Rút gọn biểu thức:
E=(-a-b+c+d)-(d+c-b-2a)
F=(a-2b-c+2d)-(3d-2c-3b+a)+15
G=a(2b-c)-b(a+c)-a(c+b)
H=(a+3b)(c-d)-(3a-d)(b+c)-2c(b-a)+2b(a+d)
help me!!!!!!!!!!!!!!!!!!!!!!
mình đang gấp giúp mình với
GIẢI ĐẦY ĐỦ NHA
ai làm đúng đầy đủ mình cho 1 tick
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{3a+2b}{a}=\dfrac{3bk+2b}{bk}=\dfrac{3k+2}{k}\)
\(\dfrac{3c+2d}{c}=\dfrac{3dk+2d}{dk}=\dfrac{3k+2}{k}\)
Do đó: \(\dfrac{3a+2b}{a}=\dfrac{3c+2d}{c}\)
b: \(\dfrac{2a-3b}{b}=\dfrac{2bk-3b}{b}=2k-3\)
\(\dfrac{2c-3d}{d}=\dfrac{2dk-3d}{d}=2k-3\)
Do đó: \(\dfrac{2a-3b}{b}=\dfrac{2c-3d}{d}\)
c: \(\dfrac{a}{a-2b}=\dfrac{bk}{bk-2b}=\dfrac{k}{k-2}\)
\(\dfrac{c}{c-2d}=\dfrac{dk}{dk-2d}=\dfrac{k}{k-2}\)
Do đó: \(\dfrac{a}{a-2b}=\dfrac{c}{c-2d}\)
a) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\left(b+d\right)c=\left(a+c\right)d\)
\(\Rightarrow dpcm\)
b) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{2a}{2b}=\dfrac{c}{d}=\dfrac{2a+c}{2b+d}=\dfrac{2a-c}{2b-d}\)
\(\Rightarrow\left(2b-d\right)\left(2a+c\right)=\left(2a-c\right)\left(2b+d\right)\)
\(\Rightarrow dpcm\)
c) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3c}{3d}=\dfrac{3a}{3b}=\dfrac{5c}{5d}=\dfrac{3a+5c}{3b+5d}=\dfrac{a-3c}{b-3d}\)
\(\Rightarrow\left(b-3d\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)
\(\Rightarrow dpcm\)
Đính chính câu c
\(\Rightarrow\left(3a+5c\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)
`#3107.101107`
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc\)
Ta có:
\(\dfrac{3b}{a}=\dfrac{3d}{c}\Rightarrow3bc=3da\Rightarrow bc=da\)
Vậy, từ tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) ta có thể suy ra tỉ lệ thức \(\dfrac{3b}{a}=\dfrac{3d}{c}\)
\(\Rightarrow B.\)
A =(a+b-2c) -(-a+b+c) -(2a-b-c)
= a+b-2c+a-b-c-2a+b+c
= b-2c
B=-(2a-b+c) + (b-2c-3a) -(-5a-3c+b)
= -2a+b-c+b-2c-3a+5a+3c-b
= b-c
C=(3a-b-2c)-( 2b+3c-a) +(2a-3b)
= a-b-2c-2b-3c+a+2a-3b
= -6b-5c
D=(5a-3b+c) +( 2a-3b+5) -( b-c+a)
= 5a-3b+c+2a-3b+5-b+c-a
= 6a-7b+2c
\(A=\left(a+b-2c\right)-\left(-a+b+c\right)-\left(2a-b-c\right)\)
\(=a+b-2c+a-b-c-2a+b+c=b-2c\)
\(B=-\left(2a-b+c\right)+\left(b-2c-3a\right)-\left(-5a-3c+b\right)\)
\(=-2a+b-c+b-2c-3a+5a+3c-b=b\)
\(C=\left(3a-b-2c\right)-\left(2b+3c-a\right)+\left(2a-3b\right)\)
\(=3a-b-2c-2b-3c+a+2a-3b=6a-6b-5c\)
\(D=\left(5a-3b+c\right)+\left(2a-3b+5\right)-\left(b-c+a\right)\)
\(=5a-3b+c+2a-3b+5-b+c-a=6a-7b+2c\)
a )\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{2a}{2c}\)
\(\frac{a-b}{c-d}=\frac{2a}{2c}\Rightarrow\frac{a-b}{2a}=\frac{c-d}{2c}\) ( đpcm)
b ) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{3a}{3c}=\frac{2b}{2d}=\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+2d}\)
\(\Rightarrow\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\) ( đpcm )
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\)
\(=\dfrac{a^2}{ab+2ac+3ad}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd}\)
\(\ge\dfrac{\left(a+b+c+d\right)^2}{4\left(ab+ad+bc+bd+ca+cd\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{3}{2}\left(a+b+c+d\right)^2}=\dfrac{2}{3}\)
*Chứng minh \(4\left(ab+ad+bc+bd+ca+cd\right)\le\dfrac{3}{2}\left(a+b+c+d\right)^2\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(a-c\right)^2+\left(c-d\right)^2\ge0\)
a.d = b.c ⇒ \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}\) = \(\dfrac{3a}{3c}=\dfrac{2b}{2d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{2a+5b}{2c+5d}\) (1)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{3a}{3c}=\dfrac{2b}{2d}=\dfrac{3a-2b}{2c-2d}\) (2)
Từ (1) và(2) ta có:
\(\dfrac{2a+5b}{2c+5d}\) = \(\dfrac{3a-2b}{3c-2d}\)(đpcm)
a.d = b.c ⇒ \(\dfrac{a}{c}=\dfrac{b}{d}\) ⇒ \(\dfrac{a.b}{c.d}\) = \(\dfrac{a^2}{c^2}\) = \(\dfrac{b^2}{d^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a.b}{c.d}=\dfrac{a^2}{c^2}\) = \(\dfrac{b^2}{d^2}\) = \(\dfrac{a^2+b^2}{c^2+d^2}\) (đpcm)
E=(-a-b+c+d)-(d+c-b-2a)
E=-a-b+c+d-d-c+b+2a
E=-a+(-)b+c+d+(-d)+(-c)+b+2a
E=-a+(-b)+c+d+(-d)+(-c)+b+2a
E=(2a-a)+(-b+b)+(-d+d)+(-c+c)=a+0+0+0=a
thanks nhiều nha ĐỨC THỊNH