=>( 2x² -12x +18 ) ( 8x² +8x +2) =0

=> x² - 6x + 9 =0 => x =3

hoặc 4x² +4x +1 =0 =>x =-1/2 

dễ lắm tick đi rồi giải cho

(3x² + 10x - 8)² = (5x² - 2x + 10)²

<=> (3x² + 10x - 8)² - (5x² - 2x + 10)² = 0

<=> (3x² + 10x - 8 - 5x² + 2x - 10)(3x² + 10x - 8 + 5x² - 2x + 10) = 0

<=> (-2x² + 12x - 18)(8x² + 8x + 2) = 0

<=> -4(x² - 6x + 9)(4x²  + 4x + 1) = 0

<=> (x - 3)²(2x + 1)² = 0

<=> \(\orbr{\begin{cases}x-3=0\\2x+1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\x=-\frac{1}{2}\end{cases}}\)

Vậy S = {3; -1/2}

a: =>-x+2x=3-7

=>x=-4

b: =>6x+2+2x-5=0

=>8x-3=0

hay x=3/8

c: =>5x+2x-2-4x-7=0

=>3x-9=0

hay x=3

d: =>10x²-10x²-15x=15

=>-15x=15

hay x=-1

a, <=> x = -4 

b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8 

c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2 

d, <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1 

a, <=> x = -4 

b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8 

c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2 

d <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1 

a) x=-4

b)4x=3

x=3/4

c)3x=9

x=3

d) 15x=15

x=1

ĐKXĐ: \(x\in R\)

\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)

=>\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}+x^2+2x-4=0\)

\(\Leftrightarrow\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}+x^2+2x+1-5=0\)

=>\(\sqrt{3x^2+6x+7}-2+\sqrt{5x^2+10x+14}-3+\left(x+1\right)^2=0\)

=>\(\dfrac{3x^2+6x+7-4}{\sqrt{3x^2+6x+7}+2}+\dfrac{5x^2+10x+14-9}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>

\(\dfrac{3x^2+6x+3}{\sqrt{3x^2+6x+7}+2}+\dfrac{5x^2+10x+5}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>\(\dfrac{3\left(x^2+2x+1\right)}{\sqrt{3x^2+6x+7}+2}+\dfrac{5\left(x^2+2x+1\right)}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

\(\Leftrightarrow\dfrac{3\left(x+1\right)^2}{\sqrt{3x^2+6x+7}+2}+\dfrac{5\left(x+1\right)^2}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>\(\left(x+1\right)^2\left(\dfrac{3}{\sqrt{3x^2+6x+7}+2}+\dfrac{5}{\sqrt{5x^2+10x+14}+3}+1\right)=0\)

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

=>x+1=0

=>x=-1(nhận)

2:

a: =>x-4>=0

=>x>=4

b: =>x+1>0

=>x>-1

a) \(\left(2x-1\right)^2=49\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x=8\\2x=-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)

jup mk mấy câu kia vs

b, \(\left(x+2\right)^2=9\left(x-4x+4\right)\)

\(\Leftrightarrow\left(x+2\right)^2=\left[3\left(x-2\right)\right]^2\)

\(\Leftrightarrow\left(x+2\right)^2-\left[3\left(x-2\right)\right]^2=0\)

\(\Leftrightarrow\left(x+2+3x-6\right)\left(x+2-3x+6\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)

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

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

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

hay x=-1/2

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

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

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

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

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

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

Vậy: \(S=\left\{3;-\dfrac{1}{2}\right\}\)