giúp mình dc ko :<<<<

Đợi mình chút

a, Để pt trên là pt bậc nhất 1 ẩn thì: \(m-1\ne0\Leftrightarrow m\ne1\)

 \(b,2x+5=3\left(x+2\right)-1\\ \Leftrightarrow2x+5=3x+6-1\\ \Leftrightarrow2x+5=3x+5\\ \Leftrightarrow x=0\)

Để pt (1) tương đương vs pt trên thì

\(2\left(m-1\right).0+3=2m-5\\ \Leftrightarrow2m-5=3\\ \Leftrightarrow2m=8\\ \Leftrightarrow m=4\)

\(a,3x-12=0\)

\(\Leftrightarrow3x=12\)

\(\Leftrightarrow x=4\)

Vậy \(S=\left\{4\right\}\)

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

\(\cdot TH1:x-2=0\)

\(\Leftrightarrow x=2\)

\(\cdot TH2:3x+3=0\)

\(\Leftrightarrow x=-1\)

vậy \(S=\left\{-1;2\right\}\)

\(c,\dfrac{x+2}{x-2}-\dfrac{6}{x+2}=\dfrac{x^2}{x^2-4}\left(ĐKXĐ:x\ne2;x\ne-2\right)\)

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

\(\Leftrightarrow x^2+2x+2x+4-6x+12=x^2\)

\(\Leftrightarrow x^2-x^2+2x+2x-6x+4+12=0\)

\(\Leftrightarrow-2x+16=0\)

\(\Leftrightarrow-2x=-16\)

\(\Leftrightarrow x=8\left(nhận\right)\)

Vậy S\(S=\left\{8\right\}\)

a) 3x-12=0
⟺3x=12⟺x=4
Vậy tập nghiệm của phương trình là S={4}
b) (x-2)(3x+3)=0
⟺ x-2=0     ⟺x=2          ⟺x=2
    3x+3=0    ⟺3x=-3      ⟺x=-1
Vậy tập nghiệm của phương trình là S={2;-1}
c)
           x-2≠0                                    x-2≠0                       x≠2
ĐKXĐ x+2≠0                        ⟺       x+2≠0             ⟺     x ≠-2  
           x²-4=(x-2)(x+2)≠0
     x+2/x-2 - 6/x+2 = x²/x²-4
⟺ (x+2)(x+2)/(x-2)(x+2) - 6(x-2)/(x-2)(x+2) = x²/(x-2)(x+2)
⟺(x+2)(x+2) - 6(x-2)=x²
⟺(x+2)(x+2-6)=x²
⟺(x+2)(x-4)-x²=0
⟺x²-4x+2x-8-x²=0
⟺-2x-8=0
⟺-2x=8
⟺x=-4
Vập tập nghiệm của phương trình là S={-4}

Xét ΔABC có MN//BC

nên AM/AB=AN/AC

=>AN/4=3/5

=>AN=2,4cm

\(\left(x-5\right)^2-6\left(1-2x\right)=\left(x-3\right)\left(x+3\right)\)

\(\Leftrightarrow x^2-10x+25-6+12x=x^2-9\)

\(\Leftrightarrow x^2-x^2-10x+12x+25-6+9=0\)

\(\Leftrightarrow2x+28=0\)

\(\Leftrightarrow2x=-28\)

\(\Leftrightarrow x=-14\)

Vậy \(S=\left\{-14\right\}\)

\(a,\dfrac{3x-2}{4}+\dfrac{x+3}{2}=\dfrac{x-1}{3}-\dfrac{-x-1}{12}\)

\(\Leftrightarrow3\left(3x-2\right)+6\left(x+3\right)=4\left(x-1\right)+x+1\)

\(\Leftrightarrow9x-6+6x+18=4x-4+x+1\)

\(\Leftrightarrow9x+6x-4x-x=1+6-18-4\)

\(\Leftrightarrow10x=-15\)

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

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

\(b,\left(3x+1\right)\left(x-2\right)=\left(x-2\right)\left(x+1\right)\)

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

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

\(\Leftrightarrow2x\left(x-2\right)=0\)

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

Vậy \(S=\left\{0;2\right\}\)

\(c,\dfrac{2}{x-1}+\dfrac{2}{x+1}-\dfrac{2x^2+2}{x^2-1}=0\left(ĐKXĐ:x\ne1;x\ne-1\right)\)

\(\Leftrightarrow2\left(x+1\right)+2\left(x-1\right)-2x^2+2=0\)

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

\(\Leftrightarrow-2x^2+4x-2=0\)

\(\Leftrightarrow-2x^2+2x+2x-2=0\)

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

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

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

Vậy \(S=\left\{2\right\}\)

\(\dfrac{x+1}{x-1}+\dfrac{1-x}{x+1}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\left(ĐKXĐ:x\ne1;x\ne-1\right)\)

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

\(\Leftrightarrow x^2+x+x+1+x-1-x^2+x=4\)

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

\(\Leftrightarrow4x-4=0\)

\(\Leftrightarrow4x=4\)

\(\Leftrightarrow x=1\left(loại\right)\)

Vậy \(S=\varnothing\)

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

⇔ \(\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(1-x\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

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

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

⇔ \(\dfrac{\left(x+1+x-1\right)\left(x+1-x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

⇔ \(\dfrac{2x.2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

⇔ \(\dfrac{4x}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

⇒ \(4x=4\)

⇔ \(x=1\) (ko thỏa mãn dktc)

Vậy phương trình vô nghiệm

Không tìm thấy tập tin

\(\Leftrightarrow x.\left(5x-4\right)-2.\left(5x-4\right)=0\\ \Leftrightarrow\left(5x-4\right)\left(x-2\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}5x-4=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{5}\\x=2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};2\right\}\)

<=> x(5x-4)-2(5x-4)=0

<=>(x-2)(5x-4)=0<=> x = 2 hoặc x=4/5

Đặt \(2y=a\)thì ta được

\(P=\frac{1}{x^2+a^2}+\frac{1}{xa}=\left(\frac{1}{x^2+a^2}+\frac{1}{2xa}\right)+\frac{1}{2xa}\)

\(\ge\frac{4}{x^2+a^2+2ax}+\frac{2}{\left(x+a\right)^2}=\frac{6}{\left(x+a\right)^2}\ge\frac{6}{4}=\frac{3}{2}\)