A=2a-5b/a-3b - 4a+b/8a-2b
biết a/b =3/4
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19 tháng 12 2021
\(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{2a-5b}{-14}=\dfrac{a-3b}{-9}=\dfrac{4a+b}{16}=\dfrac{8a-2b}{16}\\ \Leftrightarrow A=\dfrac{-14}{-9}-\dfrac{16}{16}=\dfrac{14}{9}-1=\dfrac{5}{9}\)
11 tháng 11 2023
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
TA
5
22 tháng 4 2017
a, A = 120 độ ; B = 60 độ
b, A = 108 độ ; B = 72 độ
c, A = 144 độ ; B = 36 độ
d, A = 100 độ ; B = 80 độ
e, A = 100 độ ; B = 80 độ
22 tháng 4 2017
a) Vi A = 2B nen B=1/3 180 do
B=180 do * 1/3 =60 do
A=180 do - 60 do = 120 do
b)Vi 2A = 3B nen ta co B=2/3 A
suy ra B=2/5 180 do
B=180 do * 2/5 = 72 do
A= 180 do - 72 do =126 do
c)Vi A=4B nen B=1/5 180 do
B=180 do * 1/5 = 36 do
A= 180-36=162 do
d)Vi 4A = 5B nen ta co B=4/5 A
suy ra B = 4/9 180 do
B=180 do * 4/9 =80 do
A= 180 do - 80 do = 100 do
e) B=(180 do - 20 do ):2=80 do
A= 180 do - 80 do = 100 do
d/s:a)A=120 do
B=60 do
b)A= 136 do
B= 72 do
c)A=153 do
B= 36 do
d)A=100 do
B=80 do
e)A=100 do
B= 80 do
nho k cho minh voi nha
EW
1
29 tháng 12 2017
phá ngoặc lun nà
+4a-5c+3b-2b+a-7c-7b+3c-5a=(4a+a-5a)+(3b-2b-7b)+(-5c-7c+3c)=0-6b-9c=-9c-6b
-2a+3c-b-5b-4c+12a+9b+4c-4a-6a-3b-3c+d=(-2a+12a-4a-6a)+(-b-5b+9b-3b)+(3c-4c+4c-3c)+d=0+0+0+0+d=d
H9
HT.Phong (9A5)
CTVHS
26 tháng 6 2023
a) \(\left(2a-b\right)\left(b+4a\right)+2a\left(b-3a\right)\)
\(=2ab+8a^2-b^2-4ab+2ab-6a^2\)
\(=\left(2ab+2ab-4ab\right)+\left(8a^2-6a^2\right)-b^2\)
\(=2a^2-b^2\)
b) \(\left(3a-2b\right).\left(2a-3b\right)-6a\left(a-b\right)\)
\(=6a^2-9ab-4ab+6b^2-6a^2+6ab\)
\(=\left(6a^2-6a^2\right)-\left(9ab+4ab-6ab\right)+6b^2\)
\(=-7ab+b^2\)
c) \(5b\left(2x-b\right)-\left(8b-x\right)\left(2x-b\right)\)
\(=10bx-5b^2-\left(16bx-8b^2-2x^2+bx\right)\)
\(=10bx-5b^2-16bx+8b^2+2x^2-bx\)
\(=\left(10bx-16bx-bx\right)-\left(5b^2-8b^2\right)+2x^2\)
\(=-7bx+3b^2+2x^2\)
d) \(2x\left(a+15x\right)+\left(x-6a\right)\left(5a+2x\right)\)
\(=2ax+30x^2+5ax+2x^2-30a^2-12ax\)
\(=\left(2ax+5ax-12ax\right)+\left(30x^2+2x^2\right)-30a^2\)
\(=-5ax+32x^2-30a^2\)
26 tháng 6 2023
a: =2ab+8a^2-b^2-4ab+2ab-6a^2
=2a^2-b^2
b: =6a^2-9ab-4ab+6b^2-6a^2+6ab
=-7ab+6b^2
c: =10bx-5b^2-16bx+8b^2+2x^2-xb
=3b^2+2x^2-7xb
d: =2xa+30x^2+5ax+2x^2-30a^2-12ax
=32x^2-30a^2-5ax
TG
1
29 tháng 3 2023
a: a>b
=>3a>3b
=>3a+5>3b+5
b: a>b
=>2a>2b
=>2a-3>2b-3>2b-4
Ta có : \(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}\\ Đặt\dfrac{a}{3}=\dfrac{b}{4}=k\Rightarrow\left\{{}\begin{matrix}a=3k\\b=4k\end{matrix}\right.\\ ThayvàoA,tacó:\)
\(A=\dfrac{2a-5b}{a-3b}-\dfrac{4a+b}{8a-2b}\\ \Leftrightarrow=\dfrac{2\cdot3k-5\cdot4k}{3k-3\cdot4k}-\dfrac{4\cdot3k+4k}{8\cdot3k-2\cdot4k}\\ =\dfrac{6k-20k}{3k-12k}-\dfrac{12k+4k}{24k-8k}\\ =\dfrac{14k}{9k}-\dfrac{16k}{16k}\\ =\dfrac{14}{9}-1\\ =\dfrac{5}{9}\)
giúp mk nha !!!(mk cần gấp)