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giup mik voi mik cam on!
11 tháng 3 2022
a: Xét ΔBAE và ΔBDE có
BA=BD
\(\widehat{ABE}=\widehat{DBE}\)
BE chung
Do đó: ΔBAE=ΔBDE
b: Ta có: ΔBAE=ΔBDE
nên \(\widehat{BAE}=\widehat{BDE}=90^0\)
hay ED\(\perp\)BC
c: Xét ΔAKE vuông tại A và ΔDCE vuông tại D có
EA=ED
\(\widehat{AEK}=\widehat{DEC}\)
Do đó: ΔAKE=ΔDCE
Suy ra: EK=EC
hay ΔEKC cân tại E
23 tháng 12 2021
\(x-\dfrac{1}{2}.0,5=1,15\\ \Rightarrow x-\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{23}{20}\\ \Rightarrow x-\dfrac{1}{4}=\dfrac{23}{20}\\ \Rightarrow x=\dfrac{23}{20}+\dfrac{1}{4}\\ \Rightarrow x=\dfrac{7}{5}\)
23 tháng 12 2021
\(x-\dfrac{1}{2}.0,5=1,15\)
\(x-\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{115}{100}\)
\(x-\dfrac{1}{4}\) \(=\dfrac{23}{20}\)
\(x\) \(=\dfrac{23}{20}+\dfrac{1}{4}=\dfrac{7}{5}\)
11 tháng 1 2022
\(=\dfrac{3}{5}+\dfrac{-17}{45}=\dfrac{27-17}{45}=\dfrac{10}{45}=\dfrac{2}{9}\)
11 tháng 1 2022
\(=\dfrac{\dfrac{8}{8}-\dfrac{8}{7}+\dfrac{8}{11}+\dfrac{8}{15}}{\dfrac{7}{8}-\dfrac{7}{7}+\dfrac{7}{11}+\dfrac{7}{15}}=\dfrac{8}{7}\)
12 tháng 1 2022
\(\dfrac{1-1\dfrac{1}{7}+\dfrac{8}{11}+\dfrac{8}{15}}{0,875-1+\dfrac{7}{11}+\dfrac{7}{15}}\)
\(=\dfrac{1-\dfrac{8}{7}+\dfrac{8}{11}+\dfrac{8}{15}}{\dfrac{875}{1000}-1+\dfrac{7}{11}+\dfrac{7}{15}}\)
\(=\dfrac{1-\dfrac{8}{7}+\dfrac{8}{11}+\dfrac{8}{15}}{\dfrac{7}{8}-1+\dfrac{7}{11}+\dfrac{7}{15}}\)
\(=\dfrac{\dfrac{8}{8}-\dfrac{8}{7}+\dfrac{8}{11}+\dfrac{8}{15}}{\dfrac{7}{8}-\dfrac{7}{7}+\dfrac{7}{11}+\dfrac{7}{15}}\)
\(=\dfrac{8.\left(\dfrac{1}{8}-\dfrac{1}{7}+\dfrac{1}{11}+\dfrac{1}{15}\right)}{7.\left(\dfrac{1}{8}-\dfrac{1}{7}+\dfrac{1}{11}+\dfrac{1}{15}\right)}\)
\(=\dfrac{8}{7}\)
VT
giup mik cau nay voi a
2
cac bn giup mik voi
11 tháng 7 2023
1:
a: =-1/21-3/21=-4/21
b: =-20/36-3/36=-23/36
d: =12/3-2/3=10/3
e: =10/7-4+3/14
=20/14+3/14-4
=23/14-56/14=-33/14
2:
a: =>x-3/5=2/3
=>x=2/3+3/5=19/15
b: =>3/8-x=7/5+1/8
=>3/8-x=61/40
=>x=-23/20
d: =>-37/30-x=2
=>x=-37/30-2=-97/30
cac ban giup mik voi a
5 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`-3x - 3/4 = 6/5`
`=> -3x = 6/5 + 3/4`
`=> -3x = 39/20`
`=> x = 39/20 \div (-3)`
`=> x = -13/20`
Vậy, `x=-13/20`
`b)`
`1/7 - 3/5x = 3/5`
`=> 3/5x = 1/7 - 3/5`
`=>3/5x = -16/35`
`=> x = -16/35 \div 3/5`
`=> x = -16/21`
`c)`
`3/7 - 1/2x = 5/3`
`=> 1/2x = 3/7 - 5/3`
`=> 1/2x = -26/21`
`=> x = -26/21 \div 1/2`
`=> x = -52/21`
Vậy, `x = -52/21`
`d)`
`-2/3x + 2 = 3/4`
`=> -2/3x = 3/4 - 2`
`=> -2/3x = -5/4`
`=> x = -5/4 \div (-2/3)`
`=> x = 15/8`
Vậy, `x=15/8.`
DT
Dang Tung
CTVHS
4 tháng 1
(2y-1)¹⁰=(2y-1)²⁰
→ (2y-1)²⁰-(2y-1)¹⁰=0
→ (2y-1)¹⁰.[(2y-1)¹⁰-1]=0
→ (2y-1)¹⁰=0 hay (2y-1)¹⁰-1=0
→ 2y-1=0 hay 2y-1=1 hay 2y-1=-1
→ y=½ hay y=1 hay y=0
Bài 6:
\(a)P=\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+...+\dfrac{2}{33\cdot37}+\dfrac{2}{37\cdot41}\\ =\dfrac{1}{2}\cdot\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{33\cdot37}+\dfrac{4}{37\cdot41}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{33}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{41}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{41}\right)\\ =\dfrac{1}{2}\cdot\dfrac{40}{41}\\ =\dfrac{20}{41}\\ b)Q=\dfrac{6}{2\cdot9}+\dfrac{6}{9\cdot16}+...+\dfrac{6}{114\cdot121}\\ =\dfrac{6}{7}\cdot\left(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{7}{114\cdot121}\right)\\ =\dfrac{6}{7}\cdot\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{114}-\dfrac{1}{121}\right)\\ =\dfrac{6}{7}\cdot\left(\dfrac{1}{2}-\dfrac{1}{121}\right)\\ =\dfrac{6}{7}\cdot\dfrac{119}{242}\\ =\dfrac{51}{121}\)
Bài 5:
a: Để A>0 thì \(\dfrac{2a-1}{-5}>0\)
=>2a-1<0
=>\(a< \dfrac{1}{2}\)
b: Để A<0 thì \(\dfrac{2a-1}{-5}< 0\)
=>2a-1>0
=>2a>1
=>\(a>\dfrac{1}{2}\)
c: Để A=0 thì \(\dfrac{2a-1}{-5}=0\)
=>2a-1=0
=>2a=1
=>\(a=\dfrac{1}{2}\)
Bài 6:
a: \(P=\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+...+\dfrac{2}{37\cdot41}\)
\(=\dfrac{2}{4}\cdot\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{37\cdot41}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{37}-\dfrac{1}{41}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{41}\right)=\dfrac{1}{2}\cdot\dfrac{40}{41}=\dfrac{20}{41}\)
b: \(Q=\dfrac{6}{2\cdot9}+\dfrac{6}{9\cdot16}+...+\dfrac{6}{114\cdot121}\)
\(=\dfrac{6}{7}\left(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{7}{114\cdot121}\right)\)
\(=\dfrac{6}{7}\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{114}-\dfrac{1}{121}\right)\)
\(=\dfrac{6}{7}\left(\dfrac{1}{2}-\dfrac{1}{121}\right)=\dfrac{6}{7}\cdot\dfrac{119}{242}=\dfrac{51}{121}\)