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4.a
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Từ \(a\left(y+z\right)=b\left(z+x\right)\), áp dụng t/c dãy tỉ số bằng nhau ta được
\(\dfrac{z+x}{a}=\dfrac{y+z}{b}=\dfrac{z+x-y-z}{a-b}=\dfrac{x-y}{a-b}\)
\(\Rightarrow\dfrac{z+x}{a}.\dfrac{1}{c}=\dfrac{y+z}{b}.\dfrac{1}{c}=\dfrac{x-y}{c\left(a-b\right)}\)(1)
Tương tự : từ \(b\left(z+x\right)=c\left(x+y\right)\)
\(\Rightarrow\dfrac{z+x}{c}=\dfrac{x+y}{b}=\dfrac{z+x-x-y}{c-b}=\dfrac{y-z}{c-b}\)\(\Rightarrow\dfrac{z+x}{c}.\dfrac{1}{a}=\dfrac{x+y}{b}.\dfrac{1}{a}=\dfrac{y-z}{c-b}.\dfrac{1}{a}\)
\(\Rightarrow\dfrac{z+x}{ac}=\dfrac{x+y}{ab}=\dfrac{y-z}{a\left(c-b\right)}\)(2)
từ \(a\left(y+z\right)=c\left(x+y\right)\)
\(\Rightarrow\dfrac{y+z}{c}=\dfrac{x+y}{a}=\dfrac{y+z-x-y}{c-a}=\dfrac{z-x}{c-a}\)\(\Rightarrow\dfrac{y+z}{c}.\dfrac{1}{b}=\dfrac{x+y}{a}.\dfrac{1}{b}=\dfrac{z-x}{c-a}.\dfrac{1}{b}\)
\(\Rightarrow\dfrac{y+z}{bc}=\dfrac{x+y}{ab}=\dfrac{z-x}{b\left(c-a\right)}\)(3)
Kết hợi (1);(2)(3) => ĐPCM
tik mik nha !!!
Đề nhảm.a;b;c ở đâu bạn -_-
a) Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel:
\(\left\{{}\begin{matrix}\dfrac{x}{2x+y+z}=\dfrac{x}{x+y+x+z}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)\\\dfrac{y}{2y+x+z}=\dfrac{y}{x+y+y+z}\le\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right)\\\dfrac{z}{2z+x+y}=\dfrac{z}{x+z+y+z}\le\dfrac{1}{4}\left(\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)\end{matrix}\right.\)
Cộng theo vế:
\(\dfrac{x}{2x+y+z}+\dfrac{y}{2y+x+z}+\dfrac{z}{2z+x+y}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{y+z}+\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z>0\)
b) Áp dụng bất đẳng thức AM-GM:
\(\left\{{}\begin{matrix}\left(a+b-c\right)\left(a-b+c\right)\le\dfrac{\left(a+b-c+a-b+c\right)^2}{4}=\dfrac{4a^2}{4}=a^2\\\left(a-b+c\right)\left(-a+b+c\right)\le\dfrac{\left(a-b+c-a+b+c\right)^2}{4}=\dfrac{4c^2}{4}=c^2\\\left(a+b-c\right)\left(-a+b+c\right)\le\dfrac{\left(a+b-c-a+b+c\right)^2}{4}=\dfrac{4b^2}{4}=b^2\end{matrix}\right.\)
Nhân theo vế: \(\left[\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)\right]^2\le\left(abc\right)^2\)
\(\Rightarrow\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)\le abc\)
Dấu "=" xảy ra khi: \(a=b=c>0\)
Phải chứng minh BĐT trung gian: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\forall\) a,b trước khi áp dụng chứ.
Ta có:
\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)
\(c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)
Từ (1) và (2), suy ra: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\)
Vậy \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)(đpcm)
~ Học tốt!~
Bài 1:
\(3^{-1}.3^n+4.3^n=13.3^5\)
\(\Rightarrow3^{n-1}+4.3.3^{n-1}=13.3^5\)
\(\Rightarrow3^{n-1}\left(1+4.3\right)=13.3^5\)
\(\Rightarrow3^{n-1}.13=13.3^5\)
\(\Rightarrow3^{n-1}=3^5\)
\(\Rightarrow n-1=5\)
\(\Rightarrow n=6\)
Vậy n = 6
Bài 2a: Câu hỏi của Nguyễn Trọng Phúc - Toán lớp 7 | Học trực tuyến
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-2y+4+3z-9}{2-6+12}=\dfrac{-10-6}{-8}=\dfrac{-16}{-8}=2\)\(\Rightarrow\left\{{}\begin{matrix}x=2.2+1=5\\y=2.3+2=8\\z=2.4+3=11\end{matrix}\right.\)
Theo đề bài ta có:
\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=k\)
ta có:
\(\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=k^3=\dfrac{a}{d}\)
Và \(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=k^3\)
Ta có đpcm
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.