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b) \(\frac{2\left(x+1\right)}{3x^2+x}+\frac{13\left(x+1\right)}{3x^2+x+6\left(x+1\right)}=6\) (1)
Đặt \(a=x+1;b=3x^2+x\) thì
\(\left(1\right)\Leftrightarrow\frac{2a}{b}+\frac{13a}{b+6a}=6\)
\(\Leftrightarrow4a^2-7ab-2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(4a+b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=2b\\a=-\frac{1}{4}b\end{cases}}\)
Đến đây thì dễ rồi
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
ĐK: \(x-9\ne0\Rightarrow x\ne9\)
\(\sqrt{x}\ge0\Rightarrow x\ge0\)
\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)
\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)
2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)
Bài 1:
a) đkxđ: \(x\ne0;x\ne\pm1\)
\(D=\left(\frac{1}{1-x}+\frac{1}{1+x}\right)\div\left(\frac{1}{1-x}-\frac{1}{1+x}\right)+\frac{1}{x+1}\)
\(D=\left[\frac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}\right]\div\left[\frac{1+x-1+x}{\left(1-x\right)\left(1+x\right)}\right]+\frac{1}{x+1}\)
\(D=\frac{2}{\left(1-x\right)\left(1+x\right)}\div\frac{2x}{\left(1-x\right)\left(1+x\right)}+\frac{1}{x+1}\)
\(B=\frac{1}{x}+\frac{1}{x+1}\)
\(B=\frac{2x+1}{x+1}\)
b) Ta có: \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\) đều ko thỏa mãn đkxđ
c) Khi \(D=\frac{3}{2}\)
\(\Leftrightarrow\frac{2x+1}{x+1}=\frac{3}{2}\)
\(\Leftrightarrow4x+2=3x+3\Rightarrow x=1\) không thỏa mãn đkxđ
Bài 2: (Sửa đề tí nếu sai ib t lm lại nhé:)
a) đkxđ: \(x\ne\pm1\)
\(E=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)\div\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)
\(E=\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\div\frac{x-1+x\left(x+1\right)+2}{\left(x-1\right)\left(x+1\right)}\)
\(E=\frac{x^2+2x+1-x^2+2x-1}{x-1+x^2+x+2}\)
\(E=\frac{4x}{\left(x+1\right)^2}\)
b) Ta có: \(x^2-9=0\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
+ Nếu: \(x=3\)
=> \(E=\frac{4.3}{\left(3+1\right)^2}=\frac{3}{4}\)
+ Nếu: \(x=-3\)
=> \(E=\frac{4.\left(-3\right)}{\left(-3+1\right)^2}=-3\)
c) Để \(E=-3\)
\(\Leftrightarrow\frac{4x}{\left(x+1\right)^2}=-3\)
\(\Leftrightarrow4x=-3x^2-6x-3\)
\(\Leftrightarrow3x^2+10x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-\frac{1}{3}\end{cases}}\)
d) Để \(E< 0\)
\(\Leftrightarrow\frac{4x}{\left(x+1\right)^2}< 0\) , mà \(\left(x+1\right)^2>0\left(\forall x\right)\)
=> Để E < 0 => \(4x< 0\Rightarrow x< 0\)
Vậy x < 0 thì E < 0
e) Ta có: \(E-x-3=0\)
\(\Leftrightarrow\frac{4x}{\left(x+1\right)^2}=x+3\)
\(\Leftrightarrow4x=\left(x^2+2x+1\right)\left(x+3\right)\)
\(\Leftrightarrow x^3+5x^2+7x+3-4x=0\)
\(\Leftrightarrow x^3+5x^2+3x+3=0\)
Đến đây bấm máy tính thôi, nghiệm k đc đẹp cho lắm:
\(x=-4,4798...\) ; \(x=-0,2600...+0,7759...\) ; \(x=-0,2600...-0,7759...\)
Do \(x=0\) không phải nghiệm
\(x^2+3x+1=0\Leftrightarrow x+3+\frac{1}{x}=0\Leftrightarrow x+\frac{1}{x}=-3\)
\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2=9\Rightarrow x^2+\frac{1}{x^2}=7\)
Đặt \(x_n=x^n+\frac{1}{x^n}\Rightarrow x_1=-3;x_2=7\)
\(x_1x_n=\left(x+\frac{1}{x}\right)\left(x^n+\frac{1}{x^n}\right)=x^{n+1}+\frac{1}{x^{n+1}}+x^{n-1}+\frac{1}{x^{n-1}}=x_{n+1}+x_{n-1}\)
\(\Rightarrow x_{n+1}=x_1x_n-x_{n-1}=-3x_n-x_{n-1}\)
Cho \(n=2\Rightarrow x_3=x^3+\frac{1}{x^3}=-3.x_2-x_1=-18\)
\(n=3\Rightarrow x_4=x^4+\frac{1}{x^4}=-3x_3-x_2=47\)
\(n=4\Rightarrow x_5=x^5+\frac{1}{x^5}=-3x_4-x_3=-123\)
\(n=5\Rightarrow x_6=x^6+\frac{1}{x^6}=-3x_5-x_4=322\)
Thay vào và tính, kết quả rất to