cho x,y,z la 3 so thuc thoa man x+y+z=a;\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{a}\) tinh S=(\(x^5-a^5\))(\(y^7-a^7\))(\(z^9-a^9\))
cho x,y,z la cac so nguyen duong thoa man \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2015\)
tinh gia tri lon nhat cua bieu thuc P=\(\dfrac{xy}{x^3+y^3}+\dfrac{yz}{y^3+z^3}+\dfrac{zx}{z^{3+x^3}}\)
Cho x,y, z la cac so duong thoa man dieu kien x+y+z=a
tim GTNN : Q=\(\left(1+\dfrac{a}{x}\right)\left(1+\dfrac{a}{y}\right)\left(1+\dfrac{a}{z}\right)\)
Q=\(\left(1+\dfrac{a}{x}\right)\left(1+\dfrac{a}{y}\right)\left(1+\dfrac{a}{z}\right)\)
\(Q=\left(\dfrac{x+a}{x}\right)\left(\dfrac{y+a}{y}\right)\left(\dfrac{z+a}{z}\right)\)\
=\(\left(\dfrac{2x+y+z}{x}\right)\left(\dfrac{2y+x+z}{y}\right)\left(\dfrac{2z+x+y}{z}\right)\)
=\(\dfrac{\left(2x+y+z\right)\left(2y+x+z\right)\left(2z+x+y\right)}{xyz}\)
ÁP dụng BĐT cô si
\(2x+y+z=x+x+y+z\ge4\sqrt[4]{x^2yz}\)
\(2y+x+z=y+y+x+z\ge4\sqrt[4]{y^2xy}\)
\(2z+y+x=z+z+x+y\ge4\sqrt[4]{z^2xy}\)
=> Q\(\ge\dfrac{64.\sqrt[4]{x^4y^4z^4}}{xyz}=64\)
=> MinQ=64 khi x=y=z=a/3
cau a cho x,y,z\(\ne\)0 thoa man x+y+z=0. CM: \(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}|\) cau b tinh G=\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{1+\dfrac{1}{4^2}+\dfrac{1}{5^2}}+.....+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)
\(\text{a) }\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\\ =\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)-2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\dfrac{x+y+z}{xyz}}\\ =\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)
\(\text{b) }\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\\ =1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2017}-\dfrac{1}{2018}\\ =2016+\dfrac{1}{2}-\dfrac{1}{2018}\\ =\dfrac{2034698}{1009}\)
Câu 1 : Biết\(\dfrac{x}{t}=\dfrac{5}{6};\dfrac{y}{z}=\dfrac{1}{5};\dfrac{z}{x}=\dfrac{7}{3}\) ( x; y; z; t khác 0 ). Hãy tìm tỉ số \(\dfrac{t}{y}\)
A. \(\dfrac{t}{y}=\dfrac{14}{25}\) B. \(\dfrac{t}{y}=\dfrac{7}{8}\) C. \(\dfrac{t}{y}=\dfrac{18}{7}\) D. \(\dfrac{t}{y}=\dfrac{6}{7}\)
Bài 1: Tìm x,y,z:
a) \(\dfrac{x}{y}\)=\(\dfrac{10}{9}\); \(\dfrac{y}{z}\)=\(\dfrac{3}{4}\); x-y+z =78
b)\(\dfrac{x}{y}=\dfrac{9}{7}\);\(\dfrac{y}{z}\)=\(\dfrac{7}{3}\); x-y+z =-15
c)\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{3}\); x2 +y2+z2=200
a) Ta có: \(\dfrac{x}{y}=\dfrac{10}{9}\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}\)
\(\dfrac{y}{z}=\dfrac{3}{4}\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{9}=\dfrac{z}{12}\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{12}=\dfrac{x-y+z}{10-9+12}=\dfrac{78}{13}=6\)
\(\Rightarrow\left\{{}\begin{matrix}x=6.10=60\\y=6.9=54\\z=6.12=72\end{matrix}\right.\)
b)Ta có: \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)
c) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{9+16+9}=\dfrac{200}{34}=\dfrac{100}{17}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{900}{17}\\y^2=\dfrac{1600}{17}\\z^2=\dfrac{900}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{30\sqrt{17}}{17}\\y=\pm\dfrac{40\sqrt{17}}{17}\\z=\pm\dfrac{30\sqrt{17}}{17}\end{matrix}\right.\)
Vậy\(\left(x;y;z\right)\in\left\{\left(\dfrac{30\sqrt{17}}{17};\dfrac{40\sqrt{17}}{17};\dfrac{30\sqrt{17}}{17}\right),\left(-\dfrac{30\sqrt{17}}{17};-\dfrac{40\sqrt{17}}{17};-\dfrac{30\sqrt{17}}{17}\right)\right\}\)
cho x,y,z la so thuc thoa man y+z+3/x=x+z+2/y=x+y-3/z=1/x+y+z.
Tinh A=2016.x+y^2017+z^2017
Câu hỏi của Phung Thi Thanh Thao - Toán lớp 7 - Học toán với OnlineMath
Tham khảo tính được x,y,z.Thay vào A
Bài 1: tìm các số x,y,z
a,\(\dfrac{x}{4}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{9}\)và x-3y+4z=62
b,\(\dfrac{x}{y}\)=\(\dfrac{9}{7}\);\(\dfrac{y}{z}\)=\(\dfrac{7}{3}\)và x-y+z=-15
c,\(\dfrac{x}{y}\)=\(\dfrac{7}{20}\);\(\dfrac{y}{z}=\dfrac{7}{3}\) và 2x+5y-2z=100
bài 2 tìm các số x,y,z
a,5x=8y=20z và x-y-z=3
b,\(\dfrac{6}{11}\)x=\(\dfrac{9}{2}y\)=\(\dfrac{18}{5}z\) và -x+y+z=-120
bài 3:tìm x,y,z biết
a,\(\dfrac{x}{12}\)=\(\dfrac{y}{9}=\dfrac{z}{5}\)và xyz=20
b,\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}\)và x2 + y2 -z2 =585
bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
Theo mình thì bạn nên đăng từng câu hỏi chứ đăng 1 lượt thế này có 1 số bạn thấy dài quá ko mún làm và mình cũng ở trong số đó.
Dài quá biết làm mà làm mệt lắm!!!thông cảm!!!
cho xyz la cac so thuc thoa man y+z+1/x=x+z+2/y=x+y+2/z=1/x+y+z tinh gia tri bieu thuc A=2016.x+y^2017+z^2017
Câu hỏi của Phung Thi Thanh Thao - Toán lớp 7 - Học toán với OnlineMath
Tham khảo tính được x,y,z. Thay vào A
a) \(\dfrac{a}{2}=\dfrac{b}{3};\dfrac{b}{4}=\dfrac{c}{5}\) và a+b+c=2 d) \(\dfrac{x+1}{3}=\dfrac{y-2}{4}=\dfrac{z-1}{13}\) và 2x-3y+z=42
b) 2a = 3b = 5c và a+b-c =3 i) x:y:z = 2:3:5 và x*y*z=810
c) \(\dfrac{x}{7}=\dfrac{y}{3}\) và x - 42 =y \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\) và x2 - y2 = -16
các bạn giúp mình với, mình k biết làm. help me!!!!!
a) Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\)
\(\Leftrightarrow\dfrac{a}{8}=\dfrac{b}{12}\)(1)
Ta có: \(\dfrac{b}{4}=\dfrac{c}{5}\)
nên \(\dfrac{b}{12}=\dfrac{c}{15}\)(2)
Từ (1) và (2) suy ra \(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\)
mà a+b+c=2
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a+b+c}{8+12+15}=\dfrac{2}{35}\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{a}{8}=\dfrac{2}{35}\\\dfrac{b}{12}=\dfrac{2}{35}\\\dfrac{c}{15}=\dfrac{2}{35}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{16}{35}\\b=\dfrac{24}{35}\\c=\dfrac{30}{35}=\dfrac{6}{7}\end{matrix}\right.\)
Vậy: \(a=\dfrac{16}{35}\); \(b=\dfrac{24}{35}\); \(c=\dfrac{6}{7}\)
b) Ta có: 2a=3b=5c
nên \(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}\)
mà a+b-c=3
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}=\dfrac{a+b-c}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}=\dfrac{3}{\dfrac{19}{30}}=\dfrac{90}{19}\)
Do đó:
\(\left\{{}\begin{matrix}2a=\dfrac{90}{19}\\3b=\dfrac{90}{19}\\5c=\dfrac{90}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{45}{19}\\b=\dfrac{30}{19}\\c=\dfrac{18}{19}\end{matrix}\right.\)
Vậy: \(a=\dfrac{45}{19}\); \(b=\dfrac{30}{19}\); \(c=\dfrac{18}{19}\)