\(\Delta=\left(m-1\right)^2+8\left(m+1\right)=\left(m+3\right)^2\ge0;\forall x\Rightarrow\) pt luôn có 2 nghiệm

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{m-1}{2}\\x_1x_2=-\dfrac{m+1}{2}\end{matrix}\right.\)

\(\dfrac{1}{x_1^2}+\dfrac{1}{x_2^2}=\dfrac{25}{16}\Leftrightarrow\dfrac{x_1^2+x_2^2}{\left(x_1x_2\right)^2}=\dfrac{25}{16}\)

\(\Rightarrow\left(x_1+x_2\right)^2-2x_1x_2=\dfrac{25}{16}\left(x_1x_2\right)^2\)

\(\Rightarrow\left(\dfrac{m-1}{2}\right)^2+\dfrac{2\left(m+1\right)}{2}=\dfrac{25}{16}\left(\dfrac{m+1}{2}\right)^2\)

\(\Rightarrow9m^2+18m-55=0\Rightarrow\left[{}\begin{matrix}m=\dfrac{5}{3}\\m=-\dfrac{11}{3}\end{matrix}\right.\)

làm ơn giúp 🙏🙏🙏

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9

a: \(\left[\dfrac{1}{2}x^2\left(2x-1\right)^m-\dfrac{1}{2}x^{m+2}\right]:\dfrac{1}{2}x^2=0\)

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

\(\Leftrightarrow\left(2x-1\right)^m=x^m\)

=>2x-1=x

=>x=1

b: \(\left(2x-3\right)^8=\left(2x-3\right)^6\)

\(\Leftrightarrow\left(2x-3\right)^6\cdot\left(2x-4\right)\left(2x-2\right)=0\)

hay \(x\in\left\{\dfrac{3}{2};2;1\right\}\)

c: \(\Leftrightarrow4x^2-4x+1+y^2-\dfrac{2}{3}y+\dfrac{1}{9}+\dfrac{6}{9}=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(y-\dfrac{1}{3}\right)^2+\dfrac{6}{9}=0\)(vô lý)

a. ĐK: \(x\ne\pm2\)
\(M=\left[\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x+7}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{3-x+x-2}{x-2}\)

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

\(=\dfrac{x-5}{x+2}\)

b. \(\dfrac{x-5}{x+2}< 1\Leftrightarrow\dfrac{x-5}{x+2}-1< 0\)

\(\Leftrightarrow\dfrac{-7}{x+2}< 0\Leftrightarrow x+2>0\)

\(\Leftrightarrow x>-2\)
Vậy \(x>-2,x\ne2\)

Chia nhỏ ra

a: =>1/2x=7/2-2/3=21/6-4/6=17/6

=>x=17/3

b: =>2/3:x=-7-1/3=-22/3

=>x=2/3:(-22/3)=-1/11

c: =>1/3x+2/5x-2/5=0

=>11/15x=2/5

hay x=6/11

d: =>2x-3=0 hoặc 6-2x=0

=>x=3/2 hoặc x=3

a: ĐKXĐ: x<>2; x<>0

b: \(M=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)

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

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

\(=\dfrac{x}{2}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)

c: M>=-3

=>(x+1+6x)/2x>=0

=>(7x+1)/x>=0

=>x>0 hoặc x<=-1/7

a: ta có: \(\dfrac{\left(x+2\right)^2}{2}+\dfrac{\left(2x+1\right)^2}{4}+\dfrac{\left(2x-1\right)^2}{8}-\left(x+1\right)^2=0\)

\(\Leftrightarrow4\left(x^2+4x+4\right)+2\left(4x^2+4x+1\right)+4x^2-4x+1-8\left(x+1\right)^2=0\)

\(\Leftrightarrow4x^2+16x+16+8x^2+8x+2+4x^2-4x+1-8\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow16x^2+20x+19-8x^2-16x-8=0\)

\(\Leftrightarrow8x^2+4x+11=0\)

\(\text{Δ}=4^2-4\cdot8\cdot11=-336< 0\)

Vì Δ<0 nên phương trình vô nghiệm

b.

PT \(\Leftrightarrow \frac{x^2+2x+1}{2}-\frac{4x^2-4x+1}{3}+\frac{4x^2+4x+1}{4}-\frac{x^2-10x+25}{6}=0\)

\(\Leftrightarrow \left(\frac{x^2+2x+1}{2}+\frac{4x^2+4x+1}{4}\right)-\left(\frac{4x^2-4x+1}{3}+\frac{x^2-10x+25}{6}\right)=0\)

\(\Leftrightarrow \frac{6x^2+8x+3}{4}-\frac{9x^2-18x+27}{6}=0\)

\(\Leftrightarrow \frac{3(6x^2+8x+3)-2(9x^2-18x+27)}{12}=0\)

$\Leftrightarrow 5x-\frac{15}{4}=0$

$\Leftrightarrow x=\frac{3}{4}$

c.

PT $\Leftrightarrow (x^3+9x^2+27x+27)-(3x^3+12x^2)+(x^3+6x^2+12x+8)=(-x^3+3x^2-3x+1)-8$

$\Leftrightarrow 42x+42=0$

$\Leftrightarrow x=-1$

a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)

\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)

\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)