a) Ta có: \(x^3-9x^2+19x-11=0\)

\(\Leftrightarrow x^3-x^2-8x^2+8x+11x-11=0\)

\(\Leftrightarrow x^2\left(x-1\right)-8x\left(x-1\right)+11\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-8x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-8x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{5}+4\\x=-\sqrt{5}+4\end{matrix}\right.\)

Vậy: \(S=\left\{1;\sqrt{5}+4;-\sqrt{5}+4\right\}\)

a) \(x^5-27+x^3-27x^2\) = 0

\(\Leftrightarrow x^3\left(x^2+1\right)-27\left(x^2+1\right)\)= 0

\(\Leftrightarrow\left(x^2+1\right)\left(x^3-27\right)=0\)

\(\Leftrightarrow x^3-27=0\) (Vì \(x^2+1>0\))

\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+2\dfrac{3}{2}x+\dfrac{9}{4}+\dfrac{27}{4}\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{27}{4}\right]=0\)

\(\Leftrightarrow x-3=0\) (Vì \(\left(x+\dfrac{3}{2}\right)^2+\dfrac{27}{4}>0\))

\(\Leftrightarrow x=3\)

Vậy tập nghiệm của phương trình là S = {3}

b)\(x^3-9x^2+19x-11=0\)

\(\Leftrightarrow\left(x^3-x^2\right)-\left(8x^2-8x\right)+\left(11x-11\right)=0\)

\(\Leftrightarrow x^2\left(x-1\right)-8x\left(x-1\right)+11\left(x-1\right)=0\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(x^2-8x+11\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2-\left(4+\sqrt{5}\right)x-\left(4-\sqrt{5}\right)x+11\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left\{x\left[x-\left(4+\sqrt{5}\right)\right]-\left(4-\sqrt{5}\right)\left[x-\left(4+\sqrt{5}\right)\right]\right\}=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-4-\sqrt{5}\right)\left(x-4+\sqrt{5}\right)=0\)

\(\Leftrightarrow x-1=0\) hoặc \(x-4-\sqrt{5}=0\) hoặc \(x-4+\sqrt{5}=0\)

\(\Leftrightarrow x=1\) hoặc \(x=4+\sqrt{5}\) hoặc \(x=4-\sqrt{5}\)

Vậy phương trình có tập nghiệm là \(S=\left\{1;4+\sqrt{5};4-\sqrt{5}\right\}\)

x^3 - 9X^2 +19x -11 =0

<=> (x^3 - x^2) - (8x^2 - 8x) +(11x-11)=0

<=> x^2(x-1) - 8x(x-1) + 11(x-1)=0

<=> (x-1)(x^2-8x+11) = 0

<=> x-1=0

<=> x=1

9x^3 - 6x^2 +12x=8

<=> 9x^3-6x^2+12x-8=0

<=. 3x^2(3x-2) + 4(3x-2)=0

<=> (3x-2)(3x^2 +4 ) =0

<=> 3x-2 = 0 (do 3x^2 +4 >= 4 >0)

<=> x= 2/3

a, \(x^2\)≥1

\(\Leftrightarrow\) x>1

b, \(x^2\)<1

\(\Rightarrow\) x∈∅

c, \(x^2\)+3x ≥ 0

\(\Leftrightarrow\) \(x^2\)≥-3x

\(\Leftrightarrow\) x≥-3

d, \(x^2\)+3x+3≥0

\(\Leftrightarrow\) \(\left(x+\dfrac{3}{2}\right)^2\)+\(\dfrac{3}{4}\)≥0+\(\dfrac{3}{4}\)

\(\Leftrightarrow\) \(x^2\)+\(\dfrac{3}{2}^2\)≥0

\(\Leftrightarrow\)\(x^2\)≥\(\dfrac{9}{4}\)

\(\Leftrightarrow\)x≥\(\dfrac{3}{2}\)

\(x-4\sqrt{x-2}+1=0\)(Đk x>2)

⇔\(x-2-4\sqrt{x-2}+4-1=0\)

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

\(\Leftrightarrow\left(\sqrt{x-2}-3\right)\left(\sqrt{x-2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}-3=0\\\sqrt{x-2}-1=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=9\\x-2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=3\end{matrix}\right.\)(thảo đk)

Vậy\(\left[{}\begin{matrix}x=11\\x=3\end{matrix}\right.\)là nghiệm của pt

ĐKXĐ: x≥2

x+1=\(4\sqrt{x-2}\) bình phương 2 vế ta đc:\(\left(x+1\right)^2=16\cdot\left(x-2\right)< =>x^2+2x+1=16x-32< =>x^2-14x+33=0\)

giải phương trình này ta đc:x1=11(nhận); x2=3(nhận)

vậy phương trình có 2 nghiệm: x1=11;x2=3

\(\left\{{}\begin{matrix}x^2+xy+y^2=1\\x-y-xy=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2+3xy=1\\x-y-xy=3\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x-y=u\\xy=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^2+3v=1\\u-v=3\end{matrix}\right.\)

\(\Rightarrow u^2+3\left(u-3\right)=1\)

\(\Leftrightarrow u^2+3u-10=0\Rightarrow\left[{}\begin{matrix}u=2\Rightarrow v=-1\\u=-5\Rightarrow v=-8\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}u=2\\v=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=2\\xy=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=x-2\\xy=-1\end{matrix}\right.\)

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

TH2: \(\left\{{}\begin{matrix}u=-5\\v=-8\end{matrix}\right.\) \(\Rightarrow...\) bạn tự làm tương tự