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

\(\Leftrightarrow2x^2+5x-3-4x^2+1=0\)

\(\Leftrightarrow-2x^2+5x-2=0\)

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=2\end{cases}}\)

= - 20 nha

Bạn giải hộ mk cách làm đc ko :3

Ta có:

\(\frac{AB}{BD}=\frac{AB}{BC+CD}=\frac{AB}{AB+AB}=\frac{1}{2}\)

\(\frac{AB}{AD}=\frac{AB}{AB+BC+CD}=\frac{AB}{AB+AB+AB}=\frac{1}{3}\)

\(\frac{AC}{AD}=\frac{AB+BC}{AB+BC+CD}=\frac{AB+AB}{AB+AB+AB}=\frac{2}{3}\)

(Vì AB=BC=CD) (Lưu ý 3 tỉ số ở đề bài không thể bằng nhau)

Cảm ơn bạn nhiều 

a) \(5\left(\frac{1}{x}-1\right)=\left(\frac{1}{x}-1\right)\left(x^2+1\right)\) (x khác 0)

\(\Leftrightarrow\left(x^2+1\right)\left(\frac{1}{x}-1\right)-5\left(\frac{1}{x}-1\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}-1\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}-1=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=1\\x^2=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=\pm2\end{cases}}\)

b) đk: \(x\ne-2\)

Ta có: \(\left(x^2+x+1\right)\cdot\frac{3x+1}{x+2}=\left(x^2+x+1\right)\cdot\frac{x}{2\left(x+2\right)}\)

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

\(\Leftrightarrow\left(x^2+x+1\right)\cdot\frac{5x+2}{2\left(x+2\right)}=0\)

Vì \(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow\frac{5x+2}{2\left(x+2\right)}=0\Rightarrow5x+2=0\Rightarrow x=-\frac{2}{5}\)

567-589=

M = x² + 26y² - 10xy + 14x - 76y + 59

= ( x² - 10xy + 25y² + 14x - 70y + 49 ) + ( y² - 6y + 9 ) + 1

= ( x - 5y + 7 )² + ( y - 3 )² + 1

Vì \(\hept{\begin{cases}\left(x-5y+7\right)^2\\\left(y-3\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\forall x,y\)

Dấu "=" xảy ra khi x = 8 ; y = 3

Vậy MinM = 1 <=> x = 8 . y = 3

Ta có : \(M=x^2+26y^2-10xy+14x-76y+59\)

\(=\left(x^2-10xy+25y^2\right)+14\left(x-5y\right)+49+\left(y^2-6y+9\right)+1\)

\(=\left(x-5y\right)^2+14\left(x-5y\right)+49+\left(y-3\right)^2+1\)

\(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\forall x,y\)

Dấu \("="\)xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-5y+7\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-5y+7=0\\y-3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-15+7=0\\y=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=8\\y=3\end{cases}}\)

Vậy \(MinM=1\Leftrightarrow\hept{\begin{cases}x=8\\y=3\end{cases}}\)

Ta có : \(\left(a-b\right)^2\ge0\forall a,b\)

\(\Rightarrow a^2+b^2-2ab\ge0\)

\(\Rightarrow a^2+b^2\ge2ab\)

\(\Rightarrow2\left(a^2+b^2\right)\ge2ab+a^2+b^2=\left(a+b\right)^2\left(1\right)\)

Chia cả 2 vế của \(\left(1\right)\)cho 4 , ta được :

\(\frac{a^2+b^2}{2}\ge\frac{\left(a+b\right)^2}{4}=\left(\frac{a+b}{2}\right)^2\)

\(\Rightarrowđpcm\)

Không bạn nha=))

Đặt \(a=x+4\).

Ta có: \(\left(a-1\right)^4+\left(a+1\right)^4=2\)

\(\Leftrightarrow\left(a^4-4a^3+6a^2-4a+1\right)+\left(a^4+4a^3+6a^2+4a+1\right)=2\)

\(\Leftrightarrow2a^4+12a^2+2=2\)

\(\Leftrightarrow a^4+6a^2=0\)

\(\Leftrightarrow a^2\left(a^2+6\right)=0\)

\(\Leftrightarrow a^2=0\)

\(\Rightarrow x+4=0\)

\(\Leftrightarrow x=-4\).

Đặt \(x+4=a\)

Khi đó ,PT tương đương với :

\(\left(a-1\right)^4+\left(a+1\right)^4=2\)

\(\Leftrightarrow a^4-4a^3+6a^2-4a+1+a^4+4a^3+6a^2+4a+1=2\)

\(\Leftrightarrow2a^4+12a^2+2=2\)

\(\Leftrightarrow a^4+6a^2+1=1\)

\(\Leftrightarrow a^4+6a^2=0\)

\(\Leftrightarrow a^2\left(a^2+6\right)=0\)

\(\Leftrightarrow a^2=0\)( do \(a^2+6>0\forall a\))

\(\Leftrightarrow a=0\)

\(\Leftrightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy PT có 1 nghiệm duy nhất là : \(x=-4\)