1 

Ez lắm =)

Bài 1:

Với mọi gt \(x,y\in Q\) ta luôn có: 

\(x\le\left|x\right|\) và \(-x\le\left|x\right|\) 

\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)

Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)

Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)

Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)

Dấu "=" xảy ra khi: \(xy\ge0\)

Bài 3: 

Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)

Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a² + b² + c² = 1)

Vậy: (x + y + z)² = x² + y² + z²

a.

\(\left(100+42\right)\times42+\left(200-58\right)\times58=142\times42+142\times58=142\times\left(42+58\right)=142\times100=14200\)

b.

\(\left(1998\div18-1443\div13\right)\times\left(16996-1110\div30\times305\right)=\left(111-111\right)\times\left(16996-1110\div30\times305\right)\)

\(=0\times\left(16996-1110\div30\times305\right)=0\)

c.

\(5^3_5+1,75+6^1_8+4^1_4+3,875+3,4=5,6+1,75+6,125+4,25+3,875+3,4\)

\(=\left(5,6+3,4\right)+\left(1,75+4,25\right)+\left(6,125+3,875\right)=9+6+10=25\)

Giải theo cách lớp 5

a)(100+42)x42+(200-58)x58

=142x42-142x58

=(42-58)x142

=(-16).142

=-2272

b)(1998:18-1443:13)x(16996-1110:30x305)

=(111-111)x(16996-1110:30x305)

=0x(16996-1110:30x305)

=0

c)\(5\frac{3}{5}\)+1,75+\(6\frac{1}{8}\)+\(4\frac{1}{4}\)+3,875+3,4

=(1,75+\(4\frac{1}{4}\))+(3,875+\(6\frac{1}{8}\))+(3,4+\(5\frac{3}{5}\))

=6+10+9

=25

Bạn đăng ít một thôi!

mk lỡ đăng rồi bạn ạ 

nên các bạn giải bài nào cũng được

a) x-1/3=5/14.(-7/6)

  x-1/3=-5/12

  x=-5/12+1/3

  x= -1/12

b) 3/4+1/4x=0,2

   1/4x=0,2-3/4

   1/4x=-11/20

    x=-11/20:1-4

    x=-11/5

c) 1/12.x^2=1.1/2

   1/12.x^2=1/2 

   x^2= 1/2:1/12

   x^2=6

=> x=căn bậc của 6

Chúc bn hok tốt

a)\(x-\frac{1}{3}=\frac{5}{14}\cdot\left(\frac{-7}{6}\right)\)

\(x-\frac{1}{3}=\frac{-5}{12}\)

\(x=\frac{-5}{12}+\frac{1}{3}\)

\(x=\frac{-5}{12}+\frac{4}{12}\)

\(x=\frac{-1}{12}\)

b)\(\frac{3}{4}+\frac{1}{4}\cdot x=0,2\)

\(\frac{3}{4}+\frac{1}{4}\cdot x=\frac{1}{5}\)

\(\frac{1}{4}\cdot x=\frac{1}{5}-\frac{3}{4}\)

\(\frac{1}{4}\cdot x=\frac{4}{20}-\frac{15}{20}\)

\(\frac{1}{4}\cdot x=\frac{-11}{20}\)

\(x=\frac{-11}{20}\cdot4\)

\(x=-\frac{11}{5}\)

c)\(\frac{1}{12}\cdot x^2=1\cdot\frac{1}{3}\)

\(\frac{1}{12}\cdot x^2=\frac{1}{3}\)

\(x^2=\frac{1}{3}\cdot12\)

\(x^2=4=\left(\pm2\right)^2\)

\(x=\pm2\)

Bài 1 : Thực hiện phép tính :

a, \(\frac{4}{5}+1\frac{1}{6}\cdot\frac{3}{4}\)

= \(\frac{4}{5}+\frac{7}{6}\cdot\frac{3}{4}\)

= \(\frac{4}{5}+\frac{7}{8}\)

= \(\frac{32+35}{40}=\frac{67}{40}\)

b, \(\frac{2}{3}:\left(\frac{3}{4}\cdot\frac{4}{3}\right)+2\)

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

\(=\frac{2}{3}+2=\frac{2+6}{3}=\frac{8}{3}\)

c, \(\frac{1}{2}\times\left(\frac{2}{3}+\frac{3}{5}\cdot\frac{5}{7}\right)+1\frac{1}{3}\)

\(=\frac{1}{2}\cdot\left(\frac{2}{3}+\frac{9}{35}\right)+\frac{4}{3}\)

\(=\frac{1}{2}\cdot\frac{97}{105}+\frac{4}{3}\)

\(=\frac{97}{210}+\frac{4}{3}=\frac{377}{210}\)

Bài 2 : Tìm \(x\inℤ\), biết :

a, \(\frac{2}{3}< \frac{x}{6}\le\frac{10}{3}\)

\(\Leftrightarrow\frac{4}{6}< \frac{x}{6}\le\frac{20}{6}\)

mà \(x\inℤ\Rightarrow\text{x}\in\) {\(5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20\)}

b, \(\frac{1}{3}+x=1\frac{1}{2}\)

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

\(x=\frac{3}{2}+\frac{\left(-1\right)}{3}\)

\(x=\frac{7}{6}\) (loại vì \(x\notinℤ\))

\(\Rightarrow x\in\varnothing\)

c, \(\frac{1}{7}+x=\frac{25}{14}+\frac{5}{14}\)

\(\frac{1}{7}+x=\frac{15}{7}\)

\(x=\frac{15}{7}+\frac{(-1)}{7}\)

\(x=\frac{14}{7}=2\).

a) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)

<=> \(\frac{5x+2\left(3-x\right)}{70}-\frac{5x-4\left(x-1\right)}{24}=\frac{35x+10+9-3x}{60}+\frac{2}{3}\)

<=> \(12\left(5x+6-2x\right)-35\left(5x-4x+4\right)\)

<=> \(14\left(35x+10+9-3x\right)+280.2\) <=> \(12\left(3x+6\right)-35\left(x+4\right)\)

<=> \(14\left(32x+19\right)+560\)

<=> \(36x+72-35x-140=448x+226+560\)

<=> \(-447x=894\)

<=> x = -2

giúp mk nhanh nhé

ai nhanh mk tk cho

B1

a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)

\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)

\(1-\left(x-\frac{11}{6}\right).\frac{3}{50}=0\)

\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1-0\)

\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1\)

\(x-\frac{11}{6}=1:\frac{3}{50}\)

\(x-\frac{11}{6}=\frac{50}{3}\)

\(x=\frac{50}{3}+\frac{11}{6}\)

\(x=\frac{37}{2}\)

b) \(\frac{3}{5}+\frac{5}{7}:x=\frac{1}{3}\)

\(\frac{5}{7}:x=\frac{1}{3}-\frac{3}{5}\)

\(\frac{5}{7}:x=-\frac{4}{15}\)

\(x=\frac{5}{7}:\left(-\frac{4}{15}\right)\)

\(x=-\frac{75}{28}\)

c) \(\left(4\frac{1}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)

\(\left(\frac{9}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)

\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{9}.\frac{7}{4}\)

\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{2}\)

\(\frac{2}{5}.x=\frac{9}{2}-\frac{11}{2}\)

\(\frac{2}{5}.x=-1\)

\(x=-1:\frac{2}{5}\)

\(x=-\frac{5}{2}\)

B2

a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{6}\right).24:5-\frac{9}{22}:\frac{15}{121}\)

\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{2}{6}\right).24:5-\frac{9}{22}.\frac{121}{15}\)

\(=\frac{7}{6}.24:5-\frac{33}{10}\)

\(=28:5-\frac{33}{10}\)

\(=\frac{28}{5}-\frac{33}{10}\)

\(=\frac{56}{10}-\frac{33}{10}\)

\(=\frac{23}{10}\)

b) \(\frac{5}{14}+\frac{18}{35}+\left(1\frac{1}{4}-\frac{5}{4}\right):\left(\frac{5}{12}\right)^2\)

\(=\frac{25}{70}+\frac{36}{70}+\left(\frac{5}{4}-\frac{5}{4}\right):\frac{25}{144}\)

\(=\frac{61}{70}+0:\frac{25}{144}\)

\(=\frac{61}{70}+0\)

\(=\frac{61}{70}\)