Cho a/b = c/d, biết c = 3d, c + d = 16, 2a^2 - 3ab = 9. Tính a,b,c,d.
Giúp mik nhé, mik cần gấp
tính a,b,c ,d biết a/b=c/d , c+d =16 và 2a^ 2 -3ab=9,c =3d
. Help Me ! Please :)) Mik đang gấp lắm nhé nên nếu các bạn biết thì giải giúp mik nhé :3 Cảm ơn nhiều nhiều lắm nek ~~~ Bạn nào làm đúng mik sẽ tik nhé =))
Đề bài: Cho a/b = c/d . C/minh rằng
a) 5a-3b/5a+3b = 5c-3d/5c+3d
b) 5a+11b/6a-5b = 5c + 11d/6c-5d
c) 2a^2 - 3b^2/2c^2 - 3d^2 = ab/cd
c) 22/5 + 51/9 + 11/4 + 3/5 + 1/3 + 1/4
= 22/5 +3/5 +51/9 + 1/3 +11/4+1/4
= (22/5 +3/5) +(51/9 + 3/9) +(11/4+1/4)
= 25/5 +54/9 +12/4
= 5 +6 +3
= 14
d) (1/6 + 1/10 + 1/15) : (1/6 + 1/10 - 1/15)
= (5/30 + 3/30 +2/30 ) :(5/30 +3/30 -2/30)
= 10/30 : 6/30
= 1/3 : 1/5
= 5/3
Cảm ơn pn Bexiu ^^ Nhưng đây là c/m mà bn ;) ;) Có phải tính đâu =)) Nhưng ko sao ah :3 Cảm ơn pn đã giúp <3
Giải giúp mik mấy bài này nhé ! Mơn các bn nhìu ! Cần gấp lắm !
Bài 1: a)Cho: a/c=c/b. Chứng minh rằng: a^2+c^2 / b^2+c^2=a/b
b)Cho: a/b=b/c=c/d. Chứng inh rằng: (a+b+c/b+c+d)^3=a/d
c)Tìm A biết rằng A=a/b+c=c/a+b=b/c+a
Bài 2:Chứng minh rằng nếu: a/b=c/d thì:
a) 5a+3b/3a-3b=5c+3d/5c-3d
b) 7a^2+3ab/11a^2-8b^2=7c^2+3cd/11c^2-8d^2
Bài 3 : Cho x/2=y/3=z/5 . Tìm x,y,z biết : x-2y+3z=22
Bài 4 : Cho 3x=2y;7y=5z . Tìm x,y,z biết :x-y+z=32
Bài 5 : Cho a/b=c/d (a,b,c,d ∈ Q*) CMR : 7a^2+3ab/11a^2-8b^2 = 7c^2+3cd/11a^2-8d^ .
Giúp mik vs mik cần gấp
Cho\(\frac{a}{b}=\frac{c}{d}\). Chứng minh rằng: a) \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
b) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) c) \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
Mấy bạn ơi, giúp mik vs ạ, à ko, các bật thánh nhân ơi, giúp cháu với ạ, cháu cần gấp !!!
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Ta có:
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7b^2k^2+3\cdot bk\cdot b}{11b^2k^2-8b^2}=\frac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\left(1\right)\)
\(\frac{7c^2+3cd}{11c^2-8d^2}=\frac{7d^2k^2+3dk\cdot d}{11d^2k^2-8d^2}=\frac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrowđpcm\)
Mấy bài khác tương tự
cho a/b=c/d CMR
a, a-b/a+b = c-d/c+d
b, 2a+3b/2c+3d = 4a-5b/4c-5d
AI GIÚP MIK VS Ạ
\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\)
a)\(\frac{a-b}{a+b}=\frac{c-d}{c+d}\)
\(\Leftrightarrow\left(a-b\right)\left(c+d\right)=\left(c-d\right)\left(a+b\right)\)
\(\Leftrightarrow ac-bc+ad-bd=ac-ad+bc-bd\)
\(\text{Thay }ad=bc\text{ vào}\Rightarrow ac-ad+ad-bd=ac-ad+ad-bd\)
\(\text{Đây là đẳng thức đúng }\Rightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\text{ là đúng }\)
b)\(\text{Tương tự*}\)
a) \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{b}+1=\frac{c}{d}+1\Leftrightarrow\frac{a+b}{b}=\frac{c+d}{d}\Leftrightarrow\frac{b}{a+b}=\frac{d}{c+d}\)
\(\Leftrightarrow\frac{-2b}{a+b}+1=\frac{-2d}{c+d}+1\Leftrightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\)
b) \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{4a}{b}-5=\frac{4c}{d}-5\Leftrightarrow\frac{4a-5b}{b}=\frac{4c-5d}{d}\Leftrightarrow\frac{b}{4a-5b}=\frac{d}{4c-5d}\)
\(\Leftrightarrow\frac{11b}{4a-5b}+1=\frac{11d}{4c-5d}+1\Leftrightarrow\frac{4a+6b}{4a-5b}=\frac{4c+6d}{4c-5d}\Leftrightarrow\frac{2a+3b}{4a-5b}=\frac{2c+3d}{4c-5d}\)
\(\Leftrightarrow\frac{2a+3b}{2c+3d}=\frac{4a-5b}{4c-5d}\)
1) Rút gọn biểu thức
a)A=3x-| 2x+5| -7
b)B=| x+5| -| 2x+7| -4x
2) Tìm x
a)| x+4| +| x-2| =6
b)| 4x+3| -x=5
3) Cho tỉ lệ thúc a/b=c/d. CM rằng ta có các tỉ lệ thức sau
a)\(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
b)\(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
HELP ME. Mai15/8 18:30 mik đi học rồi. Các bạn giúp mik với mik sẽ tick cho các bạn
\(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^2=\left[\dfrac{\left(bk+b\right)}{\left(dk+d\right)}\right]^2=\left[\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right]^2=\dfrac{b^2}{d^2}\)
\(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k+1\right)}{d^2\left(k+1\right)}=\dfrac{b^2}{d^2}\)
Vậy...
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)
Vậy...
cho a/b=c/d chứng minh tỉ lệ thức bằng nhau
a, ( b+d ) c = ( a+c ) d
b, ( 2x - c ) ( 2b + d) = ( 2b - d ) ( 2a + c )
c , ( 3a + 5 c ) ( b - 3d ) = ( 3b + 5d ) ( a - 3c)
mn giúp mình với ạ ! mình đang cần gấp
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
\(\Rightarrow\left(3a+5c\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)
a)cho a+b+c=2015. và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{5}\)tính A=\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
b) cho \(\frac{a}{b}=\frac{c}{d}\). CMR \(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2c^2-3cd+3d^2}{2c^2+3cd}\)
giúp mình với mình tick cho
a) \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{5}\)
\(\Leftrightarrow\frac{2015}{a+b}+\frac{2015}{b+c}+\frac{2015}{c+a}=403\)
\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=403\)
\(\Leftrightarrow3+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=403\)
\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=400\)
b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)
Thay vào rồi c/m nhé
a) Từ đẳng thức : \(A=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(\Rightarrow A+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(\Rightarrow A+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(\Rightarrow A+3=\left(a+b+c\right).\frac{1}{b+c}+\left(a+b+c\right).\frac{1}{a+c}+\left(a+b+c\right).\frac{1}{a+b}\)
\(\Rightarrow A+3=\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow A+3=2015.\frac{1}{5}\)
\(\Rightarrow A+3=403\)
\(\Rightarrow A=400\)
Vậy A = 400
b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Khi đó : \(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2\left(bk\right)^2-3b^2k+5b^2}{2\left(bk\right)^2+3b^2k}=\frac{2k^2b^2-3b^2k+5b^2}{2b^2k^2+3b^2k}=\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2k^2+3k\right)}\)
\(=\frac{2k^2-3k+5}{2k^2+3k}\left(1\right)\);
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2\left(dk\right)^2-3d^2k+5d^2}{2\left(dk\right)^2+3d^2k}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2k^2+3d^2k}=\frac{d^2.\left(2k^2-3k+5\right)}{d^2\left(2k^2+3k\right)}\)
\(=\frac{2k^2-3k+5}{2k^2+3k}\left(2\right)\)
Từ (1) và (2) => \(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2c^2-3cd+5d^2}{2c^2+3cd}\)(đpcm)