ĐKXĐ: \(x\ne-5\)

\(\dfrac{x+6}{x+5}+\dfrac{3}{2}=2\)

\(\Leftrightarrow\dfrac{x+6}{x+5}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x+6\right)=1.\left(x+5\right)\)

\(\Leftrightarrow2x+12=x+5\)

\(\Leftrightarrow2x-x=5-12\)

\(\Leftrightarrow x=-7\) (tm ĐKXĐ)

ĐKXĐ: \(x\ne-5\)

\(\dfrac{x+6}{x+5}+\dfrac{3}{2}=2\)

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

=>\(2\left(x+6\right)=x+5\)

=>\(2x+12=x+5\)

=>2x-x=5-12

=>x=-7(nhận)

\(B=\left(\dfrac{1}{2^2}-1\right)\cdot\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{1000^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{1000}-1\right)\cdot\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{1000}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-999}{1000}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot....\cdot\dfrac{1001}{1000}\)

\(=\dfrac{-1}{1000}\cdot\dfrac{1001}{2}=\dfrac{-1001}{2000}\)

Thanks

\(B=2021\cdot2025=\left(2023-2\right)\cdot\left(2023+2\right)=2023^2-4=A-4\)

=>A lớn B 4 đơn vị

\(\left|x-3,5\right|=\dfrac{4}{7}\)

=>\(\left[{}\begin{matrix}x-\dfrac{7}{2}=\dfrac{4}{7}\\x-\dfrac{7}{2}=-\dfrac{4}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{7}+\dfrac{7}{2}=\dfrac{8}{14}+\dfrac{49}{14}=\dfrac{57}{14}\\x=-\dfrac{4}{7}+\dfrac{7}{2}=\dfrac{-8}{14}+\dfrac{49}{18}=\dfrac{41}{14}\end{matrix}\right.\)

a: Ta có: \(AM=MB=\dfrac{AB}{2}\)

\(DN=NC=\dfrac{DC}{2}\)

\(BE=EC=\dfrac{BC}{2}\)

mà AB=DC=BC

nên AM=MB=DN=NC=BE=EC

Xét tứ giác AMCN có

AM//CN

AM=CN

Do đó: AMCN là hình bình hành

b: Xét ΔMBC vuông tại B và ΔECD vuông tại C có

MB=EC

BC=CD

Do đó: ΔMBC=ΔECD

=>\(\widehat{BMC}=\widehat{CED}\)

=>\(\widehat{CED}+\widehat{ECM}=90^0\)

=>CM\(\perp\)DE

c: ΔMBC=ΔECD

=>MC=ED

Bài 5:

a: \(A=23-21+19-17+15-13+11-9+7-5+3-1\)

=(23-21)+(19-17)+(15-13)+(11-9)+(7-5)+(3-1)

=2+2+2+2+2+2

=12

b: \(B=24-22+20-18+16-14+12-10+8-6+4-2\)

=(24-22)+(20-18)+(16-14)+(12-10)+(8-6)+(4-2)

=2+2+2+2+2+2

=12

c: \(C=3+5+7+9+11+13+15+17+19\)

=(3+19)+(5+17)+(7+15)+(9+13)+11

=22+22+22+22+11

=99

d: Số số hạng là \(\dfrac{101-3}{2}+1=\dfrac{98}{2}+1=50\left(số\right)\)

Tổng của dãy số là:

\(D=\left(101+3\right)\cdot\dfrac{50}{2}=104\cdot25=2600\)

Bài 4:

1: \(13\cdot58\cdot4+32\cdot26\cdot2+52\cdot10\)

\(=52\cdot58+52\cdot32+52\cdot10\)

\(=52\left(58+32+10\right)=52\cdot100=5200\)

2: \(15\cdot37\cdot4+120\cdot21+21\cdot5\cdot12\)

\(=60\cdot37+60\cdot42+60\cdot21\)

=60(37+42+21)

=60*100=6000

3: \(14\cdot35\cdot5+10\cdot25\cdot7+20\cdot70\)

\(=70\cdot35+70\cdot25+70\cdot20\)

\(=70\left(35+25+20\right)=70\cdot80=5600\)

4: \(15\cdot\left(27+18+6\right)+15\cdot\left(23+12\right)\)

\(=15\cdot\left(45+6+35\right)=15\cdot86=1290\)

5: \(24\cdot\left(15+49\right)+12\cdot\left(50+42\right)\)

\(=24\cdot64+12\cdot92=24\cdot64+24\cdot46=24\cdot\left(64+46\right)\)

\(=24\cdot110=2640\)

a: 11+13+15+17+19

=(11+19)+(13+17)+15

=30+30+15

=75

b: \(122+2116+278+84\)

=(122+278)+(2116+84)

=400+2200

=2600

c: \(12\times125\times54\)

=1500x54

=81000

d: \(27\times36+27\times64=27\times\left(36+64\right)=27\times100=2700\)

e: \(25\text{x}37+25\text{x}63-150\)

=25x(37+63)-150

=25x100-150

=2500-150

=2350

f: \(425\text{x}74-27\text{x}425-1\)

=425x(74-27-1)

=425x46=19550

g: \(8\text{x}9\text{x}14+6\text{x}17\text{x}12+19\text{x}4\text{x}18\)

=14x72+17x72+19x72

=72x(14+17+19)

=50x72=3600

a) $11+13+15+17+19$

$=(11+19)+(13+17)+15$

$=30+30+15$

$=60+15=75$

b) $122+2116+278+84$

$=(122+278)+(2116+84)$

$=400+2200=2600$

c) $12\times125\times54$

$=3\times4\times125\times2\times27$

$=(4\times2)\times125\times(3\times27)$

$=8\times125\times81$

$=1000\times81=81000$

d) $27\times36+27\times64$

$=27\times(36+64)$

$=27\times100=2700$

e) $25\times37+25\times63-150$

$=25\times(37+63)-150$

$=25\times100-150$

$=2500-150=2350$

f) $425\times74-27\times425-425$

$=425\times(74-27-1)$

$=425\times(47-1)$

$=425\times46=19550$

g) $8\times9\times14+6\times17\times12+19\times4\times18$

$=(8\times9)\times14+(6\times12)\times17+(4\times18)\times19$

$=72\times14+72\times17+72\times19$

$=72\times(14+17+19)$

$=72\times(31+19)$

$=72\times50=3600$

$\mathtt{Toru}$

Câu 1:

\(2\sin x-\sqrt{3}=0\\ \Leftrightarrow\sin x=\dfrac{\sqrt{3}}{2}=\sin\dfrac{\pi}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{\pi}{3}+k_12\pi\\x_2=\pi-\dfrac{\pi}{3}+k_22\pi=\dfrac{2\pi}{3}+k_22\pi\end{matrix}\right.\left(k_1,k_2\inℤ\right)\)

Mà: \(x\in\left[0;2\pi\right]\) do đó nên: \(k_1=0,k_2=0\)

Vậy tập nghiệm pt là: \(S=\left\{\dfrac{\pi}{3};\dfrac{2\pi}{3}\right\}\) (2 nghiệm => D)

Câu 2:

Vì: \(-1\le\cos x\le1\forall x\)

\(\Rightarrow-1\le m+1\le1\\ \Leftrightarrow-2\le m\le0\)

Mà: \(m\inℤ\Rightarrow m\in\left\{-2;-1;0\right\}\) (C)

Câu 1: \(2\cdot sinx-\sqrt{3}=0\)

=>\(sinx=\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}x=\dfrac{\Omega}{3}+k2\Omega\\x=\Omega-\dfrac{\Omega}{3}+k2\Omega=\dfrac{2}{3}\Omega+k2\Omega\end{matrix}\right.\)

Để \(x\in\left[0;2\Omega\right]\) thì \(\left[{}\begin{matrix}\dfrac{\Omega}{3}+k2\Omega\in\left[0;2\Omega\right]\\\dfrac{2}{3}\Omega+k2\Omega\in\left[0;2\Omega\right]\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2k+\dfrac{1}{3}\in\left[0;2\right]\\2k+\dfrac{2}{3}\in\left[0;2\right]\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2k\in\left[-\dfrac{1}{3};\dfrac{5}{3}\right]\\2k\in\left[-\dfrac{2}{3};\dfrac{4}{3}\right]\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k\in\left[-\dfrac{1}{6};\dfrac{5}{6}\right]\\k\in\left[-\dfrac{1}{3};\dfrac{2}{3}\right]\end{matrix}\right.\Leftrightarrow k=0\)

=>Chọn B

Câu 2:

Để phương trình cosx =m+1 có nghiệm thì -1<=m+1<=1

=>-2<=m<=0

mà m nguyên

nên \(m\in\left\{-2;-1;0\right\}\)

=>Chọn C

6:

Xét ΔMIN vuông tại I có \(tanN=\dfrac{MI}{NI}\)

=>\(NI=\dfrac{MI}{tanN}=\dfrac{11.5}{tan70}\simeq4,2\left(cm\right)\)

Xét ΔMIP vuông tại I có \(tanP=\dfrac{MI}{IP}\)

=>\(IP=\dfrac{MI}{tanP}=\dfrac{11.5}{tan38}\simeq14,7\left(cm\right)\)

NP=NI+IP=4,2+14,7=18,9(cm)

=>Chọn B

5: Xét ΔABC vuông tại A có 

\(sinC=\dfrac{AB}{BC}\)

=>\(BC=\dfrac{8}{sin30}=16\left(cm\right)\)

=>Chọn D

4:

\(\widehat{BAC}=27^0\)

Xét ΔBAC vuông tại B có \(tanBAC=\dfrac{BC}{BA}\)

=>\(BA=\dfrac{149}{tan27}\simeq292\left(m\right)\)

=>Chọn A

Bài 3:

a: \(3\in N;3\in Z;3\in Q\)

b: \(10\in N;10\in Z;10\in Q\)

c: \(-\dfrac{3}{7}\in Q\)

d: \(-2\in Z;-2\in Q\)

Bài 2:

\(\dfrac{-3}{5}=\dfrac{-27}{45};\dfrac{-5}{9}=\dfrac{-25}{45};-1\dfrac{2}{3}=\dfrac{-5}{3}=\dfrac{-75}{45};0,5=\dfrac{22,5}{45};\dfrac{10}{9}=\dfrac{50}{45}\)

mà -75<-27<-25<22,5<50

nên \(-1\dfrac{2}{3}< -\dfrac{3}{5}< -\dfrac{5}{9}< \dfrac{1}{2}< \dfrac{10}{9}\)

Bài 1:

a: \(\dfrac{9}{70}=\dfrac{9\cdot3}{70\cdot3}=\dfrac{27}{210};\dfrac{5}{42}=\dfrac{5\cdot5}{42\cdot5}=\dfrac{25}{210}\)

mà 27>25

nên \(\dfrac{9}{70}>\dfrac{5}{42}\)

b: \(\dfrac{-4}{27}=\dfrac{-4\cdot7}{27\cdot7}=\dfrac{-28}{189};\dfrac{10}{-63}=\dfrac{-10}{63}=\dfrac{-10\cdot3}{63\cdot3}=\dfrac{-30}{189}\)

mà -28>-30

nên \(\dfrac{-4}{27}>\dfrac{10}{-63}\)

c: \(\dfrac{999}{556}=1+\dfrac{443}{556};\dfrac{1000}{557}=1+\dfrac{443}{557}\)

mà \(\dfrac{443}{556}>\dfrac{443}{557}\)

nên \(\dfrac{999}{556}>\dfrac{1000}{557}\)

d: \(\dfrac{-2}{15}< 0;\dfrac{-10}{-11}=\dfrac{10}{11}>0\)

Do đó: \(\dfrac{-2}{15}< \dfrac{-10}{-11}\)