tìm số nguyên x biết: \(\dfrac{\sqrt{49}}{6}< \left|x-\dfrac{2}{3}\right|< -\dfrac{26}{\sqrt{81}}\)
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Câu 2:
\(\Leftrightarrow\dfrac{\left(n+2\right)!}{2!\cdot n!}-4\cdot\dfrac{\left(n+1\right)!}{n!\cdot1!}=2\left(n+1\right)\)
\(\Leftrightarrow\dfrac{\left(n+1\right)\left(n+2\right)}{2}-4\cdot\dfrac{n+1}{1}=2\left(n+1\right)\)
\(\Leftrightarrow\left(n+1\right)\left(n+2\right)-8\left(n+1\right)=4\left(n+1\right)\)
=>(n+1)(n+2-8-4)=0
=>n=-1(loại) hoặc n=10
=>\(A=\left(\dfrac{1}{x^4}+x^7\right)^{10}\)
SHTQ là: \(C^k_{10}\cdot\left(\dfrac{1}{x^4}\right)^{10-k}\cdot x^{7k}=C^k_{10}\cdot1\cdot x^{11k-40}\)
Số hạng chứa x^26 tương ứng với 11k-40=26
=>k=6
=>Số hạng cần tìm là: \(210x^{26}\)
\(\Rightarrow\frac{7}{6}< |x-\frac{2}{3}|< \frac{26}{9}\)
\(\Rightarrow\frac{21}{18}< |x-\frac{2}{3}|< \frac{52}{18}\)
Rùi tự thay vào
\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)
\(\Leftrightarrow\frac{7}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{9}\)
\(\Leftrightarrow\frac{7}{6}< 2\le\left|x-\frac{2}{3}\right|\le2< \frac{26}{9}\)
\(\Leftrightarrow\left|x-\frac{2}{3}\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{3}=2\\x-\frac{2}{3}=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=--\frac{4}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{8}{3};-\frac{4}{3}\right\}\)
a: ĐKXĐ: 3x^2+15/-6>=0
=>3x^2+15<=0(vô lý)
b: ĐKXĐ: -81/-x^2-12>=0
=>-x^2-12<0
=>-x^2<12
=>x^2>-12(luôn đúng)
c: ĐKXĐ: 31(x^2+21)/3>=0
=>x^2+21>=0(luôn đúng)
d: ĐKXĐ: -12/x^2+11>=0
=>x^2+11<0(vô lý)
e: ĐKXĐ: 21/-x^2-17>=0
=>-x^2-17>0
=>x^2+17<0(vô lý)
a) \(M=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{6\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-6\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)
b) \(M=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}=1-\dfrac{5}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Do \(\sqrt{x}\ge0\forall x\)
\(\Rightarrow\sqrt{x}\in\left\{3\right\}\Rightarrow x=9\left(tm\right)\)
a) Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)^2}\)
\(\left(\dfrac{1}{\sqrt{x}}-\sqrt{x}\right):\left(\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) (ĐK: \(x>0\))
\(=\left(\dfrac{1}{\sqrt{x}}-\dfrac{x}{\sqrt{x}}\right)\cdot\dfrac{-\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{-\sqrt{x}}\cdot\dfrac{-\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{-\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\left(\sqrt{x}+1\right)^2\)
c:
b;
Sửa đề: \(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
Do \(\left|x-\dfrac{2}{3}\right|\ge0;\forall x\)
Mà \(-\dfrac{26}{\sqrt{81}}< 0\)
\(\Rightarrow\) Không tồn tại x để \(\left|x-\dfrac{2}{3}\right|< -\dfrac{26}{\sqrt{81}}\)
Hay ko tồn tại số nguyên x thỏa mãn đề bài