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Bài 9:
a: Bạn chỉ cần vẽ tam giác ABC vuông tại A có AB=3cm và AC=4cm là ra cái hình rồi
Số đo góc A thì chắc chắn là 90 độ rồi
\(2^4.5-\left[31-9^2\right]=16.5-\left(31-81\right)=80-\left(-50\right)=130\)
\(2^4\).5-[1.31-(13-4)^2]
=16.5-[1.31-81]
=16.5-[31-81]
=16.5-(-50)
=80-(-50)
=130
ta nhân 3 cả hai vế, được :
\(\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{102.105}\right)x=3\)
hay
\(\left(\frac{4-1}{1.3}+\frac{7-4}{4.7}+...+\frac{105-102}{102.105}\right)x=3\) \(\Leftrightarrow\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+..+\frac{1}{102}-\frac{1}{105}\right)x=3\)
\(\Leftrightarrow\left(1-\frac{1}{105}\right)x=3\Leftrightarrow\frac{104}{105}.x=3\Leftrightarrow x=\frac{315}{104}\)
\(2-4+6-8+....+48-50\)
\(=\left(2-4\right)+\left(6-8\right)+...+\left(48-50\right)\)
\(=\left(-2\right)+\left(-2\right)+....+\left(-2\right)\)( 25 thừa số )
\(=\left(-2\right).25\)
\(=-50\)
Gọi: S=2-4+6-8+.....+48-50
=> 2S = (2-4+6-8+......+48-50)+(2-4+6-8+.........+48-50)
=(-2)+(-2)+(-2)+...........+(-2)+(-2) [25 số -2]
=(-2)x25
=-50
=> S = -25
Vậy ,,,,,,,,,,,,,,,,,,,,,,,,
a: 12h: 0 độ
10h: 60 đọ
6h: 180 độ
5h: 150 độ
b:
a: góc nhọn: góc yMz; góc tMz
b: góc vuông: góc yMt, góc xMt
c: góc tù: góc xMz
d: góc bẹt: góc xMy
\(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{19.20}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{3}-\dfrac{1}{20}< \dfrac{1}{3}\)
\(B=\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{19.20}=2\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{19.20}\right)=2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)=2\left(\dfrac{1}{3}-\dfrac{1}{20}\right)=\dfrac{2}{3}-\dfrac{2}{20}< \dfrac{2}{3}\)
\(C=\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{18.20}=\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{18.20}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{18}-\dfrac{1}{20}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{4}-\dfrac{1}{40}< \dfrac{1}{4}\)
\(D=\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{18.20}=\dfrac{1}{2}\left(\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{18.20}\right)=\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{18}-\dfrac{1}{20}\right)=\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{20}\right)=\dfrac{1}{8}-\dfrac{1}{40}< \dfrac{1}{8}\)
v: \(\dfrac{26}{-37}< 0\)
\(0< \dfrac{11}{19}\)
Do đó: \(\dfrac{26}{-37}< \dfrac{11}{19}\)
x: \(\dfrac{-7}{30}=\dfrac{-7\cdot2}{30\cdot2}=\dfrac{-14}{60}\)
\(\dfrac{13}{-20}=\dfrac{-13}{20}=\dfrac{-13\cdot3}{20\cdot3}=\dfrac{-39}{60}\)
mà -14>-39
nên \(-\dfrac{7}{30}< \dfrac{13}{-20}\)
y: \(-\dfrac{12}{16}=\dfrac{-12\cdot3}{16\cdot3}=\dfrac{-36}{48}\)
\(\dfrac{-16}{48}=\dfrac{-16}{48}\)
mà -36<-16
nên \(\dfrac{-12}{16}< -\dfrac{16}{48}\)
z: \(\dfrac{-18}{-72}=\dfrac{1}{4}>0\)
\(0>-\dfrac{5}{20}\)
Do đó: \(\dfrac{-18}{-72}>-\dfrac{5}{20}\)
z1: \(\dfrac{-36}{90}=\dfrac{-2}{5};\dfrac{-15}{25}=\dfrac{-3}{5}\)
mà -2>-3
nên \(\dfrac{-36}{90}>\dfrac{-15}{25}\)
z2: \(\dfrac{-32}{48}=\dfrac{-2}{3}=\dfrac{-4}{6}\)
\(\dfrac{-6}{12}=\dfrac{-1}{2}=\dfrac{-3}{6}\)
mà -4<-3
nên \(-\dfrac{32}{48}< -\dfrac{6}{12}\)
z3: \(\dfrac{-20}{-45}=\dfrac{4}{9}=\dfrac{40}{90}\)
\(\dfrac{35}{150}=\dfrac{7}{30}=\dfrac{21}{90}\)
mà 40>21
nên \(\dfrac{-20}{-45}>\dfrac{35}{150}\)
a: \(\Leftrightarrow9-x=6\)
hay x=3
e: \(\Leftrightarrow2^x=32\)
hay x=5