có ai on ko

🍓Từ mượn là từ vay từ ngôn ngữ khác rồi dùng trong tiếng Việt.

🌷͙֒3 ví dụ:

1. Gốc Hán:
2. • Từ mượn: Gia đình
   • Nghĩa thuần Việt: Nhà, mái ấm
3. Gốc Pháp:
4. • Từ mượn: Xà phòng
   • Nghĩa thuần Việt: Chất tẩy ℛửa (hoặc "bánh giặt" ngày xưa)
5. Gốc Anh:
6. • Từ mượn: Tivi
   • Nghĩa thuần Việt: Máy thu hình🎀

\(=-9+\left\lbrack\left(-54:256\right)+7\times4]\right.\)

\(=-9+0,2109375+28\)

\(=0,2109375+28-9\)

\(=0,2109375+19\)

\(=19,2109375\)

mik làm luôn ko viết lại đề nha

= -9 +(-54: 256 +7 x 4)

=-9 +( -27/128 + 28)

=-9 +( 3557/128)

=2405/128

Bổ sung đề: M,N∈BC

a: ΔABC cân tại A

=>\(\hat{ABC}=\hat{ACB}=\frac{180^0-\hat{BAC}}{2}=\frac{180^0-120^0}{2}=30^0\)

Ta có: \(\hat{BAM}+\hat{MAC}=\hat{BAC}\)

=>\(\hat{MAC}=120^0-90^0=30^0\)

Ta có: \(\hat{BAN}+\hat{CAN}=\hat{BAC}\)

=>\(\hat{BAN}=120^0-90^0=30^0\)

Ta có: \(\hat{BAN}+\hat{MAN}+\hat{MAC}=\hat{BAC}\)

=>\(\hat{MAN}=120^0-30^0-30^0=60^0\)

Xét ΔANB và ΔAMC có

\(\hat{NAB}=\hat{MAC}\left(=30^0\right)\)

AB=AC

\(\hat{ABN}=\hat{ACM}\left(=30^0\right)\)

Do đó: ΔANB=ΔAMC

=>AN=AM

Xét ΔANM có AN=AM và \(\hat{MAN}=60^0\)

nên ΔANM đều

b: ΔANB=ΔAMC

=>NB=MC

Xét ΔNAB có \(\hat{NAB}=\hat{NBA}\left(=30^0\right)\)

nên ΔNAB cân tại N

=>NA=NB

mà NA=NM(ΔNAM đều)

nên NB=NM

=>BN=NM=MC

Bài 1:

`a)(2x+1):3+2=5`

`(2x+1):3=5-2`

`(2x+1):3=3`

`2x+1=3*3`

`2x+1=9`

`2x=9-1`

`2x=8`

`x=8/2`

`x=4`

Vậy: `...`

`b)(2x)/3=3x-1`

`2x=3(3x-1)`

`2x=9x-3`

`9x-2x=3`

`7x=3`

`x=3/7`

Vậy: `...`

`c)(x+5)/15+(x+6)/14=(x+7)/13+(x+8)/12`

`((x+5)/15+1)+((x+6)/14+1)=((x+7)/13+1)+((x+8)/12+1)`

`(x+20)/15+(x+20)/14=(x+20)/13+(x+20)/12`

`(x+20)/15+(x+20)/14-(x+20)/13-(x+20)/12=0`

`(x+20)(1/15+1/14-1/13-1/12)=0`

`(x+20)=0`

`x=-20`

Vậy: `...`

Bài 4:

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

=>\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\cdots+\frac{3}{x\left(x+3\right)}=\frac{101}{1540}\cdot3=\frac{303}{1540}\)

=>\(\frac15-\frac18+\frac18-\frac{1}{11}+\cdots+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

=>\(\frac15-\frac{1}{x+3}=\frac{303}{1540}\)

=>\(\frac{1}{x+3}=\frac15-\frac{303}{1540}=\frac{1}{308}\)

=>x+3=308

=>x=305

Bài 3:

a: (x-2)(y+1)=3

=>(x-2;y+1)∈{(1;3);(3;1);(-1;-3);(-3;-1)}

=>(x;y)∈{(3;2);(5;0);(1;-4);(-1;-2)}

b: xy+x+y+1=5

=>x(y+1)+(y+1)=5

=>(x+1)(y+1)=5

=>(x+1;y+1)∈{(1;5);(5;1);(-1;-5);(-5;-1)}

=>(x;y)∈{(0;4);(4;0);(-2;-6);(-6;-2)}

c: xy+3y+x=2

=>y(x+3)+x+3=2+3

=>(x+3)(y+1)=5

=>(x+3;y+1)∈{(1;5);(5;1);(-1;-5);(-5;-1)}

=>(x;y)∈{(-2;4);(2;0);(-4;-6);(-8;-2)}

Bài 2:

a: \(\frac{x-2}{-2}=\frac{3x+1}{3}\)

=>-2(3x+1)=3(x-2)

=>-6x-2=3x-6

=>-6x-3x=-6+2

=>-9x=-4

=>\(x=\frac49\)

b: \(\frac{x-5}{15}+\frac{x-7}{13}=\frac{x-9}{11}+\frac{x-11}{9}\)

=>\(\left(\frac{x-5}{15}-1\right)+\left(\frac{x-7}{13}-1\right)=\left(\frac{x-9}{11}-1\right)+\left(\frac{x-11}{9}-1\right)\)

=>\(\frac{x-20}{15}+\frac{x-20}{13}=\frac{x-20}{11}+\frac{x-20}{9}\)

=>x-20=0

=>x=20

c: \(\frac{x+9}{11}+\frac{x+12}{8}+\frac{x+10}{5}=-4\)

=>\(\left(\frac{x+9}{11}+1\right)+\left(\frac{x+12}{8}+1\right)+\left(\frac{x+10}{5}+2\right)=-4+4=0\)

=>\(\frac{x+20}{11}+\frac{x+20}{8}+\frac{x+20}{5}=0\)

=>x+20=0

=>x=-20

d: \(\left(2^{3x+2}-2\right):6+22=\frac{1}{\left(-3\right)^2}\cdot3^4\)

=>\(\left(2^{3x+2}-2\right):6+22=\frac{81}{9}=9\)

=>\(2^{3x+2}-2=\left(9-22\right)\cdot6=-13\cdot6=-78\)

=>\(2^{3x+2}=-78+2=-76\) (vô lý)

=>x∈∅

a) 27 x 36 + 27 x 64

= 27 x (36 + 64)

= 27 X 100

= 2 700

b) 25 x 37 + 25 x 36 - 150

= 25 x 37 + 25 x 36 - 25 x 5

= 25 x (37 + 36 - 5)

= 25 x 68

= 1700

c) 425 x 7 x 4 - 170 x 60

= 1700 x 7 - 1700 x 6

= 1700 x (7 - 6)

= 1700 x 1

= 1700

d) 8 x 9 x 14 + 6 x 17 x 12 + 19 x 4 x 18

= 72 x 14 + 72 x 17 + 72 x 19

= 72 x (14 + 17 + 19)

= 72 x 50

= 3600

bài 3 làm như thế nào

`a)6/(x-7)=(-3)/7` (ĐK: `x\ne7)`

`x-7=6:(-3)/7`

`x-7=-14`

`x=-14+7`

`x=-7(tm)`

Vậy: `...`

`b)(-3)/5-x=1/2`

`x=(-3)/5-1/2`

`x=(-6)/10-5/10`

`x=(-11)/10`

Vậy: `...`

`c)x-3/10=7/15*3/14`

`x-3/10=1/10`

`x=3/10+1/10`

`x=4/10`

`x=2/5`

Vậy: `...`

`d)5/14-x=(-17)/36:51/18`

`5/14-x=(-17)/36*18/51`

`5/14-x=-1/6`

`x=5/14+1/6`

`x=11/21`

Vậy: `..`

`d)-3/4x+1/4(x-1)=-12/5`

`-3/4x+1/4x-1/4=-12/5`

`-1/2x-1/4=-12/5`

`1/2x+1/4=12/5`

`1/2x=12/5-1/4`

`1/2x=43/20`

`x=43/20:1/2`

`x=43/10`

Vậy: `..`

`f)(2x-4/9)(3-11x)=0`

`TH1:2x-4/9=0`

`2x=4/9`

`x=4/9:2`

`x=2/9`

`TH2:3-11x=0`

`11x=3`

`x=3/11`

Vậy: `...`

129

Đáp án: 86% của 150 bằng 129.

Gấp rưỡi=1,5 lần=3/2

Tổng số phần bằng nhau là 3+2=5(phần)

Số lớn là \(18,9:5\cdot3=11,34\)

bài 6 làm như thế nào hả các bạn

Bài 1:

\(A=-\frac25+\frac34\cdot\frac29\)

\(=-\frac25+\frac{6}{36}\)

\(=-\frac25+\frac16=-\frac{12}{30}+\frac{5}{30}=-\frac{7}{30}\)

\(B=\frac38\cdot\frac{-13}{15}-\frac38\cdot\frac{2}{15}+1\frac12\)

\(=\frac38\left(-\frac{13}{15}-\frac{2}{15}\right)+\frac32\)

\(=-\frac38+\frac32=-\frac38+\frac{12}{8}=\frac98\)

\(C=-\frac78+\frac{9}{10}\cdot\frac23\)

\(=-\frac78+\frac93\cdot\frac{2}{10}\)

\(=-\frac78+\frac35=-\frac{35}{40}+\frac{24}{40}=-\frac{11}{40}\)

\(D=\frac{5}{11}\cdot\frac{-17}{31}-\frac{5}{11}\cdot\frac{14}{31}+1\frac12\)

\(=-\frac{5}{11}\left(\frac{17}{31}+\frac{14}{31}\right)+\frac32\)

\(=-\frac{5}{11}+\frac32=-\frac{10}{22}+\frac{33}{22}=\frac{23}{22}\)

Bài 2:

a: \(\frac65-\frac15\cdot\frac{10}{3}\)

\(=\frac65-\frac{10}{5}\cdot\frac13=\frac65-\frac23\)

\(=\frac{18}{15}-\frac{10}{15}=\frac{8}{15}\)

b: \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)

\(=\left(\frac{11}{24}+\frac{13}{24}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)+0,5\)

\(=1-1+0,5=0,5\)

c: \(\frac12\cdot\frac{-3}{4}+\frac12\cdot\frac{-1}{4}+\frac23\)

\(=\frac12\left(-\frac34-\frac14\right)+\frac23\)

\(=-\frac12+\frac23=-\frac36+\frac46=\frac16\)

d: \(\frac13\cdot\frac{-5}{6}+\frac13\cdot\frac{-1}{6}+\frac12\)

\(=\frac13\left(-\frac56-\frac16\right)+\frac12\)

\(=-\frac13+\frac12=\frac16\)

Bài 3:

a: \(\frac{6}{x-7}=\frac{-3}{7}\)

=>\(\frac{6}{x-7}=\frac{-6}{14}\)

=>x-7=-14

=>x=-14+7=-7

b: \(-\frac35-x=\frac12\)

=>\(x=-\frac35-\frac12=-\frac{6}{10}-\frac{5}{10}=-\frac{11}{10}\)

c: \(x-\frac{3}{10}=\frac{7}{15}\cdot\frac{3}{14}\)

=>\(x-\frac{3}{10}=\frac{7}{14}\cdot\frac{3}{15}=\frac12\cdot\frac15=\frac{1}{10}\)

=>\(x=\frac{1}{10}+\frac{3}{10}=\frac{4}{10}=\frac25\)

d: \(\frac{5}{14}-x=-\frac{17}{36}:\frac{51}{18}\)

=>\(\frac{5}{14}-x=-\frac{17}{36}\cdot\frac{18}{51}=-\frac{17}{51}\cdot\frac{18}{36}\)

=>\(\frac{5}{14}-x=-\frac13\cdot\frac12=-\frac16\)

=>\(x=\frac{5}{14}+\frac16=\frac{15}{42}+\frac{7}{42}=\frac{22}{42}=\frac{11}{21}\)

e: \(-\frac34x+\frac14\left(x-1\right)=-\frac{12}{5}\)

=>\(-\frac34x+\frac14x-\frac14=-\frac{12}{5}\)

=>\(-\frac24x=-\frac{12}{5}+\frac14=-\frac{48}{20}+\frac{5}{20}=-\frac{43}{20}\)

=>\(\frac{x}{2}=\frac{43}{20}\)

=>\(x=\frac{43}{20}\cdot2=\frac{43}{10}\)

f: \(\left(2x-\frac49\right)\left(3-11x\right)=0\)

=>\(\left[\begin{array}{l}2x-\frac49=0\\ 3-11x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac49\\ 11x=3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac49:2=\frac29\\ x=\frac{3}{11}\end{array}\right.\)

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