a: \(\Rightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)

\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)

=>-4x=5

hay x=-5/4

b: \(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

=>42x=41

hay x=41/42

`a)(10x+9)x-(5x-1)(2x+3)=8`

`<=>10x^2+9x-10x^2-15x+2x+3=8`

`<=>-4x=5`

`<=>x=-5/4`     Vậy `S={-5/4}`

`b)(3x-5)(7-5x)+(5x+2)(3x-2)-2=0`

`<=>21x-15x^2-35+25x+15x^2-10x+6x-4-2=0`

`<=>42x=41`

`<=>x=41/42`       Vậy `S={41/42}`

a: 

=>-4x=5

hay x=-5/4

b: 

=>42x=41

hay x=41/42

\(\Leftrightarrow2\left(x^2-\dfrac{3}{2}x+\dfrac{5}{2}\right)=0\\ \Leftrightarrow2\left(x-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{31}{16}\right)=0\\ \Leftrightarrow2\left(x-\dfrac{3}{4}\right)^2+\dfrac{31}{8}=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-\dfrac{3}{4}\right)^2+\dfrac{31}{8}\ge\dfrac{31}{8}>0\right]\)

a: =>3x+3=4x-4

=>-x=-7

hay x=7(nhận)

b: (x-1)(x-3)=0

=>x-1=0 hoặc x-3=0

=>x=1 hoặc x=3

c: 2(x-1)+x=0

=>2x-2+x=0

=>3x-2=0

hay x=2/3

a, ĐKXĐ : x ≠ 1 ; x ≠ -1

\(\Rightarrow3\left(x+1\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x+3=4x-4\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\left(N\right)\)

b,

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

c,

\(\Leftrightarrow2x-2+x=0\)

\(\Leftrightarrow3x=2\)

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

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

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...

\(a,\Leftrightarrow x=7-4=3\\ b,\Leftrightarrow2x=-18+5=-13\\ \Leftrightarrow x=-\dfrac{13}{2}\\ c,\Leftrightarrow x-21=10\\ \Leftrightarrow x=31\\ d,\Leftrightarrow-12-x+19=0\\ \Leftrightarrow7-x=0\\ \Leftrightarrow x=7\)

a, <=> x=7-4

<=> x=3

b, 2x= -18 +5

<=>2x=-13

<=> x= -13/2

c, <=> x -21=-10

<=> x= -10 +21

<=> x=11

d, <=> -12+19 -x=0

<=> 7-x=0

<=> x=7

Ta có: \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\)

\(\left(y+0.4\right)^{100}\ge0\forall y\)

\(\left(z-3\right)^{678}\ge0\forall z\)

Do đó: \(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0.4\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)

Dấu '=' xảy ra khi

\(\left\{{}\begin{matrix}x-\dfrac{1}{5}=0\\y+0.4=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{2}{5}\\z=3\end{matrix}\right.\)

Vậy: (x,y,z)=\(\left(\dfrac{1}{5};-\dfrac{2}{5};3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+5x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\\ \Leftrightarrow3\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)