M
a sh q l a r
Ko‘phadlarning algebraik yig‘indisini
toping (262–267):
262.
1) 8a+(–3b+5a); 3) (6a–2b)–(5a+3b);
2) 5x–(2x–3y); 4) (4x+2)+(–x–1).
263.
1) 3x2–(4x2+2y); 3)
0,6a2–(0,5a2–0,4a);
2) 2a2–(b2–3a2); 4) 
.
264.
1) (2
)+(
);
2) (0,1c–0,4c2)–(0,1c–0,5c2);
3) (13x–11y+10z)–(–15x+10y–15z);
4) (17a+12b–14c)–(11a–10b–14c).
265.
1) (7m2–4mn–n2)–(2m2–mn+n2);
2) (5a2–11ab+8b2)+(–2b2–7a2+5ab);
3) 11ac+13bc+17b2–(10ac+10bc–3b2);
4) 41z+13az+26az2–(16z+13az–4az2).
266.
1) 
;
2) (0,3a–1,2b)+(a–b)–(1,3a–0,2b);
3) 11p3–2p2–(p3–p2)+(–5p2–3p3);
4) 5x2+6x3+(x3–x2)–(–2x3+4x2).
267.
1) (–2x3+xy2)+(x2y–1)+(x2y–xy2+3x3);
2) (3x2+5xy+7x2y)–(5xy+3x2)–(7x2y–3x2);
3) (8a2–10ab–b2)+(–6a2+2ab–b2)–(a2–8ab+4b2);
4) (4a2–2ab–b2)–(–a2–2ab+b2)+(3a2–ab+b2).
268.
Ko‘phadlarning yig’indisi va
ayirmasini toping:
1)
0,1x2+0,02y2 va 0,17x2–0,08y2;
2)
0,1x2–0,02y2 va
–0,17x2+0,08y2;
3)
a3–0,12b3 va
0,39a3–b3;
4)
a3+0,12b3 va
–0,39a3+b3;
269.
Ko‘phadlarning yig’indisini
“ustuncha” usulda toping:
1)
3ab+a2–2b2 va 2a2–3ab;
2)
3x2+2xy–4y2 va 4y2–2xy+3x2y2–x3.
270. Ko’phadlarning
ayirmasini “ustun” usulida
toping:
1)
3a2+8a–4 va 3+8a–5a2ab;
2)
b3–3b2+4b va b+2b2+b3.
271.
1) Agar P=5a2+b,
Q = –4a2 –b bo‘lsa, P+Q ifoda nimaga teng?
2)
Agar P=2p2–3q3, Q
=2p2 –4q3 bo‘lsa, P–Q
ifoda nimaga teng?
3)
Agar A=a2–b2+ab,
B =2a2 –3ab–5b2, C = –4a2+2ab–3b2
bo‘lsa, A+B+C ni toping;
4)
Agar A=2a2–3ab+4b2, B
=3a2 +4ab–b2,
C = a2+2ab+3b2 bo‘lsa, A–B+C ni toping.
272.
Isbotlang:
1)
beshta ketma-ket natural sonning yig‘indisi 5 ga bo‘linadi;
2)
to‘rtta ketma-ket natural sonning yig‘indisi 4 ga bo‘linmaydi;
3)
to‘rtta ketma-ket toq natural sonning yig‘indisi 8 ga bo‘linadi;
4)
to‘rtta ketma-ket juft natural sonning yig‘indisi 4 ga bo‘linadi.
273.
Avtobusda n nafar yo‘lovchi bor edi. Dastlabki ikki bekatning har birida m nafardan yo‘lovchi avtobusdan tushdi,
uchuinchi bekatda esa hech kim tushmadi, lekin bir necha kishi avtobusga
chiqdi, shundan so‘ng avtobusdagi yo‘lovchilar soni k nafar bo‘ldi. Uchinchi bekatda avtobusga necha kishi chiqqan?
