(\(\frac45:\frac65\) + \(\frac15\) : \(\frac{1}{x}\)) x 30 - 26 = 54

(\(\frac45\times\frac56\) + \(\frac15\times x\)) x 30 = 54 + 26

(\(\frac23+\frac15x\)) x 30 = 80

\(\frac23+\frac15x\) = 80 : 30

\(\frac23+\frac15x=\frac83\)

\(\frac15x=\frac83-\frac23\)

\(\frac15x\) = 2

\(x=2:\frac15\)

\(x=2\times5\)

\(x=10\)

Vậy \(x=10\)

6,3 x y + 3,7 x y = 100

y x (6,3 + 3,7) = 100

y x 10 = 100

y = 100 : 10

y = 10

Vậy y = 10

Bài 1: 

\(B=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{4}+\frac{3}{8}-\frac{5}{12}}+\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{8}}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)\(=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{2}\left(\frac{1}{2}+\frac{3}{4}-\frac{5}{6}\right)}+\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{8}\right)}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\) 

\(=\frac{1}{\frac{1}{2}}+3\)  \(=2+3\) \(=5\)

                                                  Vậy B=5

Bài 2:

a) x³ - 36x = 0  

=>  x(x²-36)=0

=>  x(x²+6x-6x-36)=0 

=> x[x(x+6)-6(x+6) ]=0

=> x(x+6)(x-6)=0

\(\Rightarrow\orbr{\begin{cases}^{x=0}x+6=0\\x-6=0\end{cases}}\)

 \(\Rightarrow\orbr{\begin{cases}^{x=0}x=-6\\x=6\end{cases}}\)

                                  Vậy x=0; x=-6; x=6

b)  (x - y = 4 => x=4+y)

 x−3y−2 =32  

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

=>2x-6= 3y-6

=>2x-3y=0

=>2(4+y)-3y=0

=>8+2y-3y=0

=>8-y=0

=>y=8 (thỏa mãn)

Do đó x=4+y=4+8=12 (thỏa mãn)

         Vậy x=12 và y =8

B= 1/2 + 3/4 - 5/6/1/2(1.2 + 3/4 - 5/6) + 3(1/4+ 1/5 - 1/8)/ 1/4  1/5 - 1/8 

B= 1/ 1/2 + 3

B= 2+3

B=5

B2:

a) x^3 - 36x = 0

x(x^2 - 36) = 0

=> x=0  hoặc x^2-36=0

=> x= 0 hoặc x^2=36

=> x=0 hoặc x= +- 6

Kết quả là 2870.

Giải nhanh bằng công thức tổng bình phương:

\(1^{2} + 2^{2} + \hdots + n^{2} = \frac{n \left(\right. n + 1 \left.\right) \left(\right. 2 n + 1 \left.\right)}{6}\)

Với \(n = 20\):

\(\frac{20 \cdot 21 \cdot 41}{6} = \frac{17220}{6} = 2870.\)

Ta có biểu thức:

\(1\times1+2\times2+3\times3+\ldots+20\times20=1^2+2^2+3^2+\ldots+20^2\)

Đây là tổng các số chính phương từ 1 đến 20.

Áp dụng công thức tổng bình phương:

\(1^2+2^2+3^2+\ldots+n^2=\frac{n \left(\right. n + 1 \left.\right) \left(\right. 2 n + 1 \left.\right)}{6}\)

Thay \(n = 20\):

\(\frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870\)

\(\dfrac{1}{7}=\dfrac{8}{-x}\)=> \(-x=56\)

=> \(x=56\)

2) => 18x = 18

=> x = 1

3) \(\dfrac{-4}{3}+x=\dfrac{-11}{6}\)

=> \(x=\dfrac{-11}{6}+\dfrac{4}{3}\)

=> \(x=\dfrac{-1}{2}\)

4) 45%.x =\(\dfrac{3}{5}\)

=> \(x=\dfrac{3}{5}:\dfrac{9}{20}\)

=> \(x=\dfrac{4}{3}\)

a.\(\dfrac{5}{3}x-2\dfrac{1}{3}=-\dfrac{4}{3}.1\dfrac{1}{8}-\dfrac{2}{3}\)

\(\dfrac{5}{3}x-2\dfrac{1}{3}=\dfrac{5}{6}\)

\(\dfrac{5}{3}x=\dfrac{5}{6}+2\dfrac{1}{3}\)

\(\dfrac{5}{3}x=\dfrac{19}{6}\)

\(x=\dfrac{19}{5}:\dfrac{5}{3}\)

\(x=\dfrac{19}{10}\)

b. \(2\dfrac{1}{6}:x-\dfrac{-5}{8}=\dfrac{-7}{15}:4\dfrac{1}{5}-\dfrac{-6}{7}\)

\(2\dfrac{1}{6}:x+\dfrac{5}{8}=\dfrac{47}{63}\)

\(2\dfrac{1}{6}:x=\dfrac{47}{63}-\dfrac{5}{8}\)

\(2\dfrac{1}{6}:x=\dfrac{61}{504}\)

\(x=2\dfrac{1}{6}:\dfrac{61}{504}\)

\(x=\dfrac{1092}{61}\)

a, \(\dfrac{5}{3}x-2\dfrac{1}{3}=-\dfrac{4}{3}.1\dfrac{1}{8}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{5}{3}x=-\dfrac{3}{2}-\dfrac{2}{3}+2\dfrac{1}{3}=-\dfrac{3}{2}+2=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}:\dfrac{5}{3}=\dfrac{3}{10}\)

b, \(2\dfrac{1}{6}:x-\dfrac{-5}{8}=\dfrac{-7}{15}:4\dfrac{1}{5}-\dfrac{-6}{7}\)

\(\Rightarrow\dfrac{13}{6}:x=-\dfrac{7}{15}:\dfrac{21}{5}+\dfrac{6}{7}-\dfrac{5}{8}\)

\(\Rightarrow\dfrac{13}{6}:x=\dfrac{61}{504}\Rightarrow x=\dfrac{1092}{61}\)

c, \(\left(\dfrac{5}{6}x-0,3\right):2\dfrac{1}{3}=25\%\)

\(\Rightarrow\dfrac{5}{6}x-0,3=\dfrac{1}{4}.\dfrac{7}{3}\)

\(\Rightarrow\dfrac{5}{6}x=\dfrac{7}{12}+0,3=\dfrac{53}{60}\)

\(\Rightarrow x=\dfrac{53}{60}:\dfrac{5}{6}=1,06\)

d, \(\dfrac{4}{7}-\dfrac{2}{3}x=1,5+\dfrac{4}{5}x\)

\(\Rightarrow\dfrac{4}{5}x+\dfrac{2}{3}x=\dfrac{4}{7}-1,5\)

\(\Rightarrow\dfrac{22}{15}x=-\dfrac{13}{14}\Rightarrow x=-\dfrac{195}{308}\)

Chúc bạn học tốt!!!

c: \(\left|\dfrac{7}{5}x+\dfrac{2}{3}\right|=\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|\)

=>7/5x+2/3=4/3x-1/4 hoặc 7/5x+2/3=1/4-4/3x

=>1/15x=-11/12 hoặc 41/15x=-5/12

=>x=-55/4 hoặc x=-25/164

d: |7/8x+5/6|=|1/2x+5|

=>|42x+40|=|24x+240|

=>42x+40=24x+240 hoặc 42x+40=-24x-240

=>18x=200 hoặc 66x=-280

=>x=100/9 hoặc x=-140/33

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}

\(\frac{24}{-12}\) = \(\frac{x}{5}\) = \(\frac{-y}{3}\)

- 2 = \(\frac{x}{5}\) = \(\frac{-y}{3}\)

\(x=5.\left(-2\right)\) = -10

y = -2.3:(-1) = -6:(-1) = 6

Vậy (\(x;y\)) = (-10; 6)