a,\(\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
b, 10x = 6y va 2x2 - y2 = -28
Tim x,y,z(cau a)
tim x,y ( cau b)
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a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{2\cdot2+3\cdot3-4}=\dfrac{45}{9}=5\)
Do đó: x-1=10; y-2=15; z-3=20
=>x=11; y=17; z=23
c: Ta có: 10x=6y
nên x/3=y/5
Đặt x/3=y/5=k
=>x=3k; y=5k
Ta có: \(2x^2-y^2=-28\)
\(\Leftrightarrow2\cdot9k^2-25k^2=-28\)
\(\Leftrightarrow k^2=4\)
Trường hợp 1: k=2
=>x=6; y=10
TRường hợp 2: k=-2
=>x=-6; y=-10
Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x
=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x
=> 2x+3y-1 / 12 = 2x+3y-1 / 6x
=> 12 = 6x => x =2
a)Vì \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)nên \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{x}{28}\).
Áp dụng t/c dãy tỉ số = nhau, ta có :
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)
⇒2x = 3.30 = 90 ⇒ x = 45
3y = 3.60 = 180 ⇒ y = 60
z = 3.28 = 84
Ý b) có gì đó sai sai ?
c)Ta có :
\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)
Áp dụng t/c dãy tỉ số = nhau, ta có :
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)
⇒x = 5.15 = 75
y = 5.10 = 50
z = 5.6 = 30
d)Ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\left(k\in Z\right)\)
⇒ x = 2k ; y = 3k ; z = 5k
⇒ xyz = 2k.3k.5k = 30k3 = 810
⇒ k = 3 Vậy x = 3.2 = 6; y = 3.3 = 9; z = 3.5 = 15Hình như sai đề rồi bạn :
Có phải như thế này không :
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+y}\)
Ta có\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}\)
\(=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}\)
\(=\dfrac{2x+2y+2z+1+2-3}{x+y+z}\)
\(=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Nên \(\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\)
Ta lại có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=2\)
\(\Leftrightarrow\dfrac{\left(x+y+z\right)-z+1}{x}=\dfrac{\left(x+y+z\right)-y+2}{y}=\dfrac{\left(x+y+z\right)-z-3}{z}=2\)
\(\Rightarrow\dfrac{\dfrac{1}{2}-x+1}{x}=\dfrac{\dfrac{1}{2}-y+2}{y}=\dfrac{\dfrac{1}{2}-z-3}{z}=2\)
\(\Rightarrow\dfrac{\dfrac{3}{2}-x}{x}=\dfrac{\dfrac{5}{2}-y}{y}=\dfrac{-z-\dfrac{5}{2}}{z}=2\)
\(\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{\dfrac{3}{2}-x}{x}\\\dfrac{\dfrac{5}{2}-y}{y}\\\dfrac{-z-\dfrac{5}{2}}{z}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=\dfrac{3}{2}-x\\2y=\dfrac{5}{2}-y\\2z=-z-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{5}{2}\end{matrix}\right.\)
Câu 1: Mình chỉnh sửa lại đầu bài của bạn nha. Không biết có đúng không. Nếu để đầu bài như bạn thì mình không làm ra được. Mog góp ý !!!!
Áp dụng t/c DTSBN ta có:
\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\)
\(=\dfrac{x+y+x}{y+z+1+x+z+1+x+y-2}=\dfrac{x+y+x}{2x+2y+2z}=\dfrac{1}{2}\)
=>\(\dfrac{x}{y+z+1}=\dfrac{1}{2}\left(1\right)\)
=>\(\dfrac{y}{x+z+1}=\dfrac{1}{2}\left(2\right)\)
=>\(\dfrac{z}{x+y-2}=\dfrac{1}{2}\left(3\right)\)
=> x+y+z = 1/2 (4)
Ta có : Từ (1) => 2x = y+z+1 kết hợp (4)
=> 2x = 1/2-x+1
=> 3x = 3/2 => x=1/2
Ta có: Từ (2) => 2y = x+z+1
=> 2y + y = x+y+z+1
=> 3y = 1/2+1 (theo 4) => 3y=3/2
=> y=1/2
Ta có : Từ (4) => x+y+z=1/2
=>1/2 + 1/2 +z = 1/2
=> z=-1/2
Vậy ( x;y;z)=(1/2;1/2;-1/2)
\(\text{a) }\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\\ =\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)-2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\dfrac{x+y+z}{xyz}}\\ =\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)
\(\text{b) }\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\\ =1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2017}-\dfrac{1}{2018}\\ =2016+\dfrac{1}{2}-\dfrac{1}{2018}\\ =\dfrac{2034698}{1009}\)
bình phương cả 2 vế ta được
\(A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2x^2+2y^2+2z^2\)
\(A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\) (vì x^2 +y^2 +z^2 =1)
Áp dụng BĐT cô si cho 2 số
\(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}\ge2y^2\left(1\right)\)
\(\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\ge2z^2\left(2\right)\)
\(\dfrac{x^2y^2}{z^2}+\dfrac{x^2z^2}{y^2}\ge2x^2\left(3\right)\)
(1)+(2)+(3)
=> \(2\left(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\right)\ge2\left(x^2+y^2+z^2\right)\)
<=> \(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\ge1\)
Cộng 2 vào cả 2 vế ta đc
\(A^2\ge3\)
<=> \(\ge\sqrt{3}\)
Vậy Min A= \(\sqrt{3}\) khi x=y=z =\(\dfrac{1}{\sqrt{3}}\)
Lời giải khác:
Đặt \((\frac{xy}{z}; \frac{yz}{x}; \frac{xz}{y})\mapsto (a,b,c)\)
\(\Rightarrow (x^2,y^2,z^2)=(ac,ab,bc)\)
Bài toán trở thành tìm min của $A=a+b+c$ biết $ab+bc+ac=1$ và $a,b,c>0$
Theo hệ quả quen thuộc của BĐT AM-GM:
\(A^2=(a+b+c)^2\geq 3(ab+bc+ac)=3\)
\(\Rightarrow A\geq \sqrt{3}\)
Vậy \(A_{\min}=\sqrt{3}\Leftrightarrow a=b=c\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)
a. Đặt \(\dfrac{x}{-3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=-3k\\y=5k\end{matrix}\right.\)
mà \(x.y=\dfrac{-5}{27}\)
hay \(-3k.5k=\dfrac{-5}{27}\)
\(\Rightarrow-15.k^2=\dfrac{-5}{27}\)
\(\Rightarrow k^2=\dfrac{1}{81}=\left(\pm\dfrac{1}{9}\right)^2\)
Với \(k=\dfrac{1}{9}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{5}{9}\end{matrix}\right.\)
Với \(k=\dfrac{-1}{9}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{-5}{9}\end{matrix}\right.\)
Vậy.......
b. Từ \(\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{3}=\dfrac{z}{5}\end{matrix}\) \(\Rightarrow\begin{matrix}\dfrac{x}{9}=\dfrac{y}{12}\\\dfrac{y}{12}=\dfrac{z}{20}\end{matrix}\) \(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{x-y+z}{9-12+20}=\dfrac{32}{17}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=\dfrac{32}{17}\Rightarrow x=\dfrac{32.9}{17}=\dfrac{288}{17}\\\dfrac{y}{12}=\dfrac{32}{17}\Rightarrow y=\dfrac{32.12}{17}=\dfrac{384}{17}\\\dfrac{z}{20}=\dfrac{32}{17}\Rightarrow z=\dfrac{32.20}{17}=\dfrac{640}{17}\end{matrix}\right.\)
Vậy.........
\(a)\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+x+x+z+2+x+y-3}{x+y+z}\)
\(=\dfrac{\left(x+y+z\right)+\left(x+y+z\right)+\left(1+2-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Lại có: \(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
\(\Rightarrow2=\dfrac{1}{x+y+z}\Rightarrow2\left(x+y+z\right)=1\Rightarrow x+y+z=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z+1}{x}=2\\\dfrac{x+z+2}{y}=2\\\dfrac{x+y-3}{z}=2\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y+z+x+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+1=3x\\\dfrac{1}{2}+2=3y\\\dfrac{1}{2}-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1+\dfrac{1}{2}}{3}\\y=\dfrac{\dfrac{1}{2}+2}{3}\\z=\dfrac{\dfrac{1}{2}-3}{3}\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)
Chúc bạn học tốt!