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

\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{x+2\sqrt{xy}+y}{x-y}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

28 tháng 5 2020

ĐKXĐ : \(x\ge1\)

PT đã cho tương đương với :

\(\sqrt{3x-2}+\sqrt{x-1}=\left[3x-2+2\sqrt{3x^2-5x+2}+x-1\right]-6\)

\(\Leftrightarrow\sqrt{3x-2}+\sqrt{x-1}=\left(\sqrt{3x-2}+\sqrt{x-1}\right)^2-6\)

Đặt \(\sqrt{3x-2}+\sqrt{x-1}=t\left(t\ge1\right)\)

Khi đó : \(t^2-t-6=0\Leftrightarrow\orbr{\begin{cases}t=3\\t=-2\left(loai\right)\end{cases}}\)

\(\Rightarrow\sqrt{3x-2}+\sqrt{x-1}=3\)

từ đó dễ dàng tìm được x

29 tháng 5 2020

Làm tiếp bài của @Thanh Tùng DZ

Thay t=3 vào cách đặt ta được \(\sqrt{3x-2}+\sqrt{x-1}=3\left(3a\right)\)

Ta có \(\left(3a\right)\Leftrightarrow4x-3+2\sqrt{3x^2-5x+2}=9\)

\(\Leftrightarrow\sqrt{3x^2-5x+2}=6-2x\)

\(\Leftrightarrow\hept{\begin{cases}6-2x\ge0\\3x^2-5x+2=36-24x+4x^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le3\\x=2;x=17\end{cases}\Leftrightarrow x=2}\)

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

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

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

bài 1: giải các hệ phương trình 1)\(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)=\(\dfrac{1}{2}\) x+y=9 2) \(\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\) \(\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\) 3)\(2|x|-y=3\) \(|x|+y=3\) 4)\(2\left(x+y\right)+\sqrt{x+1}=4\) \(\left(x+y\right)-3\sqrt{x+1}=-5\) 5) \(\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\) \(\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\) 6)\(\dfrac{1}{x}+\dfrac{3}{2y+1}=2\) \(\dfrac{2}{x}+\dfrac{4}{2y+1}=2\) 7)...
Đọc tiếp

bài 1: giải các hệ phương trình

1)\(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)=\(\dfrac{1}{2}\)

x+y=9

2) \(\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\)

\(\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\)

3)\(2|x|-y=3\)

\(|x|+y=3\)

4)\(2\left(x+y\right)+\sqrt{x+1}=4\)

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

5) \(\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\)

\(\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\)

6)\(\dfrac{1}{x}+\dfrac{3}{2y+1}=2\)

\(\dfrac{2}{x}+\dfrac{4}{2y+1}=2\)

7) \(\dfrac{1}{x}+\dfrac{1}{y}=2\)

\(\dfrac{3}{x}-\dfrac{1}{y}=2\)

8)\(\dfrac{1}{x+2}+\dfrac{3}{2y-1}=4\)

\(\dfrac{4}{x+2}-\dfrac{1}{2y-1}=3\)

9)\(\dfrac{4}{x+y} +\dfrac{1}{y-1}=5\)

\(\dfrac{1}{x+y}-\dfrac{2}{y-1}=-1\)

10)\(\dfrac{7}{\sqrt{2x+3}}-\dfrac{4}{\sqrt{3}-y}=\dfrac{5}{3}\)

\(\dfrac{5}{\sqrt{2x+3}}+\dfrac{3}{\sqrt{3-y}}=\dfrac{13}{6}\)

11)\(\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\)

\(\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\)

12) \(\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}\)

\(\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}2\dfrac{1}{6}\)

13) \(3\sqrt{x-1}+2\sqrt{y}=13\)

\(2\sqrt{x-1}-\sqrt{y}=4\)

14) 6x + 6y = 5xy

\(\dfrac{4}{x}-\dfrac{3}{y}=1\)

1
24 tháng 2 2018

mọi người giúp mk với gianroi

câu 6 sai nha

sửa : \(\dfrac{1}{x}+\dfrac{3}{2y+1}=2\)

\(\dfrac{2}{x}+\dfrac{4}{2y+1}=3\)

2 tháng 7 2019

a)\(\frac{3+\sqrt{3}}{1+\sqrt{3}}\)=\(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{1+\sqrt{3}}\)=\(\sqrt{3}\)

b)\(\frac{2\sqrt{3}-6}{\sqrt{8}-\sqrt{2}}\)

2 tháng 7 2019

\(\frac{y-2\sqrt{y}}{\sqrt{y}-2}\)=\(\frac{\sqrt{y}\left(\sqrt{y}-2\right)}{\sqrt{y}-2}\)=\(\sqrt{y}\)

d) \(\frac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)=\(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x+3}\right)}{\sqrt{x}-1}\)=\(\sqrt{x}\)+3

e)\(\frac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)=\(\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)=\(\sqrt{y}\)-1

g)\(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}\)=\(\frac{\sqrt{x}+1}{\sqrt{x+3}}\)

chúc bạn học tốthaha

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

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

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

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

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