tìm x nha

c, \(3x^2-7x+10=0\)

\(\Leftrightarrow3x^2+3x-10x+10=0\)

\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)

d, \(2x\left(x-10\right)-x+10=0\)

\(\Leftrightarrow2x\left(x-10\right)-\left(x-10\right)=0\)

\(\Leftrightarrow\left(x-10\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{1}{2}\end{matrix}\right.\)

1.C

2.D

\(x^2+6x+5=x^2+5x+x+5=x\left(x+5\right)+\left(x+5\right)=\left(x+1\right)\left(x+5\right)\) 

\(x^2-7x+12=x^2-4x-3x+12=x\left(x-4\right)-3\left(x-4\right)=\left(x-3\right)\left(x-4\right)\) 

\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)

\(a,x^2+6x+5=x^2+5x+x+5\)

\(=x\left(x+5\right)+\left(x+5\right)=\left(x+5\right)\left(x+1\right)\)

\(b,\)\(x^2-7x+12=x^2-3x-4x+12\)

\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)

\(c,\)\(x^2-7x+10=x^2-2x-5x+10\)

\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)

a) \(x^2+6x+5=x^2+5x+x+5\) 

 \(=x\left(x+5\right)+\left(x+5\right)=\left(x+1\right)\left(x+5\right)\) 

b)\(x^2-7x+12=x^2-4x-3x+12=x\left(x-4\right)-3\left(x-4\right)=\left(x-3\right)\left(x-4\right)\) 

c)\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)\)

  \(=\left(x-5\right)\left(x-2\right)\) 

hc tốt

1) x² -7x + 10 = x² - 2x - 5x + 10 = x(x - 2) - 5(x - 2) = (x - 5)(x - 2)

2) x² + 3x + 2 = x² + 2x + x  + 2 = x(x + 2) + (x + 2) = (x + 1)(x + 2)

3) x² - 7x + 12 = x² - 3x - 4x + 12 = x(x - 3) - 4(x - 3) = (x - 3)(x - 4)

4) x² + 7x + 12 = x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)

5) 16x - 5x² - 3 = 15x - 5x² + x - 3 = -5x(x - 3) + (x - 3) = (x - 3)(1 - 5x) 

6) 6x² + 7x - 3 = 6x² - 2x + 9x - 3 = 2x(3x - 1) + 3(3x - 1) = (2x + 3)(3x - 1)  

7) 3x² - 3x - 6 = 3x² - 6x + 3x - 6 = 3x(x - 2) + 3(x - 2) = (x - 2)(3x + 3) = 3(x - 2)(x + 1)

8) 3x² + 3x - 6 = 3x² - 3x + 6x - 6 = 3x(x - 1) + 6(x - 1) = (x - 1)(3x + 6) = 3(x - 1)(x + 2)

9) 6x² - 13x + 6 = 6x² - 9x -  4x + 6 = 3x(2x - 3) - 2(2x - 3) = (3x - 2)(2x - 3) 

10) 6x² + 15x  + 6 = 6x² + 12x + 3x + 6 = 6x(x + 2) + 3(x + 2) = (x + 2)(6x + 3) = 3(x + 2)(3x + 1)

11) 6x² - 20x + 6 = 6x² - 18x - 2x + 6 = 6x(x -3) - 2(x - 3) = (6x - 2)(x - 3) = 2(3x - 1)(x - 3)

12) 8x² + 5x - 3 = 8x² + 8x - 3x - 3 = 8x(x + 1) - 3(x + 1) = (x + 1)(8x - 3)

\(x^2-2\sqrt{x^2-7x+10}< 7x-2\)

\(ĐK:x\ge5\)

BPT \(\Leftrightarrow x^2-7x+2-2\sqrt{x^2-7x+10}< 0\)

\(\Leftrightarrow t^2-8-2t< 0\left(t=\sqrt{x^2-7x+10}\ge0\right)\)

\(\Leftrightarrow\left(t+2\right)\left(t-4\right)< 0\)

\(\Leftrightarrow-2< t< 4\Leftrightarrow-2< \sqrt{x^2-7x+10}< 4\)

\(\Leftrightarrow\sqrt{x^2-7x+10}< 4\Leftrightarrow x^2-7x-6< 0\)

\(\Leftrightarrow\orbr{\begin{cases}5\le x< \frac{7+\sqrt{73}}{2}\\\frac{7-\sqrt{73}}{2}< x\le2\end{cases}}\)

Chúc bạn học tốt !!!

\(x^2-2\sqrt{x^2-7x+10}< 7x-2\)

ĐKXĐ: \(x\ge5\)

Ta có BĐT \(\Leftrightarrow x^2-2\sqrt{x^2-7x+10}-7x+2< 0\)

\(\Leftrightarrow x^2-7x+10-2\sqrt{x^2-7x+10}+1-9< 0\)

\(\Leftrightarrow\left(\sqrt{x^2-7x+10}-1\right)^2-9< 0\)

\(\Leftrightarrow\left(\sqrt{x^2-7x+10}-4\right)\left(\sqrt{x^2-7x+10}-2\right)< 0\)

Vì \(\sqrt{x^2-7x+10}\ge0\Rightarrow\sqrt{x^2-7x+10}< 4\)

\(\Leftrightarrow x^2-7x+10< 16\)

\(\Leftrightarrow x^2-7x-6< 0\)

Chúc bạn học tốt !!!

\(x^2-2\sqrt{x^2-7x+10}< 7x-2\)

\(\Rightarrow x^2-7x+10-2\sqrt{x^2-7x+10}+1< 9\)

\(\Rightarrow\left(\sqrt{x^2-7x+10}-1\right)^2< 9\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x^2-7x+10}-1< 3\\\sqrt{x^2-7x+10}-1< -3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x^2-7x+10}< 4\\\sqrt{x^2-7x+10}< -2\left(L\right)\end{cases}}\)

\(\Rightarrow x^2-7x+10=16\)

\(\Rightarrow x^2-2x-5x+10=16\)

\(\Rightarrow\left(x-2\right)\left(x-5\right)=16\)

...........................

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

\(=\left(x^2-7x\right)^2-100\)

a) \(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow x^4-4x^2+2x^3-8x+x^2-4=0\)

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

\(\Leftrightarrow\left(x^2-4\right)\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x+1\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm2\\x=1\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{2;-2;1\right\}\)

b) \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)-72=0\)

Đặt \(t=x^2-4\), ta có :

\(t\left(t-6\right)-72=0\)

\(\Leftrightarrow t^2-6t-72=0\)

\(\Leftrightarrow\left(t-12\right)\left(t+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-12=0\\t+6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-16=0\left(tm\right)\\x^2+2=0\left(ktm\right)\end{cases}}\)

\(\Leftrightarrow x=\pm4\)

Vậy tập nghiệm của phương trình là \(S=\left\{4;-4\right\}\)

c) \(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2x^3+2x^2+5x^2+5x+2x+2=0\)

\(\Leftrightarrow2x^2\left(x+1\right)+5x\left(x+1\right)+2\left(x+1\right)=0\)

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

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

\(\Leftrightarrow\)\(x+1=0\)

hoặc \(2x+1=0\)

hoặc \(x+2=0\)

\(\Leftrightarrow\)\(x=-1\)

hoặc \(x=-\frac{1}{2}\)

hoặc \(x=-2\)

Vậy tập nghiệm của phương trình là \(S=\left\{-1;-2;-\frac{1}{2}\right\}\)

a, \(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow\left(x^3+x^2-4x-4\right)\left(x+1\right)=0\)

TH1 : \(x+1=0\Leftrightarrow x=-1\)

TH2 : \(x^3+x^2-4x-4=0\Leftrightarrow\left(x+1\right)\left(x^2-4\right)=0\)

=> \(x=-1;x=\pm2\)

b, \(\left(x+2\right)\left(x-2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow x^4-14x^2+40=72\)

\(\Leftrightarrow x^4-14x^2-32=0\) Đặt \(x^2=t\left(t\ge0\right)\)

Ta có pt mới : \(t^2-14t-32=0\) Tự xử 

1: =>|x-5|=5-7x+7x+28=33

=>x-5=33 hoặc x-5=-33

=>x=38 hoặc x=-28

3: 2|x-6|+7x-2=|x-6|+7x

=>|x-6|=2

=>x-6=2 hoặc x-6=-2

=>x=8 hoặc x=4