giúp với,khó quá
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5 tháng 9 2021
\(A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\\ \Rightarrow A^2=2-\sqrt{3}+2\sqrt{2-\sqrt{3}}\sqrt{2+\sqrt{3}}+2+\sqrt{3}\\ \Rightarrow A^2=4+2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\\ \Rightarrow A^2=4+2\sqrt{2^2-\sqrt{3^2}}\\ \Rightarrow A^2=4+2\sqrt{1}\\ \Rightarrow A^2=6\\ \Rightarrow A=\pm\sqrt{6}\)
Mà \(A>0\Rightarrow A=\sqrt{6}\)
6 tháng 11 2021
a: Xét tứ giác AKCI có
AK//CI
AK=CI
Do đó: AKCI là hình bình hành
20 tháng 2 2022
hơi mờ và còn nhiều ,mik nghĩ bạn nên cắt ra đi
DT
Đỗ Thanh Hải
CTVVIP
21 tháng 8 2021
1 not going on holiday with you
2 such easy questions that all the students got them right
3 we had had time, we would have visited the museum
4 to Daisy for breaking her vase
5 studying ENglish 5 years ago
6 are thought to be the most popular dance in Brazil
7 sooner had she received the exam result than she phoned her mom
8 It was the absence of leadership that caused most of the problems on the committee
9 you have any complants about the product, return it to the shop
10 The more fondness for the game increased , the more proficiency he has
11 been a dramatical rise in house prices this year
12 believed to have escaped in a stolen car
13 gone out with him for 2 years
14 he was going to meet his sister in front of the station
18 tháng 1 2022
đk x khác 1 ; y khác -2
\(\left\{{}\begin{matrix}\dfrac{8}{x-1}+\dfrac{15}{y+2}=1\\\dfrac{1}{x-1}+\dfrac{1}{y+2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x-1}+\dfrac{15}{y+2}=1\\\dfrac{8}{x-1}+\dfrac{8}{y+2}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{y+2}=-7\\\dfrac{1}{x-1}+\dfrac{1}{y+2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=-1\\\dfrac{1}{x-1}+\dfrac{1}{y+2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\\dfrac{1}{x-1}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\1=2x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=\dfrac{3}{2}\end{matrix}\right.\)
12 tháng 1 2022
Bài 1:
1: =>x(2x-5)(2x+5)=0
hay \(x\in\left\{0;\dfrac{5}{2};-\dfrac{5}{2}\right\}\)
2: \(\Leftrightarrow2x^2+8x-x-4=0\)
=>(x+4)(2x-1)=0
hay \(x\in\left\{-4;\dfrac{1}{2}\right\}\)
3: \(\Leftrightarrow x^3-x-6x-6=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x-6\right)=0\)
=>(x+1)(x-3)(x+2)=0
hay \(x\in\left\{-1;3;-2\right\}\)
4: \(\Leftrightarrow3x^2-12x-5x-3x^2=-34\)
=>-17x=-34
hay x=2
5: \(\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\)
=>(4x-3)(7x-3)=0
hay \(x\in\left\{\dfrac{3}{4};\dfrac{3}{7}\right\}\)
6) báo cáo