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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
\(x-\dfrac{1}{2}.0,5=1,15\\ \Rightarrow x-\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{23}{20}\\ \Rightarrow x-\dfrac{1}{4}=\dfrac{23}{20}\\ \Rightarrow x=\dfrac{23}{20}+\dfrac{1}{4}\\ \Rightarrow x=\dfrac{7}{5}\)
\(x-\dfrac{1}{2}.0,5=1,15\)
\(x-\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{115}{100}\)
\(x-\dfrac{1}{4}\) \(=\dfrac{23}{20}\)
\(x\) \(=\dfrac{23}{20}+\dfrac{1}{4}=\dfrac{7}{5}\)
\(=\dfrac{3}{5}+\dfrac{-17}{45}=\dfrac{27-17}{45}=\dfrac{10}{45}=\dfrac{2}{9}\)
\(=\dfrac{\dfrac{8}{8}-\dfrac{8}{7}+\dfrac{8}{11}+\dfrac{8}{15}}{\dfrac{7}{8}-\dfrac{7}{7}+\dfrac{7}{11}+\dfrac{7}{15}}=\dfrac{8}{7}\)
\(\dfrac{1-1\dfrac{1}{7}+\dfrac{8}{11}+\dfrac{8}{15}}{0,875-1+\dfrac{7}{11}+\dfrac{7}{15}}\)
\(=\dfrac{1-\dfrac{8}{7}+\dfrac{8}{11}+\dfrac{8}{15}}{\dfrac{875}{1000}-1+\dfrac{7}{11}+\dfrac{7}{15}}\)
\(=\dfrac{1-\dfrac{8}{7}+\dfrac{8}{11}+\dfrac{8}{15}}{\dfrac{7}{8}-1+\dfrac{7}{11}+\dfrac{7}{15}}\)
\(=\dfrac{\dfrac{8}{8}-\dfrac{8}{7}+\dfrac{8}{11}+\dfrac{8}{15}}{\dfrac{7}{8}-\dfrac{7}{7}+\dfrac{7}{11}+\dfrac{7}{15}}\)
\(=\dfrac{8.\left(\dfrac{1}{8}-\dfrac{1}{7}+\dfrac{1}{11}+\dfrac{1}{15}\right)}{7.\left(\dfrac{1}{8}-\dfrac{1}{7}+\dfrac{1}{11}+\dfrac{1}{15}\right)}\)
\(=\dfrac{8}{7}\)
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
`@` `\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.`
(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