1 - câu nào cx đúng

2 - C

\(\Rightarrow\left(x-2\right)\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\Rightarrow x\in\left\{-5;1;3;9\right\}\)

ĐKXĐ: \(n\ne-2\)

\(\dfrac{n^2+3}{n+2}=\dfrac{n\left(n+2\right)-2\left(n+2\right)+7}{n+2}=n-2+\dfrac{7}{n+2}\in Z\)

\(\Rightarrow\left(n+2\right)\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

Kết hợp ĐKXĐ:

\(\Rightarrow n\in\left\{-9;-3;-1;5\right\}\)

a) ĐKXĐ: \(a>0,a\ne1\)

b) \(Q=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+1\right)}.\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

c) \(Q=\dfrac{\sqrt{a}+1}{\sqrt{a}}=1+\dfrac{1}{\sqrt{a}}>1\)

1A,B,D

2 M=2

3 \(=\dfrac{3}{4x}\)

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

5 K rút gọn đc

6 \(=\dfrac{4\left(x-1\right)+2\left(x-1\right)}{6\left(x-1\right)}=\dfrac{6\left(x-1\right)}{6\left(x-1\right)}=1\)

Đề là \(\left(2x^2-x\right)^2+...\) hay là \(\left(2x^2-x\right)+...\) vậy bn?

a) \(A=\dfrac{3}{y+5}=\dfrac{3}{-2+5}=\dfrac{3}{-3}=-1\)

b) \(B=\dfrac{y\left(y-3\right)+2y\left(y+3\right)-3y^2-9}{\left(y-3\right)\left(y+3\right)}=\dfrac{3y-9}{\left(y-3\right)\left(y+3\right)}=\dfrac{3\left(y-3\right)}{\left(y-3\right)\left(y+3\right)}=\dfrac{3}{y+3}\)

c) \(M=-\dfrac{A}{B}=-\left(\dfrac{3}{y+5}:\dfrac{3}{y+3}\right)=-\dfrac{3\left(y+3\right)}{3\left(y+5\right)}=-\dfrac{y+3}{y+5}=-\dfrac{\left(y+5\right)-2}{y+5}=-1+\dfrac{2}{y+5}\in Z\)

\(\Rightarrow\left(y+5\right)\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Kết hợp ĐKXĐ:

\(\Rightarrow y\in\left\{-7;-6;-4\right\}\)

Xét tam giác ABM và tam giác ACM có:

AB=AC(gt)

BM=MC(M là trung điểm BC)

AM chung

=> ΔABM=ΔACM(c.c.c)

=> \(\widehat{AMB}=\widehat{AMC}\)

Mà 2 góc này kề bù

\(\Rightarrow\widehat{AMB}=\widehat{AMC}=180^0:2=90^0\Rightarrow AM\perp BC\)

\(4dm=0,4m\)

\(\left\{{}\begin{matrix}p_1=d.h_1=8000.1,8=14400\left(Pa\right)\\p_2=d.h_2=8000.\left(1,8-0,4\right)=11200\left(Pa\right)\end{matrix}\right.\)