Tính \(P=u^8+\dfrac{1}{u^8}\) biết u=\(\sqrt{2}+1\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
25 tháng 8 2019
\(x+y=3\sqrt{xy}\)
\(\Leftrightarrow\)\(\frac{x}{y}+1=3\sqrt{\frac{x}{y}}\)
\(\Leftrightarrow\)\(\frac{x}{y}-3\sqrt{\frac{x}{y}}+\frac{9}{4}=\frac{5}{4}\)
\(\Leftrightarrow\)\(\left(\sqrt{\frac{x}{y}}-\frac{3}{2}\right)^2=\frac{5}{4}\)
\(\Leftrightarrow\)\(\frac{x}{y}=\frac{7+3\sqrt{5}}{2}\)
LM
25 tháng 6 2018
Câu 1:
\(2\sqrt{\dfrac{3}{20}}+\sqrt{\dfrac{1}{60}}-\sqrt{\dfrac{1}{15}}\)
= \(\sqrt{\dfrac{2^2\cdot3}{20}}+\sqrt{\dfrac{1}{60}}-\sqrt{\dfrac{1}{15}}\)
= \(\sqrt{\dfrac{12}{20}}+\sqrt{\dfrac{1}{60}}-\sqrt{\dfrac{1}{15}}\)
= \(\dfrac{\sqrt{12}\cdot\sqrt{20}}{\left(\sqrt{20}\right)^2}+\dfrac{\sqrt{60}}{\left(\sqrt{60}\right)^2}-\dfrac{\sqrt{15}}{\left(\sqrt{15}\right)^2}\)
= \(\dfrac{\sqrt{240}}{20}+\dfrac{\sqrt{60}}{60}-\dfrac{\sqrt{15}}{15}\)
= \(\dfrac{\sqrt{15}}{5}+\dfrac{\sqrt{15}}{30}-\dfrac{\sqrt{15}}{15}\)
= \(\sqrt{15}\cdot\left(\dfrac{1}{5}+\dfrac{1}{30}-\dfrac{1}{15}\right)\)
= \(\sqrt{15}\cdot\dfrac{1}{6}\) = \(\dfrac{\sqrt{15}}{6}\)
Bài 2:
a)\(\dfrac{1}{\sqrt{18}+\sqrt{8}-2\sqrt{2}}=\dfrac{1}{\sqrt{18}+2\sqrt{2}-2\sqrt{2}}=\dfrac{1}{\sqrt{18}}=\dfrac{\sqrt{18}}{18}=\dfrac{\sqrt{2}}{6}\)
b)\(\dfrac{\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\dfrac{\sqrt{2}\cdot\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}\right)^2-3}=\dfrac{\sqrt{2}\cdot\left(1+\sqrt{2}+\sqrt{3}\right)}{1+2\sqrt{2}+2-3}=\dfrac{\sqrt{2}\cdot\left(1+\sqrt{2}+\sqrt{3}\right)}{2\sqrt{2}}=\dfrac{1}{2}\cdot\left(1+\sqrt{2}+\sqrt{3}\right)\)c) \(\dfrac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{3+2\sqrt{6}+2-5}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{6}\cdot\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)}{2\left(\sqrt{6}\right)^2}=\dfrac{\sqrt{6}}{12}\cdot\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)\)
15 tháng 7 2018
a, ta có
\(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}< 3+4< 7\) (1)
lại có \(\sqrt{65}-1>\sqrt{64}-1>8-1>7\) (2)
từ (1) và(2) =>\(\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)
bài 2
\(M=\sqrt{\frac{\left(2^3\right)^{10}-\left(2^2\right)^{10}}{\left(2^2\right)^{11}-\left(2^3\right)^4}}=\sqrt{\frac{2^{30}-2^{20}}{2^{22}-2^{12}}}=\sqrt{\frac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}=\sqrt{\frac{2^{20}}{2^{12}}}=\sqrt{2^8}=2^4\)
HN
9 tháng 4 2023
Ta có: VT
V P
Vậy
b) ĐKXĐ: .
Vậy ĐKXĐ của là , .
Với , ta có:
.
Vậy với , .
AH
Akai Haruma
Giáo viên
31 tháng 5 2018
Bài 1:
Vì $x+y+z=1$ nên:
\(Q=\frac{x}{x+\sqrt{x(x+y+z)+yz}}+\frac{y}{y+\sqrt{y(x+y+z)+xz}}+\frac{z}{z+\sqrt{z(x+y+z)+xy}}\)
\(Q=\frac{x}{x+\sqrt{(x+y)(x+z)}}+\frac{y}{y+\sqrt{(y+z)(y+x)}}+\frac{z}{z+\sqrt{(z+x)(z+y)}}\)
Áp dụng BĐT Bunhiacopxky:
\(\sqrt{(x+y)(x+z)}=\sqrt{(x+y)(z+x)}\geq \sqrt{(\sqrt{xz}+\sqrt{xy})^2}=\sqrt{xz}+\sqrt{xy}\)
\(\Rightarrow \frac{x}{x+\sqrt{(x+y)(x+z)}}\leq \frac{x}{x+\sqrt{xy}+\sqrt{xz}}=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế suy ra:
\(Q\leq \frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+ \frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+ \frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
Vậy $Q$ max bằng $1$
Dấu bằng xảy ra khi $x=y=z=\frac{1}{3}$
AH
Akai Haruma
Giáo viên
31 tháng 5 2018
Bài 2:
Vì $x+y+z=1$ nên:
\(\text{VT}=\frac{1-x^2}{x(x+y+z)+yz}+\frac{1-y^2}{y(x+y+z)+xz}+\frac{1-z^2}{z(x+y+z)+xy}\)
\(\text{VT}=\frac{(x+y+z)^2-x^2}{(x+y)(x+z)}+\frac{(x+y+z)^2-y^2}{(y+z)(y+x)}+\frac{(x+y+z)^2-z^2}{(z+x)(z+y)}\)
\(\text{VT}=\frac{(y+z)[(x+y)+(x+z)]}{(x+y)(x+z)}+\frac{(x+z)[(y+z)+(y+x)]}{(y+z)(y+x)}+\frac{(x+y)[(z+x)+(z+y)]}{(z+x)(z+y)}\)
Áp dụng BĐT AM-GM:
\(\text{VT}\geq \frac{2(y+z)\sqrt{(x+y)(x+z)}}{(x+y)(x+z)}+\frac{2(x+z)\sqrt{(y+z)(y+x)}}{(y+z)(y+x)}+\frac{2(x+y)\sqrt{(z+x)(z+y)}}{(z+x)(z+y)}\)
\(\Leftrightarrow \text{VT}\geq 2\underbrace{\left(\frac{y+z}{\sqrt{(x+y)(x+z)}}+\frac{x+z}{\sqrt{(y+z)(y+x)}}+\frac{x+y}{\sqrt{(z+x)(z+y)}}\right)}_{M}\)
Tiếp tục AM-GM cho 3 số trong ngoặc lớn, suy ra \(M\geq 3\)
Do đó: \(\text{VT}\geq 2.3=6\) (đpcm)
Dấu bằng xảy ra khi $3x=3y=3z=1$
3 tháng 6 2018
Câu 1 :
\(P=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
Câu 2 :
Ta có :
\(\Delta=m^2+16>0\)
\(=>\) phương trình có 2 nghiệm phân biệt .
Theo định lý vi-ét ta có :
\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1.x_2=-4\end{matrix}\right.\)
Thay vào ta được :
\(\dfrac{2m+7}{m^2+8}\ge-\dfrac{1}{8}\)
\(\Leftrightarrow16m+56\ge-m^2-8\)
\(\Leftrightarrow m^2+16m+64\ge0\)
\(\Leftrightarrow\left(m+8\right)^2\ge0\) ( đúng )
UV
26 tháng 12 2018
1
a,A=\(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\)
A=\(\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\)
A=\(2\sqrt{5}:\sqrt{6}=\dfrac{2\sqrt{5}}{\sqrt{6}}=\dfrac{\sqrt{30}}{3}\)
b, B=\(\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}=\dfrac{\sqrt{5.2}-\sqrt{5.3}}{\sqrt{4.2}-\sqrt{4.3}}=\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\)
B=\(\dfrac{\sqrt{5}}{2}\)
26 tháng 12 2018
Câu 1)
a) \(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}=\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}=\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}=\dfrac{2\sqrt{5}}{\sqrt{6}}=\dfrac{\sqrt{30}}{3}\)
b) \(\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}=\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}=\dfrac{\sqrt{5}}{\sqrt{4}}=\dfrac{\sqrt{5}}{2}\)
Câu 2)
ĐK: x\(\ge5\)
\(\sqrt{x-5}+\sqrt{4x-20}-\dfrac{1}{5}\sqrt{9x-45}=3\Leftrightarrow\sqrt{x-5}+\sqrt{4\left(x-5\right)}-\dfrac{1}{5}\sqrt{9\left(x-5\right)}=3\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{3}{5}\sqrt{x-5}=3\Leftrightarrow\dfrac{12}{5}\sqrt{x-5}=3\Leftrightarrow\sqrt{x-5}=\dfrac{5}{4}\Leftrightarrow x-5=\dfrac{25}{16}\Leftrightarrow x=\dfrac{105}{16}\left(tm\right)\)
HS
19 tháng 10 2018
Câu 2:
a, ĐKXĐ: x\(\ge\)0; x\(\ne\)\(\pm\)1
B=
\(\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{-2.2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =-\dfrac{4}{\sqrt{x}-1}\)
3 tháng 8 2018
\(1,\sqrt{\left(-0,3\right)^2}=\sqrt{0,09}=0,3\)
\(2,-\frac{1}{2}\sqrt{\left(0,3\right)^2}=-\frac{1}{2}.0,3=-0,15\)
\(3,\sqrt{a^{10}}=\sqrt{\left(a^5\right)^2}=a^5\left(a\ge0\right)\)
\(4,\sqrt{\left(2-x\right)^2}=\left|2-x\right|=2-x\left(x\le2\right)\)
\(5,\sqrt{x^2+2x+1}=\sqrt{\left(x+1\right)^2}=\left|x+1\right|\)
\(6,\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)(Vì \(1< \sqrt{2}\))
\(7,\sqrt{11+6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
\(8,\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)\)
\(=-2\)
\(9,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)
\(u+\dfrac{1}{u}=\dfrac{u^2+1}{u}=\dfrac{\left(\sqrt{2}+1\right)^2+1}{\sqrt{2}+1}\)
\(=\dfrac{4+2\sqrt{2}}{\sqrt{2}+1}=\dfrac{2\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\sqrt{2}\)
\(P=u^8+\dfrac{1}{u^8}\)
\(=\left(u^4+\dfrac{1}{u^4}\right)^2-2\)
\(=\left[\left(u^2+\dfrac{1}{u^2}\right)^2-2\right]^2-2\)
\(=\left\{\left[\left(u+\dfrac{1}{u}\right)^2-2\right]^2-2\right\}^2-2\)
\(=\left\{\left[\left(2\sqrt{2}\right)^2-2\right]^2-2\right\}^2-2\)
\(=1154\) (tại vì t lười tính nên khúc này bn tính tay hoặc bấm máy tính đều đc ^^)
u = \(\sqrt{2}+1\) => \(u^2=3+2\sqrt{2}\) => \(u^4=17+12\sqrt{2}\)=> \(u^8=577+408\sqrt{2}\) => \(u^{16}=665857+665857=1331714\)
P = \(u^8+\dfrac{1}{u^8}=\dfrac{u^{16}+1}{u^8}=\dfrac{1331714}{577+408\sqrt{2}}=1154\)