Cho \(f\left(x\right)=\left(x^4-3x+1\right)^{2016}\) . với \(x=9-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}+\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}\)
Giải hộ tớ ạ!!!
Tính giá trị đa thức \(f_{\left(x\right)}=\left(x^4-3x+1\right)^{2016}\) tại \(x=9-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}+\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}\)
\(x=9-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}+\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}\)
\(=9-\frac{2}{\sqrt{9-4\sqrt{5}}}+\frac{2}{\sqrt{9+4\sqrt{5}}}\)
\(=9-\frac{2}{\sqrt{\left(\sqrt{5}-2\right)^2}}+\frac{2}{\sqrt{\left(\sqrt{5}+2\right)^2}}\)
\(=9-\frac{2}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+2}\)
\(=9-\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=9-\frac{8}{5-4}\)
= 1
\(f\left(x\right)=\left(1^4-3+1\right)^{2016}=1\)
tính giá trị của f(x) \(=\left(x^4-3x+1\right)^{2016}\) tại \(x=9-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}+\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}\)
Ta có : \(\left(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\right)^2\)
\(=\frac{1}{\frac{9}{4}+\sqrt{5}}+\frac{1}{\frac{9}{4}-\sqrt{5}}-2.\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}.\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\)
\(=\frac{\frac{9}{4}-\sqrt{5}+\frac{9}{4}+\sqrt{5}}{\frac{1}{16}}-2.\frac{1}{\frac{1}{4}}\)
\(=72-8=64\)
Mà \(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}< \frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\)
\(\Rightarrow\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}< 0\)
Do đó : \(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}=-8\)
Khi đó : \(x=9-8=1\)
Với x =1 ta có ;
\(f\left(1\right)=\left(1^4-3.1+1\right)^{2016}=\left(-1\right)^{2016}=1\)
Chúc bạn học tốt !!!
tính giá trị của f(x)=\(\left(x^4-3x+1\right)^{2016}\)tại x=9-\(\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\)+\(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}\)
mọi người giúp mình nha xin đừng lướt qua vội vã
Có: \(\left(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\right)^2\)
\(=\frac{1}{\frac{9}{4}+\sqrt{5}}+\frac{1}{\frac{9}{4}-\sqrt{5}}-2\cdot\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}\cdot\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\)
\(=\frac{\frac{9}{4}-\sqrt{5}+\frac{9}{4}+\sqrt{5}}{\frac{1}{16}}-2\cdot\frac{1}{\frac{1}{4}}\)
\(=72-8=64\)
Mà; \(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}< \frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\)
\(\Rightarrow\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}< 0\)
Do đó: \(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}=-8\)
Khi đó: \(x=9-8=1\)
Với \(x=1\), ta có:
\(f\left(1\right)=\left(1^4-3\cdot1+1\right)^{2016}=\left(-1\right)^{2016}=1\)
Tìm x biết:
a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)
b) \(\left(3x-\frac{1}{4}\right).\left(x+\frac{1}{2}\right)=0\)
c) \(|x+\frac{1}{5}|-\frac{1}{2}=\frac{9}{10}\)
d) \(\sqrt{0,81}.\left(\sqrt{x}+\sqrt{\frac{16}{49}}\right)=\frac{9}{10}\)
f) \(|\frac{1}{3}.\sqrt{x+1}-\frac{2}{9}|-\frac{1}{6}=\frac{1}{9}\)
a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{2}{63}\)
b) \(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}\)
Vậy.........
c) \(\left|x+\frac{1}{5}\right|-\frac{1}{2}=\frac{9}{10}\)
\(\left|x+\frac{1}{5}\right|=\frac{7}{5}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{7}{5}\\x+\frac{1}{5}=\frac{-7}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=\frac{-8}{5}\end{cases}}\)
Vậy.........
1. \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+3-4\sqrt{x-1}}\left(2< x< 5\right)\)
2. \(\frac{6}{1-\sqrt{3}}-\frac{3\sqrt{3}-1}{\sqrt{3}+1}+\sqrt{3}\)
3. \(\sqrt{29-12\sqrt{5}+\sqrt{24-8\sqrt{3}}}\)
4. \(\sqrt{\frac{4}{9-4\sqrt{5}}}-\sqrt{\frac{4}{9+4\sqrt{5}}}\)
5. \(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{x}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\)
6. \(\frac{6-\sqrt{6}}{\sqrt{6}-1}-9\sqrt{\frac{2}{3}}-\frac{4}{2-\sqrt{6}}\)
7. \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\frac{\left(\sqrt{x}-1\right)^2}{2}\left(x\ge0,x\ne1\right)\)
Trả lời nhanh giúp mình với mình cần gấp lắm
Bài 1. Tìm điều kiện các BPT sau
a, \(\sqrt{20-x}>\sqrt{3x-6}+1\)
b, \(\frac{\sqrt{9-x^2}}{x-1}>\frac{1}{\sqrt{x}}+1\)
c, \(x+\frac{x+1}{\sqrt{x-4}}>2-\frac{2}{x^2-25}\)
d, \(\sqrt{x}>\sqrt{-x}\)
e, \(3x+\frac{4}{\sqrt{x-5}}\le9+\frac{x}{x-6}\)
f, \(\frac{x+2}{10+3x^2}\ge7+\frac{4}{\left(3x+9\right)^2}\)
g, \(\frac{\sqrt{x+2}}{\sqrt{x-2}}+\frac{1}{\left(x-4\right)\left(x+6\right)}\le\frac{3}{\sqrt{8-x}}\)
h, \(\frac{\sqrt{x+6}}{\left|x\right|-\sqrt{x+6}}\ge\sqrt{16-2x}\)
RÚT GỌN BIỂU THỨC
A=\(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)(với a>_ 0, b>_ 0, a#b)
B=\(\left(\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right).\left(\frac{\sqrt{x}+\sqrt{y}}{x-y}\right)\)(với x>_ 0, y>_ 0, x#y)
C=\(x-4-\sqrt{16-8x^2+x^4}\)(với x>4)
D=\(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}\)(với a>0, b>0, a#b)
E=\(\left(2+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right).\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\)(với a>0, a#1)
F=\(\frac{a-3\sqrt{a}}{\sqrt{a}-3}-\frac{a+4\sqrt{a}+3}{\sqrt{a}+3}\)( với a>_ 9)
G=\(\frac{9-x}{\sqrt{x}+3}-\frac{9-6\sqrt{x}+x}{\sqrt{x}-3}-6\)( với x>_ 9 )
Cho biểu thức: \(P=\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}+2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right):\left(1-\frac{3\sqrt{x}-9}{x-9}\right)\)
a)Rút gọn biểu thức
b)Tính P với \(x=\frac{\sqrt{4+2\sqrt{3}}\left(\sqrt{x}-1\right)}{\sqrt{6+2\sqrt{5}-\sqrt{5}}}\)
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
\(1.\sqrt{x-1}+\sqrt{x+3}+\sqrt{\left(x-1\right)\left(x^2-3x+5\right)}=4-2x\)
\(2.\frac{x^2}{3+\sqrt{9-x^2}}+\frac{1}{4.\sqrt[2]{9-x^2}}=1\)
1) Đk \(x\ge1\)
Ta có \(\hept{\begin{cases}\sqrt{x-1}\ge0\\\sqrt{x+3}\ge\sqrt{1+3}=2\\\sqrt{\left(x-1\right)\left(x^2-3x+5\right)}\ge0\end{cases}\Rightarrow VT\ge2}\)
Mà \(4-2x\le2\Rightarrow VT\ge VP\)
Dấu = xảy ra <=> x=1
Vậy ...^^
Ma tai sao \(\sqrt{x-1}+\sqrt{x+3}\ge2\)
=>
Đk x > 1
Ta có { x - 1 ==>0
x+3 >1+3=^2 =>:Vt>2
x-1 . (x2 - 3x+5) >0
Theo kết luận là
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