1)\(6x^2-20x+6=0\)

<=>\(6x^2-18x-2x+6=0\)

<=>6x(x-3)-2(x-3)=0

<=>(6x-2)(x-3)=0

<=>6x-2=0

hoặc x-3=0

<=>x=\(\frac{1}{3}\)

hoặc x=3

Vậy...

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

=>\(8x^2-2x+12x-3=0\)

<=>2x(4x-1)+3(4x-1)=0

<=>(2x+3)(4x-1)=0

<=>2x+3=0<=>x=\(\frac{3}{2}\)

hoặc 4x-1=0<=>x=\(\frac{1}{4}\)

Vậy ........

3)Phương trình tương đương: \(4x^2-2x+10x-5=0\)

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

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

Giải ra các trường hợp là xong

4)Phương trình tương đương:\(x^2-10x+25-1=0\)

<=>\(\left(x-5\right)^2-1^2=0\)

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

<=>(x-6)(x-4)=0 Giải các TH nữa là xong

5)\(x^2-5x-24\)=0

<=>\(x^2-8x+3x-24=0\)

<=>x(x-8)+3(x-8)=0

<=>(x+3)(x-8)=0

Giải ra các nghiệm nữa là xong

6)Phương trình tương đương :\(x^4+6x^2+9-9x^2=0\)

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

<=> \(\left(x^2+3x+3\right)\left(x^2-3x+3\right)\)=0

Đến đây tự làm nhé

7)Phương trình tương đương :\(4x^4-12x^2+9-8=0\)

<=>\(\left(2x-3\right)^2-\sqrt{8}^2\)=0

<=>(2x-3-\(\sqrt{8}\))\(\left(2x-3+\sqrt{8}\right)\)=0

Đến đây dễ rồi

a) \(x^2-16=0\Rightarrow x^2=16\Rightarrow x^2=\pm4\)

b) \(4x^2-9=0\Rightarrow\left(2x-3\right)\left(2x+3\right)=0\Rightarrow x=\pm1,5\)

c) \(25x^2-1=0\Rightarrow\left(5x-1\right)\left(5x+1\right)=0\Rightarrow x=\pm0,2\)

d) \(4\left(x-1\right)^2-9=0\Rightarrow\left(2x-2-3\right)\left(2x-2+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-5=0\Rightarrow x=2,5\\2x+1=0\Rightarrow x=-0,5\end{matrix}\right.\)

e) \(25x^2-\left(5x+1\right)^2=0\Rightarrow\left(5x+5x+1\right)\left(5x-5x-1\right)=0\Rightarrow10x+1=0\Rightarrow x=-0,1\)

f) \(\dfrac{1}{4}-9\left(x-1\right)^2=0\Rightarrow\left(\dfrac{1}{2}+3x-3\right)\left(\dfrac{1}{2}-3x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{7}{6}\end{matrix}\right.\)

g) \(\dfrac{1}{16}-\left(2x+\dfrac{3}{4}\right)^2=0\Rightarrow\left(\dfrac{1}{4}+2x+\dfrac{3}{4}\right)\left(\dfrac{1}{4}-2x-\dfrac{3}{4}\right)=0\Rightarrow\left[{}\begin{matrix}x=-0,5\\x=-0,25\end{matrix}\right.\)

h) \(\dfrac{1}{9}x^2-\dfrac{2}{3}x+1=0\Rightarrow\left(\dfrac{1}{3}x-1\right)^2=0\Rightarrow\dfrac{1}{3}x=1\Rightarrow x=3\)

k) \(4\left(x-3\right)^2-\left(2-3x\right)^2=0\Rightarrow\left(2x-6+2-3x\right)\left(2x-6-2+3x\right)=0\Rightarrow\left[{}\begin{matrix}-x-4=0\Rightarrow x=-4\\5x-8=0\Rightarrow x=1,6\end{matrix}\right.\)

l) \(x^2-x-12=0\Rightarrow x^2-4x+3x-12=0\Rightarrow x\left(x-4\right)+3\left(x-4\right)=0\Rightarrow\left(x+3\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

Cảm ơn bạn, ❤️

1.A; 2.B; 3.A

f) \(4x^2-12x+9=0\)

<=> \(\left(2x-3\right)^2\) = 0

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

<=> \(2x=3\) <=> \(x=\dfrac{3}{2}\)

Vậy ...............

g) \(3x^2+7x+2=0\)

<=> \(\left(3x^2+6x\right)+\left(x+2\right)=0\)

<=> \(3x\left(x+2\right)+\left(x+2\right)=0\)

<=> \(\left(x+2\right)\left(3x+1\right)=0\)

<=> \(\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{3}\end{matrix}\right.\)

Vậy ........................

h) \(x^2-4x+1=0\)

<=> \(\left(x^2-4x+4\right)-3=0\)

<=> \(\left(x-2\right)^2=3\)

<=> \(\left[{}\begin{matrix}x+2=\sqrt{3}\\x+2=-\sqrt{3}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3}-2\\x=-\sqrt{3}-2\end{matrix}\right.\)

Vậy .........................

i) \(2x^2-6x+1=0\)

<=> \(2\left(x^2-3x+2,25\right)-3,5=0\)

<=> \(\left(x-1,5\right)^2=1,75\)

<=> \(\left[{}\begin{matrix}x-1,5=\sqrt{1,75}\\x-1,5=-\sqrt{1,75}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{1,75}+1,5\\x=-\sqrt{1,75}+1,5\end{matrix}\right.\)

Vậy ...................

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

<=> \(\left(3x^2+6x\right)-\left(2x+4\right)=0\)

<=> \(3x\left(x+2\right)-2\left(x+2\right)\) = 0

<=> \(\left(x+2\right)\left(3x-2\right)=0\)

<=> \(\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy ....................................

f) \(4x^2-12x+9=0\)

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

\(\Rightarrow2x-3=0\)

\(\Rightarrow x=\dfrac{3}{2}\)

Vậy..

g) \(3x^2+7x+2=0\)

\(\Rightarrow3x^2+6x+x+2=0\)

\(\Rightarrow3x\left(x+2\right)+\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left(3x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{3}\end{matrix}\right.\)

Vậy..

h) \(x^2-4x+1=0\)

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

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

\(\Rightarrow\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2-\sqrt{3}=0\\x-2+\sqrt{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)

Vậy..

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

\(\Rightarrow3x^2+6x-2x-4=0\)

\(\Rightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy..

a)

a) $3x^2+12x-66=0$

=> 3(x + 2)² - 12 - 66 = 0

=> 3(x + 2)² - 78 = 0

=> 3(x + 2)² = 78

=> (x + 2)² = 26

=> x = \(\sqrt{26}-2\)

b)$9x^2-30x+225=0$

=> (3x - 5)² - 25 + 225 = 0

=> (3x - 5)² + 200 = 0

=> (3x - 5)² = -200

9x² - 30x + 225 không có ngiệmc)$x^2+3x-10=0$=> (x + 1,5)² - 2,25 - 10 = 0

=> (x + 1,5)² - 12,25 = 0

=> (x + 1,5)² = 12, 25

=> x + 1,5 = 3,5

=> x = 2

d)$3x^2-7x+1=0$=> 3(x - \(\dfrac{7}{6}\))² - \(\dfrac{49}{12}\) + 1 = 0

=> 3(x - \(\dfrac{7}{6}\))² - \(\dfrac{37}{12}\) = 0

=> 3(x - \(\dfrac{7}{6}\))² = \(\dfrac{37}{12}\)

=> (x - \(\dfrac{7}{6}\))² = \(\dfrac{37}{36}\)

=> x = \(\dfrac{\sqrt{37}}{6}+\dfrac{7}{6}=\dfrac{\sqrt{37}+7}{6}\)

e) $3x^2-7x+8=0$

=> 3(x - \(\dfrac{7}{6}\))² - \(\dfrac{49}{12}\)+ 8 = 0

=> 3(x - \(\dfrac{7}{6}\))² + \(\dfrac{47}{12}\) = 0

=> 3(x - \(\dfrac{7}{6}\))² = \(-\dfrac{47}{12}\)

KL : Không có ngiệm

a) \(x^2+x+1=\left(x^2+2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

ta có : \(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi \(x\) \(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\) với mọi \(x\) (đpcm)

b) \(2x^2+2x+1=2\left(x^2+x+\dfrac{1}{2}\right)=2\left(\left(x^2+2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{1}{4}\right)\)

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

ta có : \(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi \(x\) \(\Rightarrow2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\) với mọi \(x\) (đpcm)

c) \(-9x^2+12x-15=-\left(9x^2-12x+15\right)=-\left(9x^2-2.3.2x+4+11\right)\)

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

ta có : \(\left(3x-2\right)^2\ge0\) với mọi \(x\) \(\Rightarrow-\left(3x-2\right)^2-11\le-11< 0\) với mọi \(x\) (đpcm)

d) \(3x-x^2-4=-\left(x^2-3x+4\right)=-\left(\left(x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right)+\dfrac{7}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{7}{4}\) ta có \(\left(x-\dfrac{3}{2}\right)^2\ge0\) với mọi \(x\)

\(\Rightarrow-\left(x-\dfrac{3}{2}\right)^2-\dfrac{7}{4}\le\dfrac{-7}{4}< 0\) với mọi \(x\) (đpcm)

e) \(6x-3x^2-5=-3\left(x^2-2x+\dfrac{5}{3}\right)=-3\left(\left(x^2-2x+1\right)+\dfrac{2}{3}\right)\)

\(=-3\left(\left(x-1\right)^2+\dfrac{2}{3}\right)=-3\left(x-1\right)^2-2\)

ta có \(\left(x-1\right)^2\ge0\) với mọi \(x\) \(\Rightarrow-3\left(x-1\right)^2-2\le-2< 0\) với mọi \(x\) (đpcm)

thanks

e) \(\left(9x^2-49\right)+\left(3x+7\right)\left(7x+3\right)=0\)

\(\Rightarrow\text{[}\left(3x\right)^2-7^2\text{]}+\left(3x+7\right)\left(7x+3\right)=0\)

\(\Rightarrow\left(3x-7\right)\left(3x+7\right)+\left(3x+7\right)\left(7x+3\right)=0\)

\(\Rightarrow\left(3x+7\right)\text{[}\left(3x-7\right)+\left(7x+3\right)\text{]}=0\)

\(\Rightarrow\left(3x+7\right)\left(3x-7+7x+3\right)=0\)

\(\Rightarrow\left(3x+7\right)\left(10x-4\right)=0\)

=> 2 TH

*3x+7=0               *10x-4=0

=>3x=-7               =>10x=4

=>x=-7/3              =>x=4/10=2/5

vậy x=-7/3 hoặc x=2/5

g) \(\left(x-4\right)^2=\left(2x-1\right)^2\)

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

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

\(\Rightarrow\left(-x-3\right)\left(3x-5\right)=0\)

\(\Rightarrow-\left(x+3\right)\left(3x-5\right)=0\)

=> 2 TH

*-(x+3)=0          *3x-5=0

=>-x=-3            =>3x=5  

=x=3                =>x=5/3

h)\(x^2-x^2+x-1=0\)

\(\Rightarrow0+x-1=0\)

\(\Rightarrow x-1=0\)

=>x=0+1

=>x=1

vậy x=1

k, x(x+ 16) - 7x - 42 = 0

=>x^2+16x-7x-42=0

=>x^2+9x-42=0

vì x^2>0

do đó x^2+9x-42>0

nên o có gt nào của x t/m y/cầu đề bài

m)x^2+7x+12=0

=>x^2+3x++4x+12=0

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

=>(x+4).(x+3)=0

=>2 TH

=> *x+4=0

=>x=-4

vậy x=-4

*x+3=0

=>x=-3

vậy x=-3

n)x^2-7x+12=0

=>x^2-4x-3x+12=0

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

=>(x-3).(x-4)=0

=>2 TH

*x-3=0=>x=0+3=>x=3

*x-4=0=>x=0+4=>x=4

vậy x=3 hoặc x=4

a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1

b)(x+1)(x+2)(x+5)−x2(x+8)=27⇔x2+2x+x+2(x+5)−x3−8x2=27⇔x2(x+5)+2x(x+5)+x(x+5)+2(x+5)−x3−8x2=27⇔x3+5x2+2x2+10x+x2+5x+2x+10−x3−8x2=27⇔17x+10=27⇔17x=17⇒x=1