \(\Rightarrow\left(\dfrac{10}{2}\right)^2-1.\left(2m+7\right)\ge0\\ 25-2m-7\ge0\\ \Leftrightarrow18-2m\ge0\\ \Leftrightarrow18\ge2m\\ \Leftrightarrow m\le9\)

Vậy ...

a) x² + 10x + 16 = 0

<=> x² + 2x + 8x + 16 = 0

<=> x( x + 2 ) + 8( x + 2 ) = 0

<=> ( x + 2 )( x + 8 ) = 0

<=> \(\orbr{\begin{cases}x+2=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

b) 4x² - 12x - 7 = 0

<=> 4x² + 2x - 14x - 7 = 0

<=> 2x( 2x + 1 ) - 7( 2x + 1 ) = 0

<=> ( 2x + 1 )( 2x - 7 ) = 0

<=> \(\orbr{\begin{cases}2x+1=0\\2x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

a. \(x^2+10x+16=0\)

\(\Leftrightarrow x^2+8x+2x+16=0\)

\(\Leftrightarrow x\left(x+8\right)+2\left(x+8\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

b. \(4x^2-12x-7=0\)

\(\Leftrightarrow4x^2+2x-14x-7=0\)

\(\Leftrightarrow2x\left(2x+1\right)-7\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\2x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}2x=7\\2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}\)

             Bài làm :

 \(\text{a) }x^2+10x+16=0\)

\(\Leftrightarrow x^2+8x+2x+16=0\)

\(\Leftrightarrow x\left(x+8\right)+2\left(x+8\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

 \(\text{b) }4x^2-12x-7=0\)

\(\Leftrightarrow4x^2+2x-14x-7=0\)

\(\Leftrightarrow2x\left(2x+1\right)-7\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\2x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}2x=7\\2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}.}\)

\(ĐK:x\ge0;y\ge2;5x-y\ge0\\ PT\left(1\right)\Leftrightarrow\sqrt{y+3x}-\sqrt{5x-y}+\sqrt{2x+7y}-3\sqrt{x}=0\\ \Leftrightarrow\dfrac{2y-2x}{\sqrt{y+3x}+\sqrt{5x-y}}+\dfrac{7y-7x}{\sqrt{2x+7y}+3\sqrt{x}}=0\\ \Leftrightarrow\left(y-x\right)\left(\dfrac{2}{\sqrt{y+3x}+\sqrt{5x-y}}+\dfrac{7}{\sqrt{2x+7y}+3\sqrt{x}}\right)=0\\ \Leftrightarrow x=y\left(\dfrac{2}{\sqrt{y+3x}+\sqrt{5x-y}}+\dfrac{7}{\sqrt{2x+7y}+3\sqrt{x}}>0\right)\)

Thay vào \(PT\left(2\right)\Leftrightarrow x-4+\sqrt{x-2}=\sqrt{x^3-10x^2+33x-34}-\sqrt{x^3-9x^2+24x-16}\)

\(\Leftrightarrow\dfrac{x^2-9x+18}{x-4+\sqrt{x-2}}=\dfrac{-x^2+9x-18}{\sqrt{x^3-10x^2+33x-34}+\sqrt{x^3-9x^2+24x-16}}\\ \Leftrightarrow\left(x^2-9x+18\right)\left(\dfrac{1}{x-4+\sqrt{x-2}}+\dfrac{1}{\sqrt{x^3-10x^2+33x-34}+\sqrt{x^3-9x^2+24x-16}}\right)=0\\ \Leftrightarrow x^2-9x+18=0\left(\text{ngoặc lớn luôn }>0,\forall x\ge2\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=y=3\\x=y=6\end{matrix}\right.\)

Vậy ...