pt vô nghiệm 

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em ko bt lm

\(a+b=1\Rightarrow\left(a+b\right)^2=a^2+2ab+b^2=1\)

\(\Rightarrow ab=\frac{1-a^2-b^2}{2}\)

\(\Rightarrow M=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=1-3ab\)

\(=1-\frac{3-3a^2-3b^2}{2}\)

Để M nhỏ nhất

\(\Rightarrow\frac{3-3\left(a^2+b^2\right)}{2}\)phải có Max

=> \(3-3\left(a^2+b^2\right)\)đạt Max

Có \(3-3\left(a^2+b^2\right)\le3\left(Dấu"="xayrakhia=0;b=0\right)\)

Vậy Min M = 1-3/2=-1/2

Với a = 0 ; b = 0

M=a³+b³

=(a+b)(a² +b² + ab) ( hằng đẳng thức)

Mà a+b=1 nên:

M=a² +b² - ab

M= ( a^2 + b^2 + 2ab) - 3ab

M= ( a+b)² - 3ab

Lại có a+b=1 nên:

M= 1² - 3ab = 1 - 3ab 

3ab \(\le\)\(\frac{3\left(a+b\right)^2}{4}\)

=> M \(\ge\)\(-\)​​​​​​\(\frac{3\left(a+b\right)^2}{4}\) = 1-3/4 = 1/4

Do đó MinM = 1/4

=>a=b=1/2

\(a^2+b^2+c^2+d^2-ac-ab-ad=0\)

\(\left(b-\frac{a}{2}\right)^2+\left(c-\frac{a}{2}\right)^2+\left(d-\frac{a}{2}\right)^2+\frac{a^2}{4}=0\)

=>a=b=c=d=0

\(\frac{4x}{1-x^2}=\sqrt{5}\)   ĐKXĐ : x khác 1

\(\Rightarrow4x=\sqrt{5}\left(1-x^2\right)\)

\(\Leftrightarrow4x=\sqrt{5}-x^2\sqrt{5}\)

\(\Leftrightarrow x^2\sqrt{5}-4x-\sqrt{5}=0\)

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

\(\Leftrightarrow x\sqrt{5}\left(x-\sqrt{5}\right)+\left(x-\sqrt{5}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{5}\right)\left(x\sqrt{5}+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{5}=0\\x\sqrt{5}=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}\left(tmđk\right)\\x=-\frac{1}{\sqrt{5}}=-\frac{\sqrt{5}}{5}\left(tmđk\right)\end{cases}}}\)

\(4x=\sqrt{5}-\sqrt{5}x^2\)

\(\Rightarrow4x+\sqrt{5}x^2=\sqrt{5}\)

\(\Rightarrow x\left(4+\sqrt{5}x\right)=\sqrt{5}\)

\(\Rightarrow x.\sqrt{5}\left(\frac{4}{\sqrt{5}}+x\right)=\sqrt{5}\)

\(\Rightarrow x.\left(\frac{4}{\sqrt{5}}+x\right)=1\)

Với x = 1 \(\Rightarrow\frac{4}{\sqrt{5}}+x=1\Rightarrow x=1-\frac{4}{\sqrt{5}}=\frac{5-4\sqrt{5}}{5}\)

Với x = -1\(\Rightarrow\frac{4}{\sqrt{5}}+x=-1\Rightarrow x=-1-\frac{4}{\sqrt{5}}=-\frac{5+4\sqrt{5}}{5}\)

 ko có x thỏa mãn

a)\(ĐKXĐ:\hept{\begin{cases}x>3\\x\le-1\end{cases}}\)
TH1: \(x-3>0\)
 \(\left(x-3\right)\left(x+1\right)+4.\frac{x-3}{\sqrt{x-3}}\sqrt{x+1}=-3\)
\(\left(x-3\right)\left(x+1\right)+4\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)
Đặt \(t=\sqrt{\left(x-3\right)\left(x+1\right)}\left(t\ge0\right)\)
Phương trình trở thành:
\(t^2+4t+3=0\Leftrightarrow\orbr{\begin{cases}t=-1\\t=-3\end{cases}}\)(ktm)=> Vô Nghiệm
TH2: \(x-3< 0\)
\(\left(x-3\right)\left(x+1\right)-4.\frac{3-x}{\sqrt{3-x}}\sqrt{-x-1}=-3\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)-4\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)
Tự làm tiếp nhé
 

b)Nhân chéo chuyển vế rút gọn ta được:
\(x^3-2x^2+3x-2=0\)
\(\Leftrightarrow x\left(x^2-2x+1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)^2+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x+2\right)=0\)
\(\Rightarrow x=1\)