cho x= \(7+4\sqrt{3}\), y= \(7-4\sqrt{3}\)
tính \(x^2+y^2\), \(x^3+y^3\), \(x^7+y^7\)
lm nhanh giúp mk nhé
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NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 7 2021
\(x+y=14\) ; \(xy=\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)=1\)
\(x^2+y^2=\left(x+y\right)^2-2xy=14^2-2.1=194\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=14^3-3.1.14=2702\)
\(x^4+y^4=\left(x^2+y^2\right)^2-2\left(xy\right)^2=194^2-2.1^2=37634\)
\(x^7+y^7=\left(x^3+y^3\right)\left(x^4+y^4\right)-\left(xy\right)^3\left(x+y\right)=2702.37634-1^3.14=...\)
12 tháng 10 2022
a: \(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(VT=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{x+2\sqrt{xy}+y}{x-y}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
AH
Akai Haruma
Giáo viên
31 tháng 7 2021
Lời giải:
a. ĐKXĐ: $x\geq -9$
PT $\Leftrightarrow x+9=7^2=49$
$\Leftrightarrow x=40$ (tm)
b. ĐKXĐ: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$
$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$
$\Leftrgihtarrow 3\sqrt{2x+3}=15$
$\Leftrightarrow \sqrt{2x+3}=5$
$\Leftrightarrow 2x+3=25$
$\Leftrightarrow x=11$ (tm)
AH
Akai Haruma
Giáo viên
31 tháng 7 2021
c.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{2}{3}\)
d. ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)
\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)
\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)
\(\Leftrightarrow -1=9\) (vô lý)
Vậy pt vô nghiệm.
26 tháng 9 2018
\(\dfrac{5\left(4+\sqrt{11}\right)}{\left(4+\sqrt{11}\right)\left(4-\sqrt{11}\right)}+\dfrac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}-\dfrac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{\sqrt{7}-5}{2}\)\(=\dfrac{\left(4+\sqrt{11}\right)5}{16-11}+\dfrac{3-\sqrt{7}}{9-7}-\dfrac{6\left(\sqrt{7}+2\right)}{7-4}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}-\dfrac{3-\sqrt{7}}{2}-2\left(\sqrt{7}+2\right)-\dfrac{\sqrt{7}-5}{2}=\dfrac{8+2\sqrt{11}-3+\sqrt{7}-4\sqrt{7}-8-\sqrt{7}+5}{2}=\dfrac{2\sqrt{11}-4\sqrt{7}+2}{2}=1+\sqrt{11}-2\sqrt{7}\)
8 tháng 10 2022
a: \(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{\sqrt{7}}{2}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}\)
\(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2\left(x+2\sqrt{xy}+y\right)}{2\left(x-y\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
HT
8 tháng 4 2021
a/ \(y'=\dfrac{\left(x^3+2\sqrt{x-1}\right)'\left(x-1\right)-\left(x-1\right)'\left(x^3+2\sqrt{x-1}\right)}{\left(x-1\right)^2}\)
\(y'=\dfrac{\left(2x^2+\dfrac{1}{\sqrt{x-1}}\right)\left(x-1\right)-x^3-2\sqrt{x-1}}{\left(x-1\right)^2}=\dfrac{x^3-2x^2-\sqrt{x-1}}{\left(x-1\right)^2}\)
b/ \(y'=\dfrac{\left(4x^3+2x-3\right)'\left(\sqrt{x^2+2}\right)-\left(\sqrt{x^2+2}\right)'\left(4x^3+2x-3\right)}{x^2+2}\)
\(y'=\dfrac{\left(12x^2+2\right)\sqrt{x^2+2}-\dfrac{x}{\sqrt{x^2+2}}\left(4x^3+2x-3\right)}{x^2+2}\) (ban tu rut gon nhe)
c/ \(y'=\dfrac{\left(x^3+x+1\right)'\left(x^3+x+1\right)}{\left|x^3+x+1\right|}=\dfrac{\left(3x^2+1\right)\left(x^3+x+1\right)}{\left|x^3+x+1\right|}\)
d/ \(y'=\dfrac{3x^2-24x^3}{2\sqrt{x^3-6x^4+7}}\)
e/ \(y'=\dfrac{\left(x^5+1\right)'\left(2-\sqrt{x^2+3}\right)-\left(x^5+1\right)\left(2-\sqrt{x^2+3}\right)'}{\left(2-\sqrt{x^2+3}\right)^2}\)
\(y'=\dfrac{5x^4\left(2-\sqrt{x^2+3}\right)+\left(x^5+1\right)\dfrac{x}{\sqrt{x^2+3}}}{\left(2-\sqrt{x^2+3}\right)^2}\)
3 tháng 12 2021
e: \(f\left(-x\right)=\dfrac{\left(-x\right)^4+3\cdot\left(-x\right)^2-1}{\left(-x\right)^2-4}=\dfrac{x^4+3x^2-1}{x^2-4}=f\left(x\right)\)
Vậy: f(x) là hàm số chẵn
3 tháng 12 2021
\(c,f\left(-x\right)=\sqrt{-2x+9}=-f\left(x\right)\)
Vậy hàm số lẻ
\(d,f\left(-x\right)=\left(-x-1\right)^{2010}+\left(1-x\right)^{2010}\\ =\left[-\left(x+1\right)\right]^{2010}+\left(x-1\right)^{2010}\\ =\left(x+1\right)^{2010}+\left(x-1\right)^{2010}=f\left(x\right)\)
Vậy hàm số chẵn
\(g,f\left(-x\right)=\sqrt[3]{-5x-3}+\sqrt[3]{-5x+3}\\ =-\sqrt[3]{5x+3}-\sqrt[3]{5x-3}=-f\left(x\right)\)
Vậy hàm số lẻ
\(h,f\left(-x\right)=\sqrt{3-x}-\sqrt{3+x}=-f\left(x\right)\)
Vậy hàm số lẻ
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 12 2018
ĐKXĐ: \(x\ge3;y\ge1\)
\(\sqrt{x-3}-\sqrt{y-1}+\sqrt[3]{x^2+x+1}-\sqrt[3]{y^2+5y+7}=0\)
\(\Leftrightarrow\dfrac{x-y-2}{\sqrt{x-3}+\sqrt{y-1}}+\dfrac{x^2+x+1-y^2-5y-7}{\sqrt[3]{\left(x^2+x+1\right)}+\sqrt[3]{\left(x^2+x+1\right)\left(y^2+5y+7\right)}+\sqrt[3]{y^2+5y+7}}=0\)
Để cho gọn gàng, ta đặt:
\(\left\{{}\begin{matrix}\sqrt[3]{\left(x^2+x+1\right)}+\sqrt[3]{\left(x^2+x+1\right)\left(y^2+5y+7\right)}+\sqrt[3]{y^2+5y+7}=b>0\\\sqrt{x-3}+\sqrt{y-1}=a>0\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x-y-2}{a}+\dfrac{x^2-y^2-4y-4+x-y-2}{b}=0\)
\(\Leftrightarrow\dfrac{x-y-2}{a}+\dfrac{x^2-\left(y+2\right)^2+\left(x-y-2\right)}{b}=0\)
\(\Leftrightarrow\dfrac{x-y-2}{a}+\dfrac{\left(x-y-2\right)\left(x+y+3\right)}{b}=0\)
\(\Leftrightarrow\left(x-y-2\right)\left(\dfrac{1}{a}+\dfrac{x+y+3}{b}\right)=0\)
\(\Leftrightarrow x-y-2=0\) do \(\left\{{}\begin{matrix}x\ge3\\y\ge1\end{matrix}\right.\) \(\Rightarrow x+y+3>0\Rightarrow\dfrac{1}{a}+\dfrac{x+y+3}{b}>0\)
\(\Rightarrow x=y+2\)
Thay vào Q ta được:
\(Q=y^2-\left(y+2\right)^2+3\left(y+2\right)+4\sqrt{y}+4\)
\(\Rightarrow Q=-y+4\sqrt{y}+6=10-\left(y-4\sqrt{y}+4\right)=10-\left(\sqrt{y}-2\right)^2\le10\)
\(\Rightarrow Q_{max}=10\) khi \(\sqrt{y}-2=0\Rightarrow\left\{{}\begin{matrix}y=4\\x=6\end{matrix}\right.\)
Có \(x+y=7+4\sqrt{3}+7-4\sqrt{3}=14\)
\(xy=\left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)=1\)
\(x^2+y^2=\left(x+y\right)^2-2xy=14^2-2=194\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=14^3-3.1.14=2702\)
\(x^7+y^7=\left(x^3+y^3\right)\left(x^4+y^4\right)-x^3y^3\left(x+y\right)\)\(=2702\left[\left(x^2+y^2\right)^2-2x^2y^2\right]-14\)
\(=2702\left(194^2-2\right)-14=101687054\)
Vậy...