TH1: \(-\dfrac{1}{2}\le x\le\dfrac{5}{3}\)

\(\left(2x+1\right)+\left(5-3x\right)=6\\ =>2x+1+5-3x=6\\ =>\left(2x-3x\right)+6=6\\ =>x=0\left(tm\right)\)

TH2: \(x>\dfrac{5}{3}\)

\(\left(2x+1\right)-\left(5-3x\right)=6\\ =>2x+1-5+3x=6\\ =>2x+3x=6-1+5\\ =>5x=10\\ =>x=\dfrac{10}{5}=2\left(tm\right)\)

TH3: \(x< -\dfrac{1}{2}\)

\(-\left(2x+1\right)+\left(5-3x\right)=6\\ =>-2x-1+5-3x=6\\ =>-2x-3x+4=6\\ =>-5x=6-4=2\\ =>x=-\dfrac{2}{5}\left(ktm\right)\)

a: \(\sqrt{5}-x=\sqrt{7}\)

=>\(x=\sqrt{5}-\sqrt{7}\simeq-0,41\)

b: \(\sqrt{7}+2x=\sqrt{11}\)

=>\(2x=\sqrt{11}-\sqrt{7}\)

=>\(x=\dfrac{1}{2}\left(\sqrt{11}-\sqrt{7}\right)\simeq0,34\)

a: ĐKXĐ: x>=1

\(\sqrt{x-1}=3\)

=>\(x-1=3^2=9\)

=>x=9+1=10(nhận)

b: ĐKXĐ: x>=-1

\(\sqrt{x+1}=\sqrt{4}+\sqrt{9}\)

=>\(\sqrt{x+1}=2+3=5\)

=>x+1=25

=>x=24(nhận)

c: ĐKXĐ: x>=3

\(\sqrt{x-3}=\sqrt{25}-\sqrt{16}\)

=>\(\sqrt{x-3}=5-4=1\)

=>x-3=1

=>x=4(nhận)

sai !

a: \(\sqrt{25}-2\cdot\sqrt{36}+5\cdot\sqrt{49}\)

\(=5-2\cdot6+5\cdot7\)

=5-12+35

=40-12=28

b: \(5\sqrt{100}-\sqrt{9+16}-\sqrt{121}:11\)

\(=5\cdot10-5-11:11\)

=50-5-1=44

\(a,\left|-2\right|-2\\ =2-2\\ =0\\ b,\left|-1,6\right|\cdot\left|3,6\right|-\left|2,2\right|\\ =1,6\cdot3,6-2,2\\ =5,76-2,2\\ =3,56\\ c,\left|-5,7\right|-0,7\\ =5,7-0,7\\ =5\\ d,-\left|-\dfrac{2}{5}\right|+\left|\dfrac{2}{5}\right|-\left(-\dfrac{3}{7}\right)\\ =-\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{3}{7}\\ =\dfrac{3}{7}\)

a: \(\left|-0,25\right|+\left\{\left(4\cdot8\right)\cdot125-\left(-0,5\right)^2\right\}\)

=0,25+4000-0,25

=4000

b: \(\left(2,7+\left|-4.4\right|\right)-\left[\left(-5,6\right)-\left|-7,3\right|\right]\)

=2,7+4,4+5,6+7,3

=10+10=20

c: \(\left(-5,44\right)+4\cdot\left(1,25+0,11\right)\)

\(=-5,44+5+0,44\)

=-5+5=0

d: \(\left[\left|-6,72\right|+\left|-5,27\right|\right]-\left(0,72+1,27\right)\)

=6,72+5,27-0,72-1,27

=6+5=11

a: \(2,3\cdot\left(-6,4\right)+\left(-3,6\right)\cdot\left|-2,3\right|\)

\(=2,3\left(-6,4-3,6\right)\)

\(=2,3\cdot\left(-10\right)=-23\)

b: \(\left|-\dfrac{2}{5}+\dfrac{1}{2}\right|\cdot\left(1\dfrac{3}{4}-0,75\right)\)

\(=\left|-\dfrac{4}{10}+\dfrac{5}{10}\right|\left(1+\dfrac{3}{4}-\dfrac{3}{4}\right)\)

\(=\dfrac{1}{10}\)

c: \(\left(0,75+\left|-\dfrac{1}{2}\right|-\dfrac{1}{3}\right):0,5+1\dfrac{3}{2}\)

\(=\left(\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{1}{3}\right):\dfrac{1}{2}+\dfrac{5}{2}\)

\(=\left(\dfrac{9}{12}+\dfrac{6}{12}-\dfrac{4}{12}\right)\cdot2+\dfrac{5}{2}\)

\(=\dfrac{11}{12}\cdot2+\dfrac{5}{2}=\dfrac{11}{6}+\dfrac{5}{2}=\dfrac{11}{6}+\dfrac{15}{6}=\dfrac{26}{6}=\dfrac{13}{3}\)

d: \(\dfrac{1}{2}-\left|\left(-3\right)\cdot\dfrac{1}{6}\right|+\dfrac{5}{3}-\dfrac{9}{12}\)

\(=\dfrac{1}{2}-\left|-\dfrac{3}{6}\right|+\dfrac{20}{12}-\dfrac{9}{12}\)

\(=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{11}{12}=\dfrac{11}{12}\)

a: Xét ΔABM và ΔACM có

AB=AC

AM chung

MB=MC

Do đó: ΔAMB=ΔAMC

b: Xét ΔMAB và ΔMDC có

MA=MD

\(\widehat{AMB}=\widehat{DMC}\)(hai góc đối đỉnh)

MB=MC

Do đó: ΔMAB=ΔMDC
=>\(\widehat{MAB}=\widehat{MDC}\)

=>AB//DC

c: Ta có: ME\(\perp\)AB

AB//DC

Do đó: ME\(\perp\)DC

mà DC\(\perp\)MF

và ME,MF có điểm chung là M

nên E,M,F thẳng hàng

Xét ΔMEB vuông tại E và ΔMFC vuông tại F có

MB=MC

\(\widehat{MBE}=\widehat{MCF}\)(AB//CD)

Do đó: ΔMEB=ΔMFC

=>ME=MF

mà E,M,F thẳng hàng

nên M là trung điểm của EF

Đặt \(A=2+2^3+...+2^{2023}\)

=>\(4A=2^3+2^5+2^7+...+2^{2025}\)

=>\(4A-A=2^3+2^5+...+2^{2025}-2-2^3-...-2^{2023}\)

=>\(3A=2^{2025}-2\)

=>\(A=\dfrac{2^{2025}-2}{3}\)

Diện tích xung quanh căn phòng là:

\(\left(8+5\right)\cdot2\cdot4=13\cdot8=104\left(m^2\right)\)

Diện tích không cần quét vôi là:

\(2,2\cdot1,2\cdot2+1,4\cdot0,8\cdot4=9,76\left(m^2\right)\)

Diện tích cần quét vôi là:

\(104+8\cdot5-9,76=134,24\left(m^2\right)\)

Số tiền cần phải trả là:

\(134,24\cdot10000=1342400\left(đồng\right)\)

Cảm ơn bạn rất nhiều!