Giải giúp mình phương trình này với giải thích cho mình với ạ mình cảm ơn nhiều
\(\frac{x-4}{2016}+\frac{x-4038}{1009}+\frac{x+1004}{1008}=2\)
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a) \(4\sqrt{x}+\frac{2}{\sqrt{x}}< 2x+\frac{1}{2x}+2\)
hay \(2\sqrt{x}+\frac{1}{\sqrt{x}}< x+\frac{1}{4x}+1\)
\(\Leftrightarrow0< x+\frac{1}{4x}+1-2\sqrt{x}-\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow0< \left(\sqrt{x}\right)^2-2\sqrt{x}-2\sqrt{x}\cdot1+1+\frac{1}{\left(2\sqrt{x}\right)^2}-2\cdot\frac{1}{2\sqrt{x}}\)
\(\Leftrightarrow1< \left(\sqrt{x}-1\right)^2+\left(\frac{1}{2\sqrt{x}}-1\right)^2\)
\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}>1\\2\sqrt{x}>1\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>\frac{1}{4}\end{cases}\Rightarrow}x>1}\)
b) \(\frac{1}{1-x^2}>\frac{3}{\sqrt{1-x^2}}-1\left(1\right)\left(ĐK:-1< x< 1\right)\)
Ta có (1) <=> \(\frac{1}{1-x^2}-1-\frac{3x}{\sqrt{1-x^2}}+2>0\)\(\Leftrightarrow\frac{x^2}{1-x^2}-\frac{3x}{\sqrt{1-x^2}}+2>0\)
Đặt \(t=\frac{x}{\sqrt{1-x^2}}\)ta được
\(t^2-3t+2>0\Leftrightarrow\orbr{\begin{cases}\frac{x}{\sqrt{1-x^2}}< 1\\\frac{x}{\sqrt{1-x^2}}>2\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{1-x^2}>x\left(a\right)\\2\sqrt{1-x^2}< x\left(b\right)\end{cases}}}\)
(a) <=> \(\hept{\begin{cases}x< 0\\1-x^2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\1-x^2>x^2\end{cases}}}\)
\(\Leftrightarrow-1< x< 0\)hoặc \(\hept{\begin{cases}x\ge0\\x^2< \frac{1}{2}\end{cases}}\)
\(\Leftrightarrow-1< x< 0\)hoặc \(0\le x\le\frac{\sqrt{2}}{2}\Leftrightarrow-1< x< \frac{\sqrt{2}}{2}\)
(b) \(\Leftrightarrow\hept{\begin{cases}1-x^2>0\\x>0\\4\left(1-x^2\right)< x^2\end{cases}\Leftrightarrow\hept{\begin{cases}0< x< 1\\x^2>\frac{4}{5}\end{cases}\Leftrightarrow}\frac{2}{\sqrt{5}}< x< 1}\)
các bạn ơi mình đang cần gấp . Mình chỉ còn 20 phút thui . HUHU
Bài làm :
1) \(\frac{x+11}{4}=\frac{2x+4}{5}\)
\(\Leftrightarrow\left(x+11\right).5=4.\left(2x+4\right)\)
\(\Leftrightarrow5x+55=8x+16\)
\(\Leftrightarrow5x-8x=16-55\)
\(\Leftrightarrow-3x=-39\)
\(\Leftrightarrow x=\frac{-39}{-3}=\frac{39}{3}=13\)
2)\(\frac{x+4}{x+10}=\frac{3}{5}\)
\(\Leftrightarrow\left(x+4\right).5=\left(x+10\right).3\)
\(\Leftrightarrow5x+20=3x+30\)
\(\Leftrightarrow5x-3x=30-20\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=\frac{10}{2}=5\)
3)\(\frac{x+8}{x+14}=\frac{2}{3}\)
\(\Leftrightarrow\left(x+8\right).3=\left(x+14\right).2\)
\(\Leftrightarrow3x+24=2x+28\)
\(\Leftrightarrow3x-2x=28-24\)
\(\Leftrightarrow x=4\)
Dk: x\(\ge0\)
lien hop
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x+3}=2\Rightarrow x=1\)
@hieu nguyen Em có nhân chéo hai vế và khai triển ra nhưng cũng không ra cái gì ạ.
\(\frac{1}{x-3}=a,\frac{1}{y-4}=b\)
\(hpt\Leftrightarrow\hept{\begin{cases}a+b=\frac{5}{3}\\4a-3b=\frac{3}{2}\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{13}{14}\\b=\frac{31}{42}\end{cases}\Rightarrow}}\hept{\begin{cases}x=\frac{53}{13}\\y=\frac{166}{31}\end{cases}}\)
\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x^2-2x}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-2=x^2-2x\)
\(\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\)
Cho mình sửa lại nhé:
\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-2=x-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Đặt m = 1 / x - 3 và n = 1/y - 4
Khi đó ta có hệ m + n = 5/3
4 x x - 3 x n = 3/2
....Bạn tự giải tiếp nhé
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\left(x\ne-4;-5;-6;-7;-8\right)\)
\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{x}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow x^2+11x+28=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-13\left(tm\right)\end{cases}}}\)
vậy x=2; x=-13
Bài làm:
đkxđ: \(x\ne\left\{-4;-5;-6;-7\right\}\)
Ta có: \(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow x^2+11x+28=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-13\end{cases}}\)
Vậy tập nghiệm của PT \(S=\left\{-13;2\right\}\)
Ta có : \(\frac{x-4}{2016}+\frac{x-4038}{1009}+\frac{x+1004}{1008}=2\)
=> \(\frac{1009\left(x-4\right)}{2034144}+\frac{2016\left(x-4038\right)}{2034144}+\frac{2018\left(x+1004\right)}{2034144}=2\)
=> \(1009\left(x-4\right)+2016\left(x-4038\right)+2018\left(x+1004\right)=4068288\)
=> \(1009x-4036+2016x-8140608+2018x+2026072=4068288\)
=> \(5043x=10186860\)
=> \(x=2020\)
Vậy phương trình có nghiệm là x = 2020 .
\(\Leftrightarrow\frac{x-4}{2016}-1+\frac{x-4038}{1009}+2+\frac{x+1004}{1008}-3=0\)
\(\Leftrightarrow\frac{x-2020}{2016}+\frac{x-2020}{1009}+\frac{x-2020}{1008}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2016}+\frac{1}{1009}+\frac{1}{1008}\right)=0\)
\(\Rightarrow x=2020\)