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a: \(=\dfrac{-2}{12}-\dfrac{5}{12}+\dfrac{7}{12}=0\)
b: \(=\left(\dfrac{4}{45}-\dfrac{1}{45}+\dfrac{7}{45}+\dfrac{4}{45}-\dfrac{2}{45}-\dfrac{9}{45}\right)=\dfrac{3}{45}=\dfrac{1}{15}\)
\(=1-\dfrac{1}{2}+1-\dfrac{1}{6}+...+1-\dfrac{1}{90}\)
\(=10-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\)
\(=10-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
=9+1/10
=9,1
A = \(\dfrac{\left(\dfrac{47}{15}+\dfrac{3}{15}\right):\dfrac{5}{2}}{\left(\dfrac{38}{7}-\dfrac{9}{4}\right):\dfrac{267}{56}}=\dfrac{\dfrac{10}{3}.\dfrac{2}{5}}{\dfrac{89}{28}.\dfrac{56}{267}}=2\)
B= \(\dfrac{1,2:\left(\dfrac{6}{5}.\dfrac{5}{4}\right)}{0,32+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{4}{\dfrac{5}{\dfrac{2}{5}}}=2\)
=> A = B
\(A=\dfrac{\left(3+\dfrac{2}{15}+\dfrac{1}{5}\right):\dfrac{5}{2}}{\left(5+\dfrac{3}{7}-2-\dfrac{1}{4}\right):\left(4+\dfrac{43}{56}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{2}{5}}{\dfrac{89}{28}:\dfrac{267}{56}}=\dfrac{4}{3}:\dfrac{2}{3}=2\)
\(B=\dfrac{\dfrac{6}{5}:\left(\dfrac{6}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{2}{5}}=2\)
Do đó: A=B
a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
a) Ta có: \(\dfrac{39}{-65}=\dfrac{-39}{65}=\dfrac{-39:13}{65:13}=\dfrac{-3}{5}\)
\(\dfrac{-3}{5}=\dfrac{-3}{5}\)
Do đó: \(\dfrac{-3}{5}=\dfrac{39}{-65}\)
b) Ta có: \(\dfrac{-9}{27}=\dfrac{-9:9}{27:9}=\dfrac{-1}{3}\)
\(\dfrac{-41}{123}=\dfrac{-41:41}{123:41}=\dfrac{-1}{3}\)
Do đó: \(\dfrac{-9}{27}=\dfrac{-41}{123}\)
c) Ta có: \(\dfrac{-3}{4}=\dfrac{-3\cdot5}{4\cdot5}=\dfrac{-15}{20}\)
\(\dfrac{4}{-5}=\dfrac{-4}{5}=\dfrac{-4\cdot4}{5\cdot4}=\dfrac{-16}{20}\)
mà \(\dfrac{-15}{20}>\dfrac{-16}{20}\)
nên \(\dfrac{-3}{4}>\dfrac{4}{-5}\)
d) Ta có: \(\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot7}{3\cdot7}=\dfrac{-14}{21}\)
\(\dfrac{-5}{7}=\dfrac{-5\cdot3}{7\cdot3}=\dfrac{-15}{21}\)
mà \(\dfrac{-14}{21}>\dfrac{-15}{21}\)
nên \(\dfrac{2}{-3}>\dfrac{-5}{7}\)
a:=>0,75x-x+1,25x=0,2
=>x=0,2
b: =>1/3-x=-3/6+4/6=1/6
=>x=1/3-1/6=1/6
c: =>(x-1)/45=-6/30=-1/5
=>x-1=-9
=>x=-8
d: =>(2/5x-1)=-5/7
=>2/5x=2/7
=>x=5/7
Câu 1:
\(\Rightarrow \left[\begin{array}{} x+\frac{1}{2}=0\\ \frac{2}{3}-2x=0 \end{array} \right.\)
\(\Leftrightarrow \left[\begin{array}{} x=\frac{-1}{2}\\ x=\frac{1}{3} \end{array} \right.\)
Vậy phương trình có tập nghiệm S={\(\frac{-1}{2};\frac{1}{3}\)}
Câu 2:
\(\Rightarrow \left[\begin{array}{} 3x-10=0\\ 5-\frac{1}{2}x=0 \end{array} \right.\)
\(\Leftrightarrow \left[\begin{array}{} x-=\frac{10}{3}\\ x=10 \end{array} \right.\)
Vậy phương trình có tập nghiệm S={\(10;\frac{10}{3}\)}
Câu 3:
\(\Leftrightarrow \frac{1}{3}x=\frac{65}{4}-\frac{53}{4}\)
\( \Leftrightarrow \frac{1}{3}x=\frac{12}{4}\)
\(\Leftrightarrow x=9\)
Vậy phương trình có tập nghiệm S={9}
Câu 4:
\(\Leftrightarrow \frac{2}{3}x=\frac{2}{3}\)
\(\Leftrightarrow x=1\)
Vậy phương trình có tập nghiệm S={1}
Câu 5:
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x(x+1)}=\frac{2010}{2011}\)
\(\Leftrightarrow 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)
\(\Leftrightarrow 1-\frac{1}{x+1}=\frac{2010}{2011}\)
\(\Leftrightarrow \frac{x}{x+1}=\frac{2010}{2011}\)
\(\Rightarrow 2010x+2010=2011x\)
\(\Leftrightarrow x=2010\)
Vậy phương trình có tập nghiệm S={2010}
cảm ơn bạn Hoàng Bình Bảo nha nhưng mà đây là toán lớp 6 mà bạn
`A=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
Ai trả lời nhanh và đúng nhất thì mình sẽ tick ✅, mình chỉ chọn 3 người nha
a) \(\dfrac{-2}{4}+\dfrac{56}{89} =\dfrac{-1}{2}+\dfrac{56}{89} =\dfrac{-89}{178}+\dfrac{112}{178} =\dfrac{23}{178}\)
b) \(\dfrac{-2^2}{78}+\dfrac{65}{45}-\dfrac{5}{4}=\dfrac{-4}{78}+\dfrac{13}{9}-\dfrac{5}{4}=\dfrac{-2}{39}+\dfrac{52}{36}-\dfrac{45}{36}=\dfrac{-2}{39}+\dfrac{7}{36}=\dfrac{-24}{468}+\dfrac{91}{468}=\dfrac{67}{468}\)