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) \(\Leftrightarrow\hept{\begin{cases}1-x^2>0\\x>0\\4\left(1-x^2\right)< x^2\end{cases}\Leftrightarrow\hept{\begin{cases}0< x< 1\\x^2>\frac{4}{5}\end{cases}\Leftrightarrow}\frac{2}{\sqrt{5}}< x< 1}\)

ok đợi nấu ăn xong r làm cho

a) điều kiện x>0

khi đó

\(\left(a\right)\Leftrightarrow4\left(\sqrt{4}+\frac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)^2\)

\(\Leftrightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}>2\Leftrightarrow2x-4\sqrt{x}+1>0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}< \frac{2-\sqrt{2}}{2}\\\sqrt{x}>\frac{2+\sqrt{2}}{2}\end{cases}}\)

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\left(x\ne-4;-5;-6;-7;-8\right)\)

\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{x}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow x^2+11x+28=54\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-13\left(tm\right)\end{cases}}}\)

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 \(S=\left\{-13;2\right\}\)

x² + 9x + 20 = ( x + 4 )( x + 5 )

x² + 11x + 30 = ( x + 5 )( x + 6 )

x² + 13x + 42 = ( x + 6 )( x + 7 )

=> Phương trình đã cho đưa về dạng :

\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\left(1\right)\)

ĐKXĐ : \(x\ne\left\{-4;-5;-6;-7\right\}\)

\(\left(1\right)\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{x+7}{\left(x+4\right)\left(x+7\right)}-\frac{x+4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)

\(\Leftrightarrow x^2+11x+28-54=0\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x^2+13x\right)-\left(2x+26\right)=0\)

\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+13=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-13\\x=2\end{cases}\left(tmđk\right)}\)

Vậy S = { -13 ; 2 }

Dk: x\(\ge0\)

lien hop

\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x+3}=2\Rightarrow x=1\)

B​ạn có thể giải thích rõ hộ mình dc k???

oh ok mình hiểu r thanks =)))

ĐK: x khác 1 và - 1

\(\frac{x+1}{2\left(x-1\right)}-\frac{x-1}{2\left(x+1\right)}=\frac{2}{\left(x-1\right)\left(x+1\right)}\)

<=> \(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\frac{4}{2\left(x-1\right)\left(x+1\right)}\)

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

<=> \(\left(x+1-x+1\right)\left(x+1+x-1\right)=4\)

<=> 2.2x = 4 

<=> x = 1 loại 

Vậy phương trình vô nghiệm 

e trả lời sau đc ko ạ ? ):

\(\frac{x+1}{2x-2}-\frac{x-1}{2x+2}=\frac{2}{x^2-1}\) ĐKXĐ : \(x\ne\pm1\)

\(\frac{x+1}{2\left(x-1\right)}-\frac{x-1}{2\left(x+1\right)}=\frac{2}{\left(x-1\right)\left(x+1\right)}\)

\(\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{2\left(x+1\right)\left(x-1\right)}=\frac{4}{2\left(x+1\right)\left(x-1\right)}\)

Khử mẫu ta đc : \(\left(x+1\right)^2-\left(x-1\right)^2=4\)

\(4x=4\Leftrightarrow x=1\)Theo ĐKXĐ : ktm 

Vậy pt vô nghiệm.

Bạn chú ý cách viết phương trình.

Phương trình chỉ có dạng f(x)=g(x) thôi, không có dạng A=f(x)=g(x) như bạn viết.

\(VT=\left[8\left(x+\frac{1}{x}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\right]+4\left(x^2+\frac{1}{x^2}\right)^2\)

\(=4\left(x+\frac{1}{x}\right)^2\left(2-x^2-\frac{1}{x^2}\right)+4\left(x^2+\frac{1}{x^2}\right)^2\)

\(=-4\left(x+\frac{1}{x}\right)^2\left(x-\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2\)

\(=-4\left(x^2-\frac{1}{x^2}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2\)

\(=-4x^4+8-\frac{4}{x^4}+4x^4+8+\frac{4}{x^4}\)

\(=16\)

Phương trình đã cho trở thành

\(\left(x+4\right)^2=16\\ \Leftrightarrow\orbr{\begin{cases}x+4=-4\\x+4=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-8\\x=0\end{cases}}\)

\(giải:\)

\(1,\)\(\frac{x}{5}+\frac{2x+1}{3}=\frac{x-5}{15}\)

\(\Leftrightarrow\frac{x}{5}+\frac{2x+1}{3}-\frac{x-15}{15}=0\)

\(\Leftrightarrow\frac{3x}{15}+\frac{5\left(2x+1\right)}{15}-\frac{x-15}{15}=0\)

\(\Leftrightarrow\frac{3x+5\left(2x+1\right)-\left(x-15\right)}{15}=0\)

\(\Leftrightarrow\frac{3x+10x+5-x+15}{15}=0\)

\(\Leftrightarrow\frac{12x+20}{15}=0\)

\(\Rightarrow12x+20=0\)

\(\Leftrightarrow12x=-20\Leftrightarrow x=\frac{-5}{3}\)

vậy tập nghiệm của phương trình là \(s=\left[\frac{-5}{3}\right]\)

\(2,\)\(\left(x^3-64\right)+6x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x^3-4^3\right)+6x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+16\right)+6x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+16+6x\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2+10x+16\right)=0\)

 \(mà\)\(x^2+10x+16>0\)

\(\Rightarrow x-4=0\Rightarrow x=4\)

vậy x=4 là nghiệm của phương trình

\(3,\)\(\frac{x+2}{x-2}-\frac{x-2}{x+2}=\frac{16}{x^2-4}\)

\(\Leftrightarrow\frac{x+2}{x-2}-\frac{x-2}{x+2}=\frac{16}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{16}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow\left(x+2\right)\left(x+2\right)-\left(x-2\right)\left(x-2\right)=16\)\

\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2-16=0\)

\(\Leftrightarrow x^2+4x+4-x^2+4x-4-16=0\)

\(\Leftrightarrow8x-16=0\)

\(\Leftrightarrow8\left(x-2\right)=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

vậy x=2 là nghiệm của phương trình

Mình sẽ k cho bạn nào nhanh nhất nhé <3

\(\frac{1}{x-3}=a,\frac{1}{y-4}=b\)

\(hpt\Leftrightarrow\hept{\begin{cases}a+b=\frac{5}{3}\\4a-3b=\frac{3}{2}\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{13}{14}\\b=\frac{31}{42}\end{cases}\Rightarrow}}\hept{\begin{cases}x=\frac{53}{13}\\y=\frac{166}{31}\end{cases}}\)

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)

Dễ thấy: \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\Rightarrow x+2004=0\Leftrightarrow x=-2014\)

x = -2014

ti-ck nha

.........

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\left(x+1\right)=0\)

\(\Rightarrow x+1=0\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)

\(x=-1\)

Vậy \(x=-1\)

\(pt\Leftrightarrow\frac{6\left(x+1\right)+3\left(x+3\right)}{4.3}=\frac{3.4.3-4\left(x+2\right)}{4.3}\)

\(\Leftrightarrow6x+6+3x+9=36-4x-8\)

\(\Leftrightarrow13x=13\)

\(\Leftrightarrow x=1\)