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Bài 1:
a) \(\frac{16}{15}.\frac{\left(-5\right)}{14}.\frac{54}{24}.\frac{56}{21}\)
\(=\frac{4.2.2}{5.3}.\frac{\left(-5\right)}{2.7}.\frac{3.3}{4}.\frac{8}{3}\)
\(=\frac{4.2.2.\left(-5\right).3.3.8}{5.3.2.7.4.3}\)
\(=\frac{-16}{7}\)
b) \(\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{15}{21}.\frac{4}{\left(-5\right)}\)
\(=\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{5}{7}.\frac{2.2}{\left(-5\right)}\)
\(=\frac{7.\left(-5\right).5.2.2}{3.2.7.\left(-5\right)}\)
\(=\frac{10}{3}\)
Bài 2:
a) \(\frac{21}{24}.\frac{11}{9}.\frac{5}{7}=\frac{7}{8}.\frac{11}{9}.\frac{5}{7}=\frac{11.5}{8.9}=\frac{55}{72}\)
b) \(\frac{5}{23}.\frac{17}{26}+\frac{5}{23}.\frac{9}{26}\)
\(=\frac{5}{23}.\left(\frac{17}{26}+\frac{9}{26}\right)=\frac{5}{23}.1=\frac{5}{23}\)
c) \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}=\frac{3}{29}.\frac{29}{3}-\frac{1}{5}.\frac{29}{3}\)
\(=1-1\frac{14}{15}=\frac{14}{15}\)
Bài 3:
a) x/5 = 2/5
=> x =2
b) -4/x = 20/14 = 10/7
=> -4/x = 10/7
=> x.10 = (-4).7
x.10 = - 28
x= -28 :10
x= -2,8
c) 4/7 = 12/x = 12/ 21
=> 12/x = 12/21
=> x = 21
d) 3/7 = x / 21 = 9/21
=> x/21 = 9/21
=> x= 9
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
Sửa đề:
A=/x+5/+10
Ta có: /x+5/>= 0 với mọi x>=0
=> A=/x+5/+10 >= 10
=> Amin=10. Dấu "=" xảy ra <=> x+5=0<=> x=-5
Vậy...
\(\text{a) }A=\left|x+5\right|+10\)
\(\text{Vì }\left|x+5\right|\ge0\forall x\)
\(\Rightarrow A=\left|x+5\right|+10\ge10\)
\(\text{Dấu ''='' xảy ra khi :}\)
\(\left|x+5\right|=0\)
\(\Rightarrow x=-5\)
\(\text{Vậy Min}_A=10\Leftrightarrow x=-5\)
\(\text{b) }\left|3-x\right|+5\)
\(\text{Vì }\left|3-x\right|\ge0\forall x\)
\(\Rightarrow\left|3-x\right|+5\ge5\)
\(\text{Dấu ''='' xảy ra khi :}\)
\(\left|3-x\right|=0\)
\(\Rightarrow x=3\)
\(\text{Vậy Min}_B=5\Leftrightarrow x=3\)
\(\text{d) }D=\left(x+2\right)^2+15\)
\(\text{Vì ( x + 2 )}^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+15\ge15\)
\(\text{Dấu ''='' xảy ra khi :}\)
\(\left(x+2\right)^2=0\)
\(\Rightarrow x+2=0\)
\(\Rightarrow x=-2\)
bài 1 công thức
SCSH: ( cuối - đầu ) : khoảng cách + 1 = ?
Tổng: ( cuối + đầu ) . SCSH : 2 = ?
Hk tốt
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