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7 tháng 7 2021

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...

NV
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

\(\dfrac{5\left(4+\sqrt{11}\right)}{\left(4+\sqrt{11}\right)\left(4-\sqrt{11}\right)}+\dfrac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}-\dfrac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{\sqrt{7}-5}{2}\)\(=\dfrac{\left(4+\sqrt{11}\right)5}{16-11}+\dfrac{3-\sqrt{7}}{9-7}-\dfrac{6\left(\sqrt{7}+2\right)}{7-4}-\dfrac{\sqrt{7}-5}{2}\)

\(=4+\sqrt{11}-\dfrac{3-\sqrt{7}}{2}-2\left(\sqrt{7}+2\right)-\dfrac{\sqrt{7}-5}{2}=\dfrac{8+2\sqrt{11}-3+\sqrt{7}-4\sqrt{7}-8-\sqrt{7}+5}{2}=\dfrac{2\sqrt{11}-4\sqrt{7}+2}{2}=1+\sqrt{11}-2\sqrt{7}\)

26 tháng 9 2018

Mk lam sai oy

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: \(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}\)

\(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)

\(=\dfrac{2\left(x+2\sqrt{xy}+y\right)}{2\left(x-y\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

NV
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 \(\sqrt{y}-2=0\Rightarrow\left\{{}\begin{matrix}y=4\\x=6\end{matrix}\right.\)

9 tháng 12 2018

Nguyễn Việt Lâm Mashiro Shiina Akai Haruma

a: \(=\dfrac{2\sqrt{7}-10-6+\sqrt{7}}{4}+\dfrac{24+6\sqrt{7}-20+5\sqrt{7}}{9}\)

\(=\dfrac{3\sqrt{7}-16}{4}+\dfrac{4+11\sqrt{7}}{9}\)

\(=\dfrac{27\sqrt{7}-144+16+44\sqrt{7}}{36}=\dfrac{71\sqrt{7}-128}{36}\)

b: \(=\dfrac{\sqrt{y}\left(x+y\right)}{\sqrt{xy}}\cdot\dfrac{\sqrt{x}-\sqrt{y}}{x+y}\)

\(=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}}\)

c: \(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)+3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right)\cdot\dfrac{3\sqrt{x}-1}{3\sqrt{x}-5}\)

\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+3\sqrt{x}-1}{3\sqrt{x}+1}\cdot\dfrac{1}{3\sqrt{x}-5}\)

\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)}\cdot\dfrac{1}{3\sqrt{x}-5}\)

\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-5\right)}\)