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111matematika togarak konspekti 10 11 si
3. Mustahkamlash. Testyechiladi.
TESTLAR.
a=Ab va с+ЪЬ=0 (Ъф 0) bo‘Isa, - ni toping.
c
A)-i- ’ 3
2.
f
1
B)lf
1
C).i D)-i E) -f
^ a 2 +2a
K a(a + 1) (a + l)(a + 2) ;
8
ni soddalashtiring.
A)
B)
1
C)
8 ' 4
\ За + 16 6(a-6)
a + 6 a 2 +12a + 36j a 2 -36 a + 6
D)
1
E)
3a
2a
4 ' 8
ni soddalashtiring.
6
1
D) a+ 6 E) a-6
A) 6
( 5m
B)
C)
a + 6 ' a -6
14 m \ 5m + \ 3(m-3)
A)
m + 3 m 2 +6m + 9J m 2 -9 m + 3
3
B)3
5.
m + 3
a 2 +ab + b 2 a 2 -ab + b 2
ni soddalashtiring.
C) m-3 D) 1
E)
m-3 m + 3
A)
a 3 -b 3
2b
a +b
ni soddalashtiring.
6 .
b 2 -a 2 7 a 2 -b 2
x 3 -S x 2 -4
n \ 2ci p\ \ 2ci
2 / 2 2 / 2
a* + 2a* + 4 a* 2
a 2 -b 2
ni soddalashtiring.
b 2 -a 2
E)
a 2 -b 2
A) 4
В) 2x C) —2x D) 0
E) -4
7.
42
1 -b l + b 2
ni soddalashtiring.
C)b-\ D)
l
f a3 +b3 ^
A) 1 В) -1
8. {a 3 -3a 2 b + 3ab 2 -b 3 \a + by.
A) b 2 -a 2 В )a 2 -b 2 C) {a-bf D ){a+bf E ) a 2 +b 2
a + b
-ab
b +1
ni soddalashtiring.
E) -b -1
l
l да .+ m ni soddalashtiring.
да -да да -lJ да + 2 да -4
А)
2да-2 да 2 -4
В)
да
т-2
С)
да -4
D)
1
да + 2
Е)
2да + 1 4-да 2
Darsni yakunlash.
Uyga vazifa: test yechish tematik axborotnomalardan
Tayyorladi:
Tekshirdi: O’TIBDCE :
20 y.
Sana:
21 -mashs ( ulot
Dars mavzusi. Bir noma’lumli tenglamalar.
Dars maqsadlari: o‘quvchilarga bir noma’lumli tenglamalarni yechishni o‘rgatish, ulaming fanga qiziqishlarini oshirish.
Darsning borishi:
Tashkiliy qism.
Bir noma’lumli tenglamalar.
Bir noma’lumli tenglamalar.
ax=b ifodaga bir noma’lumli chiziqli tenglama deyiladi va uning ildizi x = - bo‘lib, u
a
uch xil ko‘rinishga ega boMishi mumkin:
a^0,b^R, x = — yechimyagona
a
a=0, Ьф 0, xe0 yechim yo‘q
<3=0, b= 0, xeR yechim cheksiz ko‘p.
misol : Tenglamani yeching. 5v + 3(3v + 7)=35.
Yechish : 5v + 9v + 21=35 => 1 4x = 35-21 => \4x = 14 x = 1.
misol : a ning qanday qiymatida ax+a=5x+8 tenglama yagona yechimga ega?
8 — <з
Yechish: ax-5x = 8-a => x{a -5) = 8- < 2 =>jc = => а ф 5.
<з-5
misol : к ning qanday qiymatida k+4x=k 2 x+2 tenglama yechimga ega emas? Yechish : 4 x-k 2 x = 2 -k => x{4- k 2 ) = 2-k: 1) 4- k 2 =0 к = ±2
2) 2-k ф 0 к ф 2 ■
Javob: k=-2.
misol : m ning qanday qiymatida 6+m 2 x=36x+m tenglama cheksiz ko‘p yechimga ega?
Yechish: m 2 x - 36x = m - 6 => x(m 2 - Зб) = m - 6':
nr -36 = 0 =>m = ± 6.
m-6 = 0 => in = 6
Javob: m=6.
3. Mustahkamlash. Test yechiladi.
TESTLAR.
(з^ + л-] : 41 = 5 tenglamani yeching.
A) 21 B)17— C)18— D)17— E) 18 —
y y 22 22 22 У 11
2. n ning qanday qiymatlarida nx+\=n+x tenglama cheksiz ko‘p yechimga ega boTadi?
А) пф! В) n=2 C) n=\ D) n=0 E) n=-2 3
3. a ning qanday qiymatlarida ax=2x+3 tenglama yechimga ega boMmaydi?
А) аФ-2 В) аФ 2 С) а= 2 D) аф\ Е) а= 0
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