\(34,257\simeq34,3\)

\(37,359\simeq34,3\)

\(34,075\simeq34,1\)

\(34,175\simeq34,2\)

\(37,303\simeq37,3\)

ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

\(\dfrac{x+1}{x-3}+\dfrac{x+1}{x+3}=\dfrac{2\left(x^2+5\right)}{x^2-9}\)

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

=>\(\dfrac{x^2+4x+3+x^2-2x-3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x^2+10}{\left(x-3\right)\left(x+3\right)}\)

=>\(2x^2+2x=2x^2+10\)

=>2x=10

=>x=5(nhận)

Câu 16:

a: \(\left|x\right|=\dfrac{1}{3}\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b: \(1-2x=\dfrac{9}{8}+\dfrac{7}{5}:\dfrac{2}{5}\)

=>\(1-2x=\dfrac{9}{8}+\dfrac{7}{2}=\dfrac{9}{8}+\dfrac{28}{8}=\dfrac{37}{8}\)

=>\(2x=1-\dfrac{37}{8}=-\dfrac{29}{8}\)

=>\(x=-\dfrac{29}{8}:2=-\dfrac{29}{16}\)

c: \(\dfrac{1}{3}+x=2+\dfrac{6}{9}\)

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

=>\(x=2+\dfrac{2}{3}-\dfrac{1}{3}=2+\dfrac{1}{3}=\dfrac{7}{3}\)

Câu 18:

a: Xét ΔAMB và ΔAMC có

AM chung

MB=MC

AB=AC

Do đó: ΔAMB=ΔAMC

b: ΔAMB=ΔAMC

=>\(\widehat{MAB}=\widehat{MAC}\)

Xét ΔAEM vuông tại E và ΔAFM vuông tại F có

AM chung

\(\widehat{EAM}=\widehat{FAM}\)

Do đó: ΔAEM=ΔAFM

=>AE=AF

c: Xét ΔABC có \(\dfrac{AE}{AB}=\dfrac{AF}{AC}\)

nên EF//BC

Ta có: \(8=2^3;12=2^2\cdot3;15=3\cdot5\)

=>\(BCNN\left(8;12;15\right)=2^3\cdot3\cdot5=8\cdot15=120\)

a: 

b:

c: Ta có: AB//CD

=>\(\widehat{AEF}=\widehat{EFD}\)(hai góc so le trong)

=>\(\widehat{EFD}=90^0\)

=>MN\(\perp\)CD tại F

Gọi d=ƯCLN(5n+3;2n-1)

=>\(\left\{{}\begin{matrix}5n+3⋮d\\2n-1⋮d\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}10n+6⋮d\\10n-5⋮d\end{matrix}\right.\)

=>\(10n+6-10n+5⋮d\)

=>\(11⋮d\)

=>ƯCLN(5n+3;2n-1) chưa chắc bằng 1 nha bạn

=>5n+3 và 2n-1 chưa chắc là hai số nguyên tố cùng nhau

Ta có: \(\left(\dfrac{1}{3}+\left|x\right|\right)\cdot\left(3\sqrt{x-1}-\dfrac{9}{4}\right)=0\)

mà \(\left|x\right|+\dfrac{1}{3}>=\dfrac{1}{3}>0\forall x\)

nên \(3\sqrt{x-1}-\dfrac{9}{4}=0\)

=>\(3\sqrt{x-1}=\dfrac{9}{4}\)

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

=>\(x-1=\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\)

=>\(x=\dfrac{9}{16}+1=\dfrac{25}{16}\left(nhận\right)\)

Đây là dạng bài tìm bội chung nha bạn

Ta có: \(P=\left(x+1\right)^3-\left(x-2\right)\left(x^2+2x+4\right)-3x^2\)

\(=x^3+3x^2+3x+1-\left(x^3-8\right)-3x^2\)

\(=x^3+3x+1-x^3+8=3x+9\)

Thay \(x=-\dfrac{1}{3}\) vào P, ta được:

\(P=3\cdot\dfrac{-1}{3}+9=9-1=8\)

Để \(\dfrac{a^2+a+3}{a+1}\) là số nguyên thì \(a^2+a+3⋮a+1\)

=>\(a\left(a+1\right)+3⋮a+1\)

=>\(3⋮a+1\)

=>\(a+1\in\left\{1;-1;3;-3\right\}\)

=>\(a\in\left\{0;-2;2;-4\right\}\)