Ta có \(\sqrt{3x^2+6x+7}=\sqrt{3\left(x+1\right)^2+4}\ge\sqrt{4}=2\)

Dấu"=" xảy ra khi x=-1

Tương tự \(\sqrt{5x^2+10x+14}=\sqrt{5\left(x+1\right)^2+9}\ge\sqrt{9}=3\)

Dấu"=" xảy ra khi x=-1

\(\Rightarrow4-2x-x^2\ge5\)

\(\Rightarrow-\left(x+1\right)^2+5\ge5\)

\(\Rightarrow\left(x+1\right)^2\le0\)

mà \(\left(x+1\right)^2\ge0\)

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

Vậy....................

mn ơi giúp em vs ạ !!!

giúp e vs

a , \(x=2\left(\sqrt{x-1}-\sqrt{x-2}\right)\)

suy ra x =2 

b, x=3 

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ĐKXĐ: \(-1\le x\le\dfrac{5}{2}\)

\(\Leftrightarrow\sqrt{3x+3}-3+1-\sqrt{5-2x}=x^3-3x^2-10x+24\)

\(\Leftrightarrow\dfrac{3\left(x-2\right)}{\sqrt{3x+3}+3}+\dfrac{2\left(x-2\right)}{1+\sqrt{5-2x}}=\left(x-2\right)\left(x-4\right)\left(x+3\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{3}{\sqrt{3x+3}+3}+\dfrac{2}{1+\sqrt{5-2x}}=\left(x-4\right)\left(x+3\right)\left(1\right)\end{matrix}\right.\)

Xét (1), ta có:

\(\dfrac{3}{\sqrt{3x+3}+3}+\dfrac{2}{1+\sqrt{5-2x}}>0\)

\(-1\le x\le\dfrac{5}{2}\Rightarrow\left\{{}\begin{matrix}x+3>0\\x-4< 0\end{matrix}\right.\) \(\Rightarrow\left(x+3\right)\left(x-4\right)< 0\)

\(\Rightarrow\left(1\right)\) vô nghiệm hay pt có nghiệm duy nhất \(x=2\)

1. \(x^3-6x^2+10x-4=0\)

<=> \(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)

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

<=> \(\orbr{\begin{cases}x=2\\x^2-4x+2=0\left(1\right)\end{cases}}\)

Giải pt (1): \(\Delta=\left(-4\right)^2-4.2=8>0\)

=> pt (1) có 2 nghiệm: \(x_1=\frac{4+\sqrt{8}}{2}=2+\sqrt{2}\)

\(x_2=\frac{4-\sqrt{8}}{2}=2-\sqrt{2}\)

1) Ta có: \(x^3-6x^2+10x-4=0\)

       \(\Leftrightarrow\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)

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

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

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

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

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

                                             \(\Leftrightarrow\)\(x-2=\pm\sqrt{2}\)

                                             \(\Leftrightarrow\)\(\orbr{\begin{cases}x=2+\sqrt{2}\approx3,4142\left(TM\right)\\x=2-\sqrt{2}\approx0,5858\left(TM\right)\end{cases}}\)

Vậy \(S=\left\{0,5858;2;3,4142\right\}\)

4) \(x^4+5x^3-12x^2+5x+1=0\)

<=> \(\left(x^4-x^3\right)+\left(6x^3-6x^2\right)-\left(6x^2-6x\right)-\left(x-1\right)=0\) 

<=> \(\left(x^3+6x^2-6x-1\right)\left(x-1\right)=0\)

<=> \(\left[\left(x-1\right)\left(x^2+x+1\right)-6x\left(x-1\right)\right]\left(x-1\right)=0\)

<=> \(\left(x-1\right)^2\left(x^2-5x+1\right)=0\)

<=> \(\orbr{\begin{cases}x=1\\x^2-5x+1=0\left(1\right)\end{cases}}\)

Giải pt (1) ta có: \(\Delta=\left(-5\right)^2-4=21>0\)

=> pt có 2 nghiệm

\(x_1=\frac{5+\sqrt{21}}{2}\); \(x_2=\frac{5-\sqrt{21}}{2}\)