giải phương trình
x-4\(\sqrt{x-2}\)+1=0
Giải phương trình
X +\(\sqrt{9-X^2}-X\sqrt[]{9-X^2}\)=3
\(x+\sqrt{9-x^2}-x\sqrt{9-x^2}=3\left(-3\le x\le3\right)\)
\(\Leftrightarrow\sqrt{9-x^2}-x\sqrt{9-x^2}=3-x\\ \Leftrightarrow9-x^2+x^2\left(9-x^2\right)-2x\sqrt{\left(9-x^2\right)^2}=9-6x+x^2\\ \Leftrightarrow9+8x^2-x^4-2x\left(9-x^2\right)=x^2-6x+9\\ \Leftrightarrow-x^4+2x^3+7x^2-12x=0\\ \Leftrightarrow-x\left(x^3-2x^2-7x+12\right)=0\Leftrightarrow-x\left(x^3-3x^2+x^2-3x-4x+12\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x^2+x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=3\left(N\right)\\x^2+x-4=0\left(1\right)\end{matrix}\right.\)
\(\Delta\left(1\right)=1-4\left(-4\right)=17>0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1-\sqrt{17}}{2}\left(N\right)\\x=\dfrac{-1+\sqrt{17}}{2}\left(N\right)\end{matrix}\right.\)
Vậy \(S=\left\{0;3;\dfrac{-1-\sqrt{17}}{2};\dfrac{-1+\sqrt{17}}{2}\right\}\)
Tick ✔
giải phương trình
x+1/2021 +x+2/2020=x+3/2019+x+4/2018
\(\dfrac{x+1}{2021}+\dfrac{x+2}{2020}=\dfrac{x+3}{2019}+\dfrac{x+4}{2018}\)
=>\(\dfrac{x+1}{2021}+1+\dfrac{x+2}{2020}+1=\dfrac{x+3}{2019}+1+\dfrac{x+4}{2018}+1\)
=>\(\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}=\dfrac{x+2022}{2019}+\dfrac{x+2022}{2018}\)
=> (x+2022)(\(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2019}-\dfrac{1}{2018}\))=0
=>x+2022=0
=> x=-2022
giải phương trình
x\(^3\) - 9x\(^2\) + 19x-11=0
\(\Leftrightarrow x^3-x^2-8x^2+8x+11x-11=0\)
\(\Leftrightarrow x^2\left(x-1\right)-8x\left(x-1\right)+11\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-8x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-8x+11=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4-\sqrt{5}\\x=4+\sqrt{5}\end{matrix}\right.\)
Giải bất phương trình
x^2>= 1
x^2 < 1
x^2+3x>=0
x^2+3x+3 >=0
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≥\(\dfrac{3}{2}\)
Giải bất phương trình
x-1/x-3>1 (đk: x#3)
giải giúp mình với ạ
\(\dfrac{x-1}{x-3}>1\left(x\ne3\right)\)
\(\Leftrightarrow\dfrac{x-1-x+3}{x-3}>0\)
\(\Leftrightarrow2>0\)
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}
Giải hệ phương trình
x - 1/y = 1
y - 1/z = 1
z - 1/x = 1
Mn giúp mik bài này ạ
Giải phương trình
x + 1 phần x - 1 trừ x -1 phần x + 1 cộng x^2 + 3 phần 1 - x^2 bằng 0
ĐKXĐ ; \(x\ne\pm1\)
Ta có : \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2+3}{1-x^2}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}+\dfrac{-x^2-3}{x^2-1}=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2-x^2-3=0\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1-x^2-3=0\)
\(\Leftrightarrow-x^2+4x-3=0\)
\(\Leftrightarrow-x^2+3x+x-3=0\)
\(\Leftrightarrow-x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=1\left(L\right)\end{matrix}\right.\)
=> X = 3
Vậy ..
Giải hệ phương trình
x^2+y^2=1
x^3+y^3=1