c: \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\\x=-5\\x=5\end{matrix}\right.\)

mik lớp 6 bạn

a: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)

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

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

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

cảm ơn nhìu ạ!!!!

ĐKXĐ: \(x\ne-\dfrac{5}{6}\)

a) Để A>0 thì \(\left[{}\begin{matrix}7x-8>0\\6x+5< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x>8\\6x< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{8}{7}\\x< -\dfrac{5}{6}\end{matrix}\right.\)

b) Để A<0 thì \(\left\{{}\begin{matrix}6x+5>0\\7x-8< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-\dfrac{5}{6}\\x< \dfrac{8}{7}\end{matrix}\right.\Leftrightarrow-\dfrac{5}{6}< x< \dfrac{8}{7}\)

c) Để A=0 thì \(\dfrac{7x-8}{6x+5}=0\)

\(\Leftrightarrow7x-8=0\)

\(\Leftrightarrow x=\dfrac{8}{7}\)

Cảm ơn bạn nhiều !

b: \(\Delta=\left\lbrack2\left(m+1\right)\right\rbrack^2-4\cdot1\cdot\left(2m+1\right)\)

\(=4m^2+8m+4-8m-4=4m^2\) >=0∀m

=>Phương trình luôn có hai nghiệm

Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=2\left(m+1\right)=2m+2;x_1x_2=\frac{c}{a}=2m+1\)

\(x_1-x_2=8\)

=>\(\left(x_1-x_2\right)^2=8^2=64\)

=>\(\left(x_1+x_2\right)^2-4x_1x_2=64\)

=>\(\left(2m+2\right)^2-4\left(2m+1\right)=64\)

=>\(4m^2+8m+4-8m-4=64\)

=>\(4m^2=64\)

=>\(m^2=16\)

=>m=4 hoặc m=-4

\(x_1x_2-2\left(x_1+x_2\right)\le5\)

=>2m+1-2(2m+2)<=5

=>2m+1-4m-2<=5

=>-2m-1<=5

=>-2m<=6

=>m>=-3

`a)(2x-1)^2-0,25=0`

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

`<=>(2x-1,5)(2x-0,5)=0`

`<=>[(x=0,75)(x=0,25):}`

`b)x^2+9=6x`

`<=>(x-3)^2=0`

`<=>x-3=0`

`<=>x=3`

`c)(x^2-4)-3x-6=0`

`<=>(x-2)(x+2)-3(x+2)=0`

`<=>(x+2)(x-2-3)=0`

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

`<=>[(x=-2),(x=5):}`

a: =>(2x-1-0,5)(2x-1+0,5)=0

=>(2x-1,5)(2x-0,5)=0

=>x=0,25 hoặc x=0,75

b: =>x^2-6x+9=0

=>(x-3)^2=0

=>x-3=0

=>x=3

c: =>(x-2)(x+2)-3(x+2)=0

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

=>x=5 hoặc x=-2

Giải nhanh giúp mình,please

Ta có: \(x^2+4x+3\left|x+2\right|=0\)

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

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

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

=>|x+2|-1=0

=>|x+2|=1

=>\(\left[\begin{array}{l}x+2=1\\ x+2=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\\ x=-3\end{array}\right.\)