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

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

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

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

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

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

\(\dfrac{x+1}{2021}+\dfrac{x+2}{2020}=\dfrac{x+3}{2019}+\dfrac{x+4}{2018}\)

=>\(\dfrac{x+1}{2021}+1+\dfrac{x+2}{2020}+1=\dfrac{x+3}{2019}+1+\dfrac{x+4}{2018}+1\)

=>\(\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}=\dfrac{x+2022}{2019}+\dfrac{x+2022}{2018}\)

=> (x+2022)(\(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2019}-\dfrac{1}{2018}\))=0

=>x+2022=0

=> x=-2022

\(\left\{{}\begin{matrix}x+y=1\\2x-y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y=2\\2x-y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\\left(2x-2x\right)+\left(2y+y\right)=2-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\3y=-6\end{matrix}\right.\)

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

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

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

Vậy hệ pt có nghiệm duy nhất \(\left(x;y\right)=\left(3;-2\right)\)

\(\dfrac{x-1}{x-3}>1\left(x\ne3\right)\)

\(\Leftrightarrow\dfrac{x-1-x+3}{x-3}>0\)

\(\Leftrightarrow2>0\)

Vậy \(S=\left\{2\right\}\)

-ĐKXĐ: \(x\ne3\)

\(\dfrac{x-1}{x-3}>1\)

\(\Leftrightarrow\dfrac{x-1}{x-3}-\dfrac{x-3}{x-3}>0\)

\(\Leftrightarrow\dfrac{x-1-x+3}{x-3}>0\)

\(\Leftrightarrow\dfrac{2}{x-3}>0\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\)

-Vậy tập nghiệm của BĐT là {x l x>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