c) pt <=> \(x-\frac{21}{5}=\frac{23}{7}< =>x=\frac{23}{7}+\frac{21}{5}=\frac{262}{35}\)

vậy x = \(\frac{262}{35}\) 

d) \(x-\frac{3}{4}=\frac{51}{8}< =>x=\frac{51}{8}+\frac{3}{4}=\frac{57}{8}\) 

vậy x = \(\frac{57}{8}\) 

e) pt <=> \(\frac{7}{8}:x=\frac{7}{2}< =>\frac{7}{8}.\frac{1}{x}=\frac{7}{2}< =>\frac{7}{8x}=\frac{7}{2}< =>56x=14< =>x=\frac{14}{56}=\frac{1}{4}\)

vậy x = \(\frac{1}{4}\)

a) pt <=> \(x+\frac{11}{4}=\frac{17}{3}< =>x=\frac{17}{3}-\frac{11}{4}=\frac{35}{12}\)

vậy x = \(\frac{35}{12}\)

b) pt <=> \(\frac{x.7}{2}=\frac{19}{4}< =>x=\frac{19.2}{4.7}=\frac{38}{28}=\frac{19}{14}\)

vậy x = \(\frac{19}{14}\) 

a) \(|x+1|=3\)

\(\Rightarrow x+1=\pm3\)

+) \(x+1=3\)                                            +) \(x+1=-3\)

\(\Rightarrow x=2\)                                                    \(\Rightarrow x=-4\)

Vậy \(x\in\left\{2;-4\right\}\)

b) \(3^2x+2^4=5^2\)

     \(9x+16=25\)

                 \(9x=25-16\)

                  \(9x=9\)

                     \(x=1\)

c) \(\frac{4+x}{7+y}=\frac{4}{7}\)

\(\Rightarrow\left(4+x\right).7=\left(7+y\right).4\)

\(\Rightarrow28+7x=28+4y\)

\(\Rightarrow7x=4y\)

Mà \(\left(7,4\right)=1\) và \(x+y=11\)

Vậy \(x=4;y=7\)

a) Ta có: \(\left|x+1\right|=3\)

\(\Rightarrow x+1=\pm3\)

Nếu x + 1 = 3 => x = 2

Nếu x + 1 = -3 => x = -4

Vậy x = {2;-4}

b) \(3^2x+2^4=5^2\)

\(\Rightarrow9x+16=25\)

\(\Rightarrow9x=9\)

^{\(\Rightarrow x=1\)}

Vậy x = 1

c) \(\frac{4+x}{7+x}=\frac{4}{7}\)

\(\Rightarrow7\left(4+x\right)=4\left(7+x\right)\)

\(\Rightarrow28+7x=28+4x\)

\(\Rightarrow7x-4x=0\)

\(\Rightarrow x=0\)

Vậy x = 0

a) (x + 1/2) . (2/3 − 2x) = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)

\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)

\(\Rightarrow x=-\frac{2}{11}\)

c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)

\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)

\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)

\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)

\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)

\(\Rightarrow x=1\)

d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)

\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)

\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)

\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)

\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)

\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)

\(\Rightarrow x\approx28,7\) (số hơi lẻ)

e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)

a, \(x:\left(\frac{-1}{2}\right)^3=\frac{-1}{2}\)

=> \(x=\frac{-1}{2}\cdot\left(\frac{-1}{2}\right)^3\)

=>\(x=\left(\frac{-1}{2}\right)^4\)

=>\(x=\frac{1}{16}\)

b,\(\frac{3^5}{4}\cdot x=\frac{3^7}{4}\)

=> \(x=\frac{3^7}{4}:\frac{3^5}{4}\)

=> \(x=\frac{3^2}{1}=9\)

a) \(\left(4,5-2x\right)\cdot1\frac{4}{7}=\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)\cdot\frac{11}{7}=\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{7}\div\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)=1\)

\(2x=\frac{9}{2}-1\)

\(x=\frac{7}{2}\div2\)

\(x=\frac{7}{4}\)

b) \(|\frac{3}{4}\cdot x-\frac{1}{2}|-1=\frac{1}{4}\)

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

\(|\frac{3}{4}\cdot x|=\frac{5}{4}+\frac{1}{2}\)

\(x=\frac{7}{4}\div\frac{3}{4}\)

\(x=\frac{7}{3}\)

c) \(\frac{1}{4}-|3-x|=-\frac{3}{4}\)

\(|3-x|=\frac{1}{4}-\left(-\frac{3}{4}\right)\)

\(|3-x|=1\)

\(x=3-1\)

\(\Rightarrow x=2\)

d) \(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=1,4\)

\(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=\frac{7}{5}\)

\(4\cdot\left(x-\frac{6}{7}\right)=\frac{7}{5}+\frac{3}{5}\)

\(4\cdot\left(x-\frac{6}{7}\right)=2\)

\(\left(x-\frac{6}{7}\right)=2\div4\)

\(x=\frac{1}{2}+\frac{6}{7}\)

\(x=\frac{19}{14}\)

\(\)

k cho mình nhé 

chúc bạn học giỏi

\(\frac{4}{7}\times x=\frac{1}{5}+\frac{2}{3}\)

\(\frac{4}{7}x=\frac{13}{15}\)

\(\Rightarrow x=\frac{91}{60}\)

các bài còn lại tương tự nha 

mấy cái này dễ mà toán tìm x này là cơ bản!!

67865785685685785785774677567568568

Câu a:

\(\frac{-8}{3x-1}\) = \(\frac{4}{-7}\)

-8.(-7) = 4.(3\(x\) - 1)

56 = 12\(x\) - 4

12\(x\) = 56+ 4

12\(x\) = 60

\(x\) = 60 : 12

\(x\) = 5

Vậy \(x\) = 5

Câu b:

\(\frac{x}{-3}\) = \(\frac{-3}{x}\)

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

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

Vậy \(x\in\left\lbrace-3;3\right\rbrace\)

Câu c:

\(-\frac{4}{y}=\frac{x}{2}\)

-4.2 = \(x.y\)

\(xy=-8\)

Ư(8) = (-8; -4; -2; -1; 1; 2; 4; 8}

Vậy (\(x;y\)) = (-8; 1); (-4; 2); (-2; 4); (-1; 8); (1; -8); (2; -4); (4; -2); (8; -1)

Câu 2:

(\(x-1)\)(y + 2) = 7

Ư(7) = {-7; -1; 1; 7}

Lập bảng ta có:

Theo bảng trên ta có:

(\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; - 1)

Vậy (\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; -1)

\(a,3^4:3^x=3^3\)

\(3^{4-x}=3^3\)

=> 4 -x = 3

=> x = 1

Nguyễn Tấn Tài bn làm hết nhé

Câu a:

33 ⋮ (x+ 1)

(x+ 1) ∈ Ư(33) = {-33; -11; -3; -1; 1; 3; 11}

x ∈ {-34; -12; -4; -2; 0; 2; 10}

Vậy: {-34; -12; -4; -2; 0; 2; 10}

Câu b:

x ∈ ƯC(250; 48)

250 = 2.5^3; 48 = 2^4.3

ƯCLN(250; 48) = 2

x ∈ ƯC(2) = {-2; -1}

Vậy x ∈ {-2; -1}