Cho tỉ lệ thức: \(\frac{a}{b}=\frac{c}{d}\). Chứng minh rằng rằng \(\frac{2012a+2013b}{2012a-2013b}=\frac{2012c+2013d}{2012c-2013d}\)
cho tỉ lệ thức a/b=c/d.chứng minh rằng 2012a+2013b/2012a-2013b=2012c+2013d/2012c2013d
Giải:
Ta có : a/b = c/d => a/c = b/d
Đặt a/c = b/d = k => a = ck ; b = dk
Khi đó, ta có : \(\frac{2012.ck+2013.dk}{2012.ck-2013.dk}=\frac{\left(2012c+2013d\right).k}{\left(2012c-2013d\right).k}=\frac{2012c+2013d}{2012c-2013d}\)(đpcm)
Cho a/b=c/d.CM 2010a+2011b/2010c+2011d=2012a-2013b/2012c-2013d
Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\) (a, b, c, d > 0). Tính:
A=\(\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)
cho\(\frac{a}{2b}\)=\(\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\)(a, b, c, d > 0). Tính:
A=\(\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)
Vì a ; b ; c ; d > 0
=> a + b + c + d > 0
=> 2(a + b + c + d) > 0
=> 2a + 2b + 2c + 2d > 0
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)
=> \(\frac{a}{2b}=\frac{1}{2}\Rightarrow2a=2b\Rightarrow a=b\)
Tương tự,ta được a = b = c = d
Khi đó A = \(\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)
= \(\frac{2013a-2012a}{2a}+\frac{2013b-2012b}{2b}+\frac{2013c-2012c}{2c}+\frac{2013d-2012d}{2d}\)(Vì a = b = c = d)
= \(\frac{a}{2a}+\frac{b}{2b}+\frac{c}{2c}+\frac{d}{2d}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
\(a,b,c,d>0\text{ nên : }a+b+c+d>0\Rightarrow\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)
do đó: a=b=c=d hay A=1/2+1/2+1/2+1/2=2
Tìm x biết
a)\(||3x-\frac{7}{3}|-2|=7\)
b) Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\)(a, b, c, d > 0). Tính
A = \(\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)
Bài giải
a, \(\left| |3x-\frac{7}{3} | -2\right|=7\)
\(\Rightarrow\orbr{\begin{cases}|3x-\frac{7}{3}|-2=-7\\|3x-\frac{7}{3}|-2=7\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}|3x-\frac{7}{3}|=-5\text{ ( loại) }\\|3x-\frac{7}{3}|=9\end{cases}}\) \(\Rightarrow\text{ }\left|3x-\frac{7}{3}\right|=9\) \(\Rightarrow\orbr{\begin{cases}3x-\frac{7}{3}=-9\\3x-\frac{7}{3}=9\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}3x=\frac{-20}{3}\\3x=\frac{34}{3}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-\frac{20}{9}\\x=\frac{34}{9}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-\frac{20}{9}\text{ ; }\frac{34}{9}\right\}\)
Cho\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\left(a,b,c,d>0\right)\)
Tính A=\(\dfrac{2013a-2012b}{c+d}+\dfrac{2013b-2012c}{a+d}+\dfrac{2013c-2012d}{a+b}+\dfrac{2013d-2012a}{b+c}\)
CHO \(\frac{a}{2b}\)=\(\frac{b}{2c}\)=\(\frac{c}{2d}\)=\(\frac{d}{2a}\) (a,b,c,d > 0).TÍNH:
A=\(\frac{2013a-2012b}{c+d}\)+\(\frac{2013b-2012c}{a+d}\)+\(\frac{2013c-2012d}{a+b}\)+\(\frac{2013d-2012a}{b+c}\)
Cho dãy tỉ số bằng nhau : \(\frac{2012a+b+c+d}{a}=\frac{a+2012b+c+d}{b}=\frac{a+b+2012c+d}{c}=\frac{a+b+c+2012d}{d}\).Tính giá trị biểu thức: M=\(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
Vào câu hỏi tương tự nhé bạn, tham khảo link này :
https://olm.vn/hoi-dap/detail/94049096720.html
họ bảo ko có đường dẫn
Cho a, b, c là các số không âm thỏa mãn a + b + c = 1006. Chứng minh rằng:
\(\sqrt{2012a+\frac{\left(b-c\right)^2}{2}}+\sqrt{2012b+\frac{\left(c-a\right)^2}{2}}+\sqrt{2012c+\frac{\left(a-b\right)^2}{2}}\le2012\sqrt{2}\)
\(\sqrt{2012a+\frac{\left(b-c\right)^2}{2}}=\sqrt{2a\left(a+b+c\right)+\frac{\left(b-c\right)^2}{2}}\)
\(=\sqrt{\frac{4a^2+4ab+4ac+b^2+c^2-2bc}{2}}=\sqrt{\frac{\left(2a+b+c\right)^2-4bc}{2}}\le\sqrt{\frac{\left(2a+b+c\right)^2}{2}}=\frac{1}{\sqrt{2}}\left(2a+b+c\right)\)
Tương tự:
\(\sqrt{2012b+\frac{\left(c-a\right)^2}{2}}\le\frac{1}{\sqrt{2}}\left(a+2b+c\right)\) ; \(\sqrt{2012c+\frac{\left(a-b\right)^2}{2}}\le\frac{1}{\sqrt{2}}\left(a+b+2c\right)\)
Cộng vế với vế:
\(VT\le\frac{1}{\sqrt{2}}\left(4a+4b+4c\right)=2\sqrt{2}\left(a+b+c\right)=2012\sqrt{2}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1006;0;0\right)\) và hoán vị