Dương Thanh Trúc
Giới thiệu về bản thân
a, tam giac ABC cân tại A nên B^=C^
=>BAM
BAM^=CAN
kết hợp(1);(2);(3)ta có tam giác BAM= tam giác CAN(g-c
b,BM=CN(tam giac BAM=tam giacCA
BM=BN+MN=MN+MC
=>BN=CM
c,BAN^+NAC^=BAC^=120⁰
=>BAN^=120⁰-NAC^=120⁰-90⁰=30⁰
ABN^=(180⁰-120⁰:2=30⁰
=>BAN^ =ABN^=30⁰=tan giacANB can tai N
a, tam giac ABC cân tại A nên B^=C^
=>BAM
BAM^=CAN
kết hợp(1);(2);(3)ta có tam giác BAM= tam giác CAN(g-c
b,BM=CN(tam giac BAM=tam giacCA
BM=BN+MN=MN+MC
=>BN=CM
c,BAN^+NAC^=BAC^=120⁰
=>BAN^=120⁰-NAC^=120⁰-90⁰=30⁰
ABN^=(180⁰-120⁰:2=30⁰
=>BAN^ =ABN^=30⁰=tan giacANB can tai N
Goi so may va thoi gian hoan thanh cong viec la2 dai luong ti le nghich
Nen ta co:a.2=b.4=c6
=> 2a/48=4b/48=6c/48
=>a/24=b/12=c_8
Theo bai ra ta co:
a+b+c= 33
ap dung tinh chat day ti so bang nhau
a/24=b/12=c/8=a+b+c/24+12+8=33/44=3/4
=>3/4.24=18
3/4.12=9
3/4.8=6
vay doi1:18 may
Doi 2:may
Doi 3:6 may
Goi so may va thoi gian hoan thanh cong viec la2 dai luong ti le nghich
Nen ta co:a.2=b.4=c6
=> 2a/48=4b/48=6c/48
=>a/24=b/12=c_8
Theo bai ra ta co:
a+b+c= 33
ap dung tinh chat day ti so bang nhau
a/24=b/12=c/8=a+b+c/24+12+8=33/44=3/4
=>3/4.24=18
3/4.12=9
3/4.8=6
vay doi1:18 may
Doi 2:may
Doi 3:6 may
Goi so may va thoi gian hoan thanh cong viec la2 dai luong ti le nghich
Nen ta co:a.2=b.4=c6
=> 2a/48=4b/48=6c/48
=>a/24=b/12=c_8
Theo bai ra ta co:
a+b+c= 33
ap dung tinh chat day ti so bang nhau
a/24=b/12=c/8=a+b+c/24+12+8=33/44=3/4
=>3/4.24=18
3/4.12=9
3/4.8=6
vay doi1:18 may
Doi 2:may
Doi 3:6 may
P(x)-Q(x)=-X³-2X²-2X+5;b)thayx=1vaoP(x);Q(x)
Ta co: P(1)=1³-3.1²+1+1=0;Q(1)=2.1³-1²+3.1-4=0; vay1la nghiem cua ca hai da thucP(x)va Q(x)
P(x)-Q(x)=-X³-2X²-2X+5;b)thayx=1vaoP(x);Q(x)
Ta co: P(1)=1³-3.1²+1+1=0;Q(1)=2.1³-1²+3.1-4=0; vay1la nghiem cua ca hai da thucP(x)va Q(x)
P(x)-Q(x)=-X³-2X²-2X+5;b)thayx=1vaoP(x);Q(x)
Ta co: P(1)=1³-3.1²+1+1=0;Q(1)=2.1³-1²+3.1-4=0; vay1la nghiem cua ca hai da thucP(x)va Q(x)
P(x)-Q(x)=-X³-2X²-2X+5;b)thayx=1vaoP(x);Q(x)
Ta co: P(1)=1³-3.1²+1+1=0;Q(1)=2.1³-1²+3.1-4=0; vay1la nghiem cua ca hai da thucP(x)va Q(x)
P(x)-Q(x)=-X³-2X²-2X+5;b)thayx=1vaoP(x);Q(x)
Ta co: P(1)=1³-3.1²+1+1=0;Q(1)=2.1³-1²+3.1-4=0; vay1la nghiem cua ca hai da thucP(x)va Q(x)