so snh
can 2+can 3 va can 10(loi giai co n ko hieu)
(can2 + can3)^2 = 2 + 2can6 + 3 = 5+2can6
(can10)^2 = 10
ta co 24<25
=> (2can6)^2<5^2
=> 2can6 < 5
=>5 + 2can 6 <10
=> (can 2 + can3)^2 < (can10)^2
=>can2 + can3 < can10
(â2 + â3)^2 = 2 + 2â6 + 3 = 5+2â6
â10^2 = 10
Ta so sánh 5+2â6 vá»i 10
hay 2â6 vá»i 5
ta thấy: (2â6)^2 = 24 còn 5^2 =25
suy ra 2â6 < 5
váºy â2 + â3 < â10
gia su can2+can3<can10
binh phuong 2 ve khong am ta co
(can2+can3)^2<10
(=)5+2can6<10
(=)2can6<5
binh phuong tiep 2 ve khong am
(=)24<25=>dung
vay dieu gia su la dung
vay can2+can3<can10
can2+can3 vs can10
bình phương 2 vế la ra thôi
2+3+2cÄn6 vs 10
2 cÄn 6 vs 5 bình phÆ°Æ¡ng nữa
4 *6 vs 25
24<25
Lay may tinh ra bam:
can 2+can 3=3,15
can 10=3,16
Ta co: 3,15 < 3,16
=>can 2+can 3 < can 10
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Answers & Comments
Verified answer
(can2 + can3)^2 = 2 + 2can6 + 3 = 5+2can6
(can10)^2 = 10
ta co 24<25
=> (2can6)^2<5^2
=> 2can6 < 5
=>5 + 2can 6 <10
=> (can 2 + can3)^2 < (can10)^2
=>can2 + can3 < can10
(â2 + â3)^2 = 2 + 2â6 + 3 = 5+2â6
â10^2 = 10
Ta so sánh 5+2â6 vá»i 10
hay 2â6 vá»i 5
ta thấy: (2â6)^2 = 24 còn 5^2 =25
suy ra 2â6 < 5
váºy â2 + â3 < â10
gia su can2+can3<can10
binh phuong 2 ve khong am ta co
(can2+can3)^2<10
(=)5+2can6<10
(=)2can6<5
binh phuong tiep 2 ve khong am
(=)24<25=>dung
vay dieu gia su la dung
vay can2+can3<can10
can2+can3 vs can10
bình phương 2 vế la ra thôi
2+3+2cÄn6 vs 10
2 cÄn 6 vs 5 bình phÆ°Æ¡ng nữa
4 *6 vs 25
24<25
Lay may tinh ra bam:
can 2+can 3=3,15
can 10=3,16
Ta co: 3,15 < 3,16
=>can 2+can 3 < can 10