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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
1. \(x=\frac{61}{42}\)
2. \(x=\frac{-36}{5}\)
3. \(x=\frac{13}{11}\)
4. \(x=\frac{1}{12}\)
5.\(x=\frac{-5}{2}\)
b; \(3\frac{1}{4}.x-1\frac{1}{6}.x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x.\left(3\frac{1}{4}-1\frac{1}{6}-1\frac{2}{3}\right)=\frac{5}{12}\)
\(\Rightarrow x.\left(\frac{13}{4}-\frac{7}{6}-\frac{5}{3}\right)=\frac{5}{12}\)
\(\Rightarrow x.\left(\frac{39}{12}-\frac{14}{12}-\frac{20}{12}\right)=\frac{5}{12}\)
\(\Rightarrow x.\frac{5}{12}=\frac{5}{12}\)
\(\Rightarrow x=\frac{5}{12}\div\frac{5}{12}=1\)
Vậy x=1
a) \(\frac{x}{5}=\frac{2}{5}\Rightarrow x=\frac{2\cdot5}{5}=2\)
b)\(\frac{3}{8}=\frac{6}{x}\Rightarrow x=\frac{6\cdot8}{3}=16\)
c)\(\frac{1}{9}=\frac{x}{27}\Rightarrow x=\frac{27}{9}=3\)
d) \(\frac{4}{x}=\frac{8}{6}\Rightarrow x=\frac{4\cdot6}{8}=3\)
e)
\(\frac{3}{x-5}=-\frac{4}{x+2}\\ \Rightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Leftrightarrow3x+6=-4x+20\\ \Leftrightarrow3x+6+4x-20=0\\ \Leftrightarrow7x-14=0\\ \Leftrightarrow x=2\)
\(g,\frac{x}{-2}=-\frac{8}{x}\Rightarrow x^2=16\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy có 2 giá trị của x là 4 ; -4
a) Ta có: \(\frac{x}{5}=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{2\cdot5}{5}=2\)
Vậy: x=2
b) Ta có: \(\frac{3}{8}=\frac{6}{x}\)
\(\Leftrightarrow x=\frac{6\cdot8}{3}=\frac{48}{3}=16\)
Vậy: x=16
c) Ta có: \(\frac{1}{9}=\frac{x}{27}\)
\(\Leftrightarrow x=\frac{1\cdot27}{9}=\frac{27}{9}=3\)
Vậy: x=3
d) Ta có: \(\frac{4}{x}=\frac{8}{6}\)
\(\Leftrightarrow x=\frac{4\cdot6}{8}=\frac{24}{8}=3\)
Vậy: x=3
e) Ta có: \(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Leftrightarrow3\cdot\left(x+2\right)=-4\cdot\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow3x+6+4x-20=0\)
\(\Leftrightarrow7x-14=0\)
\(\Leftrightarrow7x=14\)
hay x=2
Vậy: x=2
g) Sửa đề: \(\frac{x}{-2}=\frac{-8}{x}\)
Ta có: \(\frac{x}{-2}=\frac{-8}{x}\)
\(\Leftrightarrow x^2=\left(-8\right)\cdot\left(-2\right)=16\)
hay x∈{4;-4}
Vậy: x∈{4;-4}
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 = \(\frac{19}{14}\)