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

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

Bạn ơi, đây là phần ghi cảm xúc của riêng bạn, điều này không thể vay mượn hay tham khảo được

ò kiểu í là mình biết ( trừ phần cuối) í nhưng mà ko bt cách trình bày á bn

|\(x+2\)| = 4 - 2\(x\)

|\(x+2\)| ≥ 0 ∀ \(x\) ⇒ 4 -2\(x\) ≥ 0 ⇒ 4 > 2\(x\) ⇒ \(\frac42>x\) ⇒ 2 > \(x\)

Với \(x<2\) ta có:

\(\left[\begin{array}{l}x+2=4-2x\\ x+2=-4+2x\end{array}\right.\)

\(\left[\begin{array}{l}x+2x=4-2\\ 2x-x=4+2\end{array}\right.\)

\(\left[\begin{array}{l}3x=2\\ x=6\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac23\\ x=6\end{array}\right.\)

Vì 2 > \(x\) nên \(x=6\) loại

Vậy \(x=\frac23\)

|x+2|=4-2x

|x+2|-(4-2x)=0

x+2-4+2x=0

(2-4)+(x+2x)=0

-2+3x=0

3x=-2

x=-2/3

\(\frac{2x-3}{4}\) = \(\frac{x+1}{-3}\)

-3.(\(2x-3\)) = 4.(\(x+1\))

-6\(x+9\) = 4\(x+\) 4

6\(x\) + 4\(x\) = 9 - 4

10\(x\) = 5

\(x\) = 5 : 10

\(x\) = \(\frac{5}{10}\)

Vậy \(x=\frac12\)

2x-3=-3

4=x+1

ta có

\(\frac{2 x + 3}{4} = \frac{x + 1}{- 3}\)

\(\left(\right. 2 x + 3 \left.\right) \left(\right. - 3 \left.\right) = 4 \left(\right. x + 1 \left.\right)\) \(- 6 x - 9 = 4 x + 4\) \(- 6 x - 4 x = 4 + 9\) \(-10x=13\textrm{ }\Rightarrow\textrm{ }x=-\frac{13}{10}\)

vậy

\(x = - \frac{13}{10}\).

\(\frac{2x+3}{4}\) = \(\frac{x+1}{-3}\)

-3.(\(2x+3\)) = 4.(\(x+1\))

- 6\(x-9\) = 4\(x+4\)

4\(x+6x\) = -9 - 4

10\(x\) = - 13

\(x=-13:10\)

\(x=-\frac{13}{10}\)

Vậy \(x=-\frac{13}{10}\)

Olm chào em, cảm ơn đánh giá của em về chất lượng bài giảng của Olm, cảm ơn em đã đồng hành cùng Olm trên hành trình tri thức. Chúc em học tập hiệu quả và vui vẻ cùng Olm em nhé!

ừ

a)1/4 - 5/6 + 7/12

= -7/12 + 7/12

= 0

\(a.\frac14-\frac56+\frac{7}{12}\)

\(=\frac{3}{12}-\frac{10}{12}+\frac{7}{12}\)

\(=\frac{0}{12}=0\)

\(b.6\frac27\cdot\frac15-1\frac27\cdot\frac15+\frac45\)

\(=\frac{44}{7}\cdot\frac15-\frac97\cdot\frac15+\frac45\)

\(=\frac15\cdot\left(\frac{44}{7}-\frac97\right)+\frac45\)

\(=\frac15\cdot\frac{35}{7}+\frac45\)

\(=\frac15\cdot5+\frac45\)

\(=1+\frac45=\frac95\)

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

P ∈ Z ⇔ (3\(x-2\) )⋮ (2\(x+1\))

[2.(3\(x\) - 2)] ⋮ (2\(x\) + 1)

[3.(2\(x\) + 1) - 7] ⋮ (2\(x\) + 1)

7 ⋮ (2\(x+1\))

(2\(x+1\)) ∈ Ư(7) = {-7; -1 ;1; 7}

Lập bảng ta có:

Theo bảng trên ta có: \(x\in\) {-4; -1; 0; 3}

Vậy \(x\in\) {-4; -1; 0; 3}

ĐKXĐ: \(x<>-\frac12\)

Để p là số nguyên thì 3x-2⋮2x+1

=>6x-4⋮2x+1

=>6x+3-7⋮2x+1

=>-7⋮2x+1

=>2x+1∈{1;-1;7;-7}

=>2x∈{0;-2;6;-8}

=>x∈{0;-1;3;-4}

\(\left(\frac{y}{3}-5\right)^{2000}=\left(\frac{y}{3}-5\right)^{2008}\)

\(\frac{y}{3}-5=0\) hoặc \(\frac{y}{3}-5=1\) hoặc \(\frac{y}{3}-5=-1\)

\(\frac{y}{3}=5\) hoặc \(\frac{y}{3}=6\) hoặc \(\frac{y}{3}=4\)

y=15 hoặc y=18 hoặc y=12

(\(\frac{y}{3}\) - 5)\(^{2000}\) = (\(\frac{y}{3}\) - 5)\(^{2008}\)

(\(\frac{y}{3}\) - 5)\(^{2000}\) - (\(\frac{y}{3}\) - 5)\(^{2008}\) = 0

(\(\frac{y}{3}\) - 5)\(^{2000}\).[1 - (\(\frac{y}{3}\) - 5)\(^8\)] = 0

\(\left[\begin{array}{l}\frac{y}{3}-5=0\\ \frac{y}{3}-5=\pm1\end{array}\right.\)

\(\left[\begin{array}{l}y=5\times3\\ y=\left(1+5\right)\times3\\ y=\left(-1+5\right)\times3\end{array}\right.\)

\(\left[\begin{array}{l}y=15\\ y=18\\ y=12\end{array}\right.\)

Vậy y ∈ {12; 15; 18}