Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
x,y,z tỉ lệ với 3; 4; 5
\(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=n\) (n>0)
\(\Rightarrow\left\{{}\begin{matrix}x=3n\\y=4n\\z=5n\end{matrix}\right.\)\(\Rightarrow x+y+z=3n+4n+5n=12n\)
a, b, c tỉ lệ với 4; 5; 6
\(\Rightarrow\frac{a}{4}=\frac{b}{5}=\frac{c}{6}=m\) (m>0)
\(\Rightarrow\left\{{}\begin{matrix}a=4m\\b=5m\\c=6m\end{matrix}\right.\)\(\Rightarrow a+b+c=4m+5m+6m=15m\)
Mà \(x+y+z=a+b+c\)
\(\Rightarrow12n=15m\Rightarrow4n=5m\)
\(\Rightarrow n=\frac{5}{4}m\)
\(\Rightarrow\left\{{}\begin{matrix}x=3n=3.\frac{5}{4}m=\frac{15}{4}m\\y=4n=4.\frac{5}{4}m=5m\\z=5n=5.\frac{5}{4}m=\frac{25}{4}m\end{matrix}\right.\)
Ta có:
\(\left\{{}\begin{matrix}a=4m\\x=\frac{15}{4}m=3,75m\end{matrix}\right.\)mà m>0 nên \(a>x\left(đpcm\right)\)
\(\left\{{}\begin{matrix}b=5m\\y=5m\end{matrix}\right.\)\(\Rightarrow y=b\left(đpcm\right)\)
\(\left\{{}\begin{matrix}z=\frac{25}{4}m=6,25m\\c=6m\end{matrix}\right.\) mà m>0 nên \(z>c\left(đpcm\right)\)
Giải:
Vì \(a,b,c\) tỉ lệ thuân với \(x,y,z\) nên: \(\dfrac{x}{a}=\dfrac{y}{y}=\dfrac{z}{c}.\)
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=\dfrac{x+y+z}{1}=x+y+z.\)
Lại có: \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\Rightarrow\left(\dfrac{x}{a}\right)^2=\left(\dfrac{y}{b}\right)^2=\left(\dfrac{z}{c}\right)^2\Rightarrow\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\left(x+y+z\right)^2_{\left(1\right)}.\)
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=\dfrac{x^2+y^2+z^2}{1}=x^2+y^2+z^2_{\left(2\right)}.\)
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2\left(đpcm\right).\)
Vì a;b;c tỉ lệ thuận với x;y;z \(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
\(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=x+y+z\)
\(\Rightarrow\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\left(x+y+z\right)^2\)
Ta lại có :
\(\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\\ \Rightarrow x^2+y^2+z^2=\left(x+y+z\right)^2\left(đpcm\right)\)
a) Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> ad = bc
Ta có : (a + 2c)(b + d)
= a(b + d) + 2c(b + d)
= ab + ad + 2cb + 2cd (1)
Ta có : (a + c)(b + 2d)
= a(b + 2d) + c(b + 2b)
= ab + a2d + cb + c2b
= ab + c2d + ad + c2b (Vì ad = cd) (2)
Từ (1),(2) => (a + 2c)(b + d) = (a + c)(b + 2d) (ĐPCM)
Sửa đề bài : P = \(\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\)
Ta có : \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)
=> \(\dfrac{y+z+t}{x}=\dfrac{z+t+x}{y}=\dfrac{t+x+y}{z}=\dfrac{x+y+z}{t}\)
=> \(\dfrac{y+z+t}{x}+1=\dfrac{z+t+x}{y}+1=\dfrac{t+x+y}{z}+1=\dfrac{x+y+z}{t}+1\)=> \(\dfrac{y+z+t+x}{x}=\dfrac{z+t+x+y}{y}=\dfrac{t+x+y+z}{z}=\dfrac{x+y+z+t}{t}\)TH1: x + y + z + t # 0
=> x = y = z = t
Ta có : P = \(\dfrac{x+y}{z+t}=\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)
P = \(\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}\)
P = 1 + 1 + 1 + 1 = 4
TH2 : x + y + z + t = 0
=> x + y = -(z + t)
y + z = -(t + x)
z + t = -(x + y)
t + x = -(y + z)
Ta có : P = \(\dfrac{x+y}{z+t}=\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)
P = \(\dfrac{-\left(z+t\right)}{z+t}=\dfrac{-\left(t+x\right)}{t+x}=\dfrac{-\left(x+y\right)}{x+y}=\dfrac{-\left(y+z\right)}{y+z}\)
P = (-1) + (-1) + (-1) + (-1)
P = -4
Vậy ...
Ta có:
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}.\)
\(\Rightarrow\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}=\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0.\)
\(\Rightarrow\left\{{}\begin{matrix}bz=cy\\cx=az\\ay=bx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\frac{y}{b}=\frac{z}{c}\\\frac{x}{a}=\frac{z}{c}\\\frac{x}{a}=\frac{y}{b}\end{matrix}\right.\)
\(\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\left(đpcm\right).\)
Chúc bạn học tốt!
Ta có: 2x=4y=3z
\(\frac{a+b-c}{6}=\frac{b+c-a}{10}=\frac{c+a-b}{2}=\frac{a}{4}=\frac{b}{8}=\frac{c}{6}\)
\(\Rightarrow\frac{2ãx}{4}=\frac{4by}{8}=\frac{3cz}{6}=\frac{ax}{2}=\frac{by}{2}=\frac{cz}{2}\)
\(\Rightarrowãx=by=cz\)
+)Vì x<y
Suy ra a/b<c/d
Suy ra a.b+a.d<b.c+b.a
Suy ra a.(b+d)<b.(c+a)
Suy ra a/b<c+a/b+d
Suy ra a/b<c+a/b+d<c/d
Suy ra x<z<y