Tìm x
\(x+\frac{2}{3}=8:4-1\)
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Câu hỏi của Vũ Mai Linh - Toán lớp 7 - Học toán với OnlineMath
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
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 = \(\frac{19}{14}\)
\(\left(-\frac{2}{3}:-\frac{1}{3}\right).\left(-\frac{9}{2}\right)-\frac{1}{4}< \frac{x}{8}< -\frac{1}{2}.\frac{3}{4}:\frac{1}{8}+1\)
\(2.\left(-\frac{9}{2}\right)-\frac{1}{4}< \frac{x}{8}< \left(-3\right)+1\)
\(\left(-9\right)-\frac{1}{4}< \frac{x}{8}< \left(-2\right)\)
\(\left(-\frac{37}{4}\right)< \frac{x}{8}< \left(-2\right)\)
\(-\frac{74}{8}< \frac{x}{8}< -\frac{16}{8}\)
Vậy -74<x<-16
ĐKXXD : \(x\ne20;8;3;1\)
\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{\left(x-1\right)-\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\frac{\left(x-3\right)-\left(x-8\right)}{\left(x-3\right)\left(x-8\right)}+\frac{\left(x-8\right)-\left(x-20\right)}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-8}-\frac{1}{x-3}+\frac{1}{x-20}-\frac{1}{x-8}+\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow-\frac{1}{x-1}=-\frac{3}{4}\Leftrightarrow x-1=\frac{4}{3}\Rightarrow x=\frac{7}{3}\)
\(x+\frac{2}{3}=8:4-1\)
\(\Rightarrow x+\frac{2}{3}=2-1=1\)
\(\Rightarrow x=1-\frac{2}{3}=\frac{1}{3}\)
x=1/3