chịu bạn e

I

1 in

2 sound

3 has helped

4 wrote

5 when

6 been

7 yet

8 The

9 his

10 going

\(\left(\frac58-\frac37\right)-\left(\frac38-\frac27\right)\)

\(=\frac58-\frac37-\frac38+\frac27\)

\(=\left(\frac58-\frac38\right)-\left(\frac37-\frac27\right)\)

\(=\frac28-\frac17\)

\(=\frac14-\frac17\)

\(=\frac{7}{28}-\frac{4}{28}\)

\(=\frac{3}{28}\)

\(\frac{3}{28}\)

1. This river does not run through my home town.

2. My drawing class does not start at 8 a.m. every Sunday.

3. They do not enjoy collecting stamps.

4. I do not do judo every Tuesday.

5. My brother does not love making model cars.

1.This river doesn't run through my home town .

2.My drawing class doesn't start at 8 a.m every Sunday .

3.They don't enjoy collecting stamps .

4.I don't do judo every Tuesday .

5.My brother doesn't love making model cars .

Ở trong SGK nào vậy bạn? Khi đăng câu hỏi bạn nên đăng thêm văn bản liên quan nhé.

21A, 22B, 23D, 24D, 25C, 26B, 27D, 28D, 29A, 30C, 31B, 32D, 33A, 34D, 35A

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

(x – 1)(x + 1)(x + 2)

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

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

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

=\(x^3+2x^2-x-2\)

\(\frac{7}{15}\) - \(\frac57\) + \(\frac{23}{15}\) + \(\frac57\) - \(\frac35\)

= (\(\frac{7}{15}\) + \(\frac{23}{15}\)) - (\(\frac57\) - \(\frac57\)) - \(\frac35\)

= 2 - 0 - \(\frac35\)

= 2 - \(\frac35\)

= \(\frac{10}{5}\) - \(\frac35\)

= \(\frac75\)

\(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)+\left(-\frac57+\frac57\right)-\frac35\)

\(=\frac{30}{15}-\frac35\)

\(=\frac{30}{15}-\frac{9}{15}\)

\(=\frac{21}{15}\)

Ta có: \(\left|x-0,25\right|+\left|2x-1\right|+\left|x-2,5\right|=x-3\)

=>x-3>=0

=>x>=3

=>x-0,25>=2,75>0; 2x-1>=5>0; x-2,5>=0,5>0

Phương trình sẽ trở thành:

x-0,25+2x-1+x-2,5=x-3

=>4x-3,75=x-3

=>3x=-3+3,75=0,75

=>x=0,25(loại)

Vậy: Phương trình vô nghiệm

x−0,25∣+∣2x−1∣+∣x−2,5∣=x−3

=>x-3>=0

=>x>=3

=>x-0,25>=2,75>0; 2x-1>=5>0; x-2,5>=0,5>0

Phương trình sẽ trở thành:

x-0,25+2x-1+x-2,5=x-3

=>4x-3,75=x-3

=>3x=-3+3,75=0,75

=>x=0,25(loại)

a: Ta có: \(3x+\left(x-\frac{9}{20}\right)=-\frac{13}{40}\)

=>\(3x+x-\frac{9}{20}=-\frac{13}{40}\)

=>\(4x=-\frac{13}{40}+\frac{9}{20}=-\frac{13}{40}+\frac{18}{40}=\frac{5}{40}=\frac18\)

=>\(x=\frac18:4=\frac{1}{32}\)

b: \(x+\left(\frac14x-2,5\right)=-\frac{11}{20}\)

=>\(x+\frac14x-2,5=-\frac{11}{20}\)

=>\(1,25x=-0,55+2,5=1,95\)

=>\(x=\frac{1.95}{1.25}=\frac{195}{125}=\frac{39}{25}\)

c: \(\frac35x+\left(x+0,5\right)=-\frac{13}{15}\)

=>\(\frac35x+x+0,5=-\frac{13}{15}\)

=>\(\frac85x=-\frac{13}{15}-0,5=-\frac{26}{30}-\frac{15}{30}=-\frac{41}{30}\)

=>\(x=-\frac{41}{30}:\frac85=-\frac{41}{30}\cdot\frac58=\frac{-41}{6\cdot8}=-\frac{41}{48}\)

d: \(-\frac23x+\left(4x-\frac67\right)=\frac{9}{21}\)

=>\(-\frac23x+4x-\frac67=\frac37\)

=>\(\frac{10}{3}x=\frac37+\frac67=\frac97\)

=>\(x=\frac97:\frac{10}{3}=\frac97\cdot\frac{3}{10}=\frac{27}{70}\)

bài 11: câu a:

\(3x+\left(x-\frac{9}{20}\right)=-\frac{13}{40}\)

\(3x+x-\frac{9}{20}=-\frac{13}{40}\)

\(4x=-\frac{13}{40}+\frac{9}{20}\)

\(4x=-\frac{13}{40}+\frac{18}{40}\)

\(4x=\frac{5}{40}\)

\(4x=\frac18\)

\(x=\frac18:4=\frac18\cdot\frac14=\frac{1}{32}\)

b. \(x+\left(\frac14x-2,5\right)=-\frac{11}{20}\)

\(x+\frac14x-2,5=-\frac{11}{20}\)

\(\frac54x-2,5=-\frac{11}{20}\)

\(\frac54x=-\frac{11}{20}+2,5\)

\(\frac54x=\frac{39}{20}\)

\(x=\frac{39}{20}:\frac54=\frac{39}{20}\cdot\frac45=\frac{39}{25}\)

c. \(\frac35x+\left(x+0,5\right)=-\frac{13}{15}\)

\(\frac35x+x+0,5=-\frac{13}{15}\)

\(\frac85x+\frac12=-\frac{13}{15}\)

\(\frac85x=-\frac{13}{15}-\frac12\)

\(\frac85x=-\frac{41}{30}\)

\(x=-\frac{41}{30}:\frac85=-\frac{41}{30}\cdot\frac58=-\frac{41}{48}\)

\(d.-\frac23x+\left(4x-\frac67\right)=\frac{9}{21}\)

\(-\frac23x+4x-\frac67=\frac{9}{21}\)

\(\frac{10}{3}x=\frac97\)

\(x=\frac97:\frac{10}{3}=\frac97\cdot\frac{3}{10}=\frac{27}{70}\)

a: \(\left(-\frac54x+3,25\right)\left\lbrack\frac35-\left(-\frac52x\right)\right\rbrack=0\)

=>\(\left(\frac54x-\frac{13}{4}\right)\left(\frac52x+\frac35\right)=0\)

=>\(\left[\begin{array}{l}\frac54x-\frac{13}{4}=0\\ \frac52x+\frac35=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac54x=\frac{13}{4}\\ \frac52x=-\frac35\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{13}{4}:\frac54=\frac{13}{5}\\ x=-\frac35:\frac52=-\frac{6}{25}\end{array}\right.\)

b: \(\left(-\frac72x+1,75\right)\left\lbrack\frac45-\left(-\frac53x\right)\right\rbrack=0\)

=>\(\left[\begin{array}{l}-\frac72x+1,75=0\\ \frac45-\left(-\frac53x\right)=0\end{array}\right.\Longrightarrow\left[\begin{array}{l}-\frac72x=-1,75=-\frac74\\ \frac53x=-\frac45\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{-7}{4}:\frac{-7}{2}=\frac24=\frac12\\ x=-\frac45:\frac53=-\frac45\cdot\frac35=-\frac{12}{25}\end{array}\right.\)

c: \(\left(x^2-4\right)\left(x+\frac27\right)=0\)

=>\(\left[\begin{array}{l}x^2-4=0\\ x+\frac27=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=4\\ x=-\frac27\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-2\\ x=-\frac27\end{array}\right.\)

d: \(\left(25-x^2\right)\left(5x-\frac59\right)=0\)

=>\(\left[\begin{array}{l}25-x^2=0\\ 5x-\frac59=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=25\\ 5x=\frac59\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-5\\ x=\frac19\end{array}\right.\)